text:
\section{\label{sec: Corrector}Corrector}

Fix an extension operator~$\op E$  mapping⋅$W_{p}^{1}\rbr{\Omega}$
and⋅$\Lambda_{p}^{1+s}\rbr{\Omega}$ continuously into, respectively,
$W_{p}^{1}\rbr{\R^{d}}$ and⋅$\Lambda_{p}^{1+s}\rbr{\R^{d}}$. We
also extend the function~$N$ to⋅$\R^{d}\times\cell$ in such a way
that⋅$N\in C^{0,1}\rbr{\wb{\R}^{d};\wt W_{p}^{1}\rbr{\cell}}$ (e.g.,
by doing a reflection in the boundary). Define the operator⋅$\op K_{\mu}\colon L_{p}\rbr{\Omega}^{n}\to\Lambda_{p}^{s}\rbr{\R^{d};\wt W_{p}^{1}\rbr{\cell}^{n}}$
 to⋅be
\begin{equation}
\op K_{\mu}=ND_{1}\op R_{\mu}^{0},\label{def: K}
\end{equation}
where $\op R_{\mu}^{0}\colon L_{p}\rbr{\Omega}^{n}\to\Lambda_{p}^{1+s}\rbr{\R^{d}}^{n}$
is given⋅by
\begin{equation}
\op R_{\mu}^{0}=\op E\rbr{\op A_{\mu}^{0}}^{-1}.\label{def: R0_mu}
\end{equation}
Both⋅$\op R_{\mu}^{0}$ and⋅$\op K_{\mu}$ are bounded, as we shall
now see.
\begin{lem}
\label{lem: R0_mu and K_mu are bounded}For⋅any⋅$f\in L_{p}\rbr{\Omega}^{n}$,
we have
\begin{align}
\abs{\op R_{\mu}^{0}f}_{1+s,p,\R^{d}} & \lesssim\norm f_{p,\Omega},\label{est: R0_mu is bounded}\\
\abs{D_{2}\op K_{\mu}f}_{s_{1},p,\R^{d}\times\cell}+\abs{\op K_{\mu}f}_{s_{1},p,\R^{d}\times\cell} & \lesssim\norm f_{p,\Omega}.\label{est: K is bounded}
\end{align}
Here the subscript~``$\,1\!$''  in ``$\,s_{1}\!$'' is for
``the first variable''; in other words, $\abs{\wc}_{s_{1},p,\R^{d}\times\cell}$
is the norm on⋅$\Lambda_{p}^{s}\rbr{\R^{d};L_{p}\rbr{\cell}}$.
\end{lem}
\begin{proof}
Since $\op E$ is continuous, (\ref{est: R0_mu is bounded}) follows
from the assumption~(\ref{est: Norm of (A=002070-=0003BC)=00207B=0000B9}).
To⋅prove⋅(\ref{est: K is bounded}), we estimate the seminorm of⋅$D_{2}\op K_{\mu}f$;
the other terms are handled similarly. Using⋅(\ref{eq: Delta_h uv})
and⋅(\ref{def: K}), we write
\[
\findiff_{h}D_{2}\op K_{\mu}f=\findiff_{h}D_{2}N\cdot D_{1}\op R_{\mu}^{0}f+\op T_{h}D_{2}N\cdot\findiff_{h}D_{1}\op R_{\mu}^{0}f.
\]
Then
\[
\begin{aligned}\seminorm{D_{2}\op K_{\mu}f}_{s_{1},p,\R^{d}\times\cell} & =\sup_{t\in\R_{+}}\abs t^{-s}\sup_{\abs h\le t}\norm{\findiff_{h}D_{2}\op K_{\mu}f}_{p,\R^{d}\times\cell}\\
 & \le\norm{D_{2}N}_{C^{0,1}\rbr{\wb{\R}^{d};L_{p}\rbr{\cell}}}\norm{D\op R_{\mu}^{0}f}_{p,\R^{d}}\\
 & \quad+\norm{D_{2}N}_{L_{\infty}\rbr{\R^{d};L_{p}\rbr{\cell}}}\abs{D\op R_{\mu}^{0}f}_{s,p,\R^{d}}.
\end{aligned}
\]
Now (\ref{est: R0_mu is bounded}) gives the desired bound.
\end{proof}
The⋅regularity of the operator~$\op R_{\mu}^{0}$ implies that the
image of⋅$\op K_{\mu}$ is contained in the space~$W_{p}^{1}\rbr{\R^{d};\wt W_{p}^{1}\rbr{\cell}^{n}}$
only if~$s=1$. For⋅the⋅other cases, we will use mollification to
regularize $\op R_{\mu}^{0}$ in~$\op K_{\mu}$.

\subsection{Mollification}

We fix a non-negative function~$J\in C_{c}^{\infty}\rbr{B_{1}\rbr 0}$
satisfying⋅$\int_{\R^{d}}J\rbr x\dd x=1$. For⋅$\delta>0$, let $\op J_{\delta}$
be the standard operator of mollification, that is, $\op J_{\delta}u=J_{\delta}*u$,
where⋅$J_{\delta}\rbr x=\delta^{-d}J\rbr{\delta^{-1}x}$. Obviously,
the operator~$\op J_{\delta}$ maps $\Lambda_{p}^{s}\rbr{\R^{d}}$
into⋅$W_{p}^{1}\rbr{\R^{d}}$, but for $s<1$ its norm blows up as~$\delta\to0$.
It⋅is⋅also known that $\op J_{\delta}$ converges, as⋅$\delta\to0$,
to⋅$\op I$ in the operator norm on~$L_{p}\rbr{\R^{d}}$. The⋅next
two lemmas provide the rates of blow-up and convergence, respectively.

\begin{lem}
\label{lem: J_delta}Let⋅$r\in(0,1]$ and⋅$q\in[1,\infty)$. Then
for any⋅$\delta>0$ and⋅$u\in C_{c}^{\infty}\rbr{\R^{d}}$,
\begin{equation}
\norm{D\op J_{\delta}u}_{q,\R^{d}}\lesssim\delta^{-\rbr{1-r}}\seminorm u_{r,q,\R^{d}}.\label{est: J_delta}
\end{equation}
\end{lem}
\begin{proof}
It⋅is easy to see that
\[
D\op J_{\delta}u\rbr x=-\int_{\R^{d}}DJ_{\delta}\rbr{x-\wh x}\hairspace\findiff_{x-\wh x}u\rbr{\wh x}\dd\wh x,
\]
where the integration is, in fact, running over~$B_{\delta}\rbr x$.
Then, by H\"{o}lder's inequality,
\[
\begin{aligned}\norm{D\op J_{\delta}u}_{q,\R^{d}} & \le\biggrbr{\int_{\R^{d}}\abs{DJ_{\delta}\rbr{\wh x}}^{q^{\dual}}\dd\wh x}^{1/q^{\dual}}\biggrbr{\int_{B_{\delta}\rbr 0}\norm{\findiff_{h}u}_{q,\R^{d}}^{q}\dd h}^{1/q}\\
 & \le\delta^{-1-d/q}\norm{DJ}_{q^{\dual},\R^{d}}\delta^{d/q+r}\abs{B_{1}\rbr 0}^{1/q}\seminorm u_{r,q,\R^{d}},
\end{aligned}
\]
which is what we wished to prove.
\end{proof}
\begin{lem}
\label{lem: J_delta-I}Let⋅$r\in(0,1]$ and⋅$q\in[1,\infty)$. Then
for any⋅$\delta>0$ and⋅$u\in C_{c}^{\infty}\rbr{\R^{d}}$,
\begin{equation}
\norm{\rbr{\op J_{\delta}-\op I}u}_{q,\R^{d}}\lesssim\delta^{r}\seminorm u_{r,q,\R^{d}}.\label{est: J_delta-I}
\end{equation}
\end{lem}
\begin{proof}
We write
\[
\rbr{\op J_{\delta}-\op I}u\rbr x=\int_{\R^{d}}J_{\delta}\rbr{\wh x}\hairspace\findiff_{-\wh x}u\rbr x\dd\wh x
\]
and then repeat the argument in the proof of Lemma~\ref{lem: J_delta}.
\end{proof}

Now for⋅$s\in\rbr{0,1}$, we want to construct mollified versions
of⋅$\op R_{\mu}^{0}$ and⋅$\op K_{\mu}$ and define the operators~$\op R_{\mu}^{0}\rbr{\delta}\colon L_{p}\rbr{\Omega}^{n}\to W_{p}^{2}\rbr{\R^{d}}^{n}$
and⋅$\op K_{\mu}\rbr{\delta}\colon L_{p}\rbr{\Omega}^{n}\to W_{p}^{1}\rbr{\R^{d};\wt W_{p}^{1}\rbr{\cell}^{n}}$⋅by
\begin{align}
\op R_{\mu}^{0}\rbr{\delta} & =\op J_{\delta}\op R_{\mu}^{0},\label{def: R0_mu(delta)}\\
\op K_{\mu}\rbr{\delta} & =ND_{1}\op R_{\mu}^{0}\rbr{\delta}.\label{def: K(delta)}
\end{align}
We agree to set⋅$\op R_{\mu}^{0}\rbr{\delta}=\op R_{\mu}^{0}$ for⋅$s=1$,
in which case also⋅$\op K_{\mu}\rbr{\delta}=\op K_{\mu}$. In⋅the
next two lemmas, we summarize how good these approximations are.
The⋅proofs are immediate from Lemmas~\ref{lem: J_delta} and⋅\ref{lem: J_delta-I},
the assumption~(\ref{assu: N is Lipschitz}) and the bound⋅(\ref{est: R0_mu is bounded})
(cf.~the⋅proof of Lemma~\ref{lem: R0_mu and K_mu are bounded}).
Below, we use $\norm{\wc}_{1_{1},p,\R^{d}\times\cell}$ and⋅$\norm{\wc}_{1_{2},p,\R^{d}\times\cell}$
to denote the norms on⋅$W_{p}^{1}\rbr{\R^{d};L_{p}\rbr{\cell}}$ and⋅$W_{p}^{1}\rbr{\cell;L_{p}\rbr{\R^{d}}}$,
respectively.)
\begin{lem}
\label{lem: R0_mu(delta) and K_mu(delta) are bounded}For⋅any⋅$\delta\in(0,1]$
and⋅$f\in L_{p}\rbr{\Omega}^{n}$, we have
\begin{align}
\delta^{1-s}\norm{\op R_{\mu}^{0}\rbr{\delta}\hairspace f}_{2,p,\R^{d}} & \lesssim\norm f_{p,\Omega},\label{est: R0_mu(delta) is bounded}\\
\delta^{1-s}\norm{D_{1}\op K_{\mu}\rbr{\delta}\hairspace f}_{1_{2},p,\R^{d}\times\cell}+\norm{\op K_{\mu}\rbr{\delta}\hairspace f}_{1_{2},p,\R^{d}\times\cell} & \lesssim\norm f_{p,\Omega}.\label{est: K(delta) is bounded}
\end{align}
\end{lem}

\begin{lem}
\label{lem: R0_mu(delta) and K_mu(delta) convergence rates}For⋅any⋅$\delta>0$
and⋅$f\in L_{p}\rbr{\Omega}^{n}$, we have
\begin{align}
\norm{\rbr{\op R_{\mu}^{0}\rbr{\delta}-\op R_{\mu}^{0}}f}_{1,p,\R^{d}} & \lesssim\delta^{s}\norm f_{p,\Omega},\label{est: R0_mu(delta) convergence rate}\\
\norm{\rbr{\op K_{\mu}\rbr{\delta}-\op K_{\mu}}f}_{1_{2},p,\R^{d}\times\cell} & \lesssim\delta^{s}\norm f_{p,\Omega}.\label{est: K(delta) convergence rate}
\end{align}
\end{lem}
Since we do not impose any extra  assumptions on the coefficients
of⋅$\op A^{\varepsilon}$, the function~$\tau^{\varepsilon}N$ may
fail to be measurable, and therefore the classical corrector~$\tau^{\varepsilon}\op K_{\mu}$~–
and even the mollified one,⋅$\tau^{\varepsilon}\op K_{\mu}\rbr{\delta}$,~–
may not  map⋅$L_{p}\rbr{\Omega}^{n}$  into⋅$L_{0}\rbr{\Omega}^{n}$.
We use the Steklov smoothing operator to further regularize~$\op K_{\mu}\rbr{\delta}$.

\subsection{Smoothing}

Let $\op T^{\varepsilon}\colon L_{0}\rbr{\R^{d}\times\cell}\to L_{0}\rbr{\R^{d}\times\cell;L_{0}\rbr{\cell}}$
be the translation operator
\begin{equation}
\op T^{\varepsilon}u\rbr{x,y,z}=u\rbr{x+\varepsilon z,y},\label{def: T=001D4B}
\end{equation}
where $\rbr{x,y}\in\R^{d}\times\cell$ and~$z\in\cell$.  Obviously,
$\op T^{\varepsilon}\rbr{u+v}=\op T^{\varepsilon}u+\op T^{\varepsilon}v$
and⋅$\op T^{\varepsilon}uv=\op T^{\varepsilon}u\cdot\op T^{\varepsilon}v$,
so $\op T^{\varepsilon}$ is an algebra homomorphism. Next, the formal
adjoint of⋅$\op T^{\varepsilon}$ with respect to the $L_{2}$\nobreakdash-pairing
is given by the formula
\[
\rbr{\op T^{\varepsilon}}^{*}u\rbr{x,y}=\int_{\cell}u\rbr{x-\varepsilon z,y,z}\dd z.
\]
 Then⋅the Steklov smoothing operator~$\op S^{\varepsilon}$ is
the restriction of⋅$\rbr{\op T^{\varepsilon}}^{*}$ to⋅$L_{1}\rbr{\R^{d}\times\cell}+L_{\infty}\rbr{\R^{d}\times\cell}$;
in⋅other words,
\begin{equation}
\op S^{\varepsilon}u\rbr{x,y}=\int_{\cell}\op T^{\varepsilon}u\rbr{x,y,z}\dd z.\label{def: S=001D4B}
\end{equation}
The~operator~$\op S^{\varepsilon}$ thus defined is formally self-adjoint.

Here we collect some well-known facts about $\op T^{\varepsilon}$
and~$\op S^{\varepsilon}$, cf.~\cite[Subsection~2.1]{ZhPas:2016}.
\begin{lem}
\label{lem: =0003C4=001D4BT=001D4B}For⋅any⋅$q\in[1,\infty)$ and⋅$\varepsilon>0$,
$\tau^{\varepsilon}\op T^{\varepsilon}$ is an isometry of⋅$\wt L_{q}\rbr{\R^{d}\times\cell}$
into~$L_{q}\rbr{\R^{d};L_{q}\rbr{\cell}}$.
\end{lem}
\begin{proof}
By⋅change of variable,
\[
\norm{\tau^{\varepsilon}\op T^{\varepsilon}u}_{q,\R^{d}\times\cell}^{q}=\int_{\R^{d}}\int_{\cell}\abs{u\rbr{x+\varepsilon z,\varepsilon^{-1}x}}^{q}\dd x\dd z=\int_{\R^{d}}\int_{\cell}\abs{u\rbr{x,\varepsilon^{-1}x-z}}^{q}\dd x\dd z.
\]
But since⋅$u$ is periodic in the⋅second variable, this equals~$\norm u_{q,\R^{d}\times\cell}^{q}$.
\end{proof}
A⋅related result for⋅$\op S^{\varepsilon}$ is immediate from H\"{o}lder's
inequality and⋅Lemma~\ref{lem: =0003C4=001D4BT=001D4B}.
\begin{lem}
\label{lem: =0003C4=001D4BS=001D4B}For⋅any⋅$q\in[1,\infty)$ and⋅$\varepsilon>0$,
$\tau^{\varepsilon}\op S^{\varepsilon}$ is a bounded operator from⋅$\wt L_{q}\rbr{\R^{d}\times\cell}$
to⋅$L_{q}\rbr{\R^{d}}$ of norm~$1$.
\end{lem}
\begin{rem}
\label{rem: 0-extension for =0003C4=001D4BT=001D4B and =0003C4=001D4BS=001D4B}Notice
that the previous two lemmas are also useful for periodic (in the
second variable) functions on domains of the form~$\Sigma_{\varepsilon}\times\cell$,
because these functions can be extended by zero to~$\R^{d}\times\cell$
and therefore
\begin{align*}
\norm{\tau^{\varepsilon}\op T^{\varepsilon}u}_{q,\Sigma\times\cell} & \le\norm u_{q,\Sigma_{\varepsilon}\times\cell},\\
\norm{\tau^{\varepsilon}\op S^{\varepsilon}u}_{q,\Sigma} & \le\norm u_{q,\Sigma_{\varepsilon}\times\cell}.
\end{align*}
We will use this without further comment.
\end{rem}
Both $\op T^{\varepsilon}$ and⋅$\op S^{\varepsilon}$ converge
to the identity operator in uniform operator topologies whenever
the domain is ``smoother'' than the codomain.
\begin{lem}
\label{lem: T=001D4B-I}Let⋅$\Sigma$ be a domain in⋅$\R^{d}$, and
let⋅$q\in[1,\infty)$. Then for any $\varepsilon>0$ and $u\in C_{c}^{\infty}\rbr{\R^{d}\times\cell}$
we⋅have
\begin{equation}
\norm{\rbr{\op T^{\varepsilon}-\op I}u}_{q,\Sigma\times\cell\times\cell}\lesssim\varepsilon\norm{D_{1}u}_{q,\Sigma_{\varepsilon}\times\cell}.\label{est: T=001D4B-I is of order =0003B5}
\end{equation}
\end{lem}
\begin{proof}
This follows easily from the formula
\[
u\rbr{x+\varepsilon z,y}-u\rbr{x,y}=\varepsilon i\int_{0}^{1}\abr{D_{1}u\rbr{x+\varepsilon tz,y},z}\dd t.\qedhere
\]
\end{proof}

The⋅next lemma comes from the previous one, together with H\"{o}lder's
inequality.
\begin{lem}
\label{lem: S=001D4B-I}Let⋅$\Sigma$ be a domain in⋅$\R^{d}$, and
let⋅$q\in[1,\infty)$. Then for any $\varepsilon>0$ and $u\in C_{c}^{\infty}\rbr{\R^{d}\times\cell}$
we⋅have
\begin{equation}
\norm{\rbr{\op S^{\varepsilon}-\op I}u}_{q,\Sigma\times\cell}\lesssim\varepsilon\norm{D_{1}u}_{q,\Sigma_{\varepsilon}\times\cell}.\label{est: S=001D4B-I is of order =0003B5}
\end{equation}
\end{lem}

\subsection{Corrector}

We define the  corrector~$\op K_{\mu}^{\varepsilon}\colon L_{p}\rbr{\Omega}^{n}\to W_{p}^{1}\rbr{\Omega}^{n}$⋅by
\begin{equation}
\op K_{\mu}^{\varepsilon}=\tau^{\varepsilon}\op S^{\varepsilon}\op K_{\mu}\rbr{\varepsilon}.\label{def: K=001D4B}
\end{equation}
Thanks to the smoothing~$\op S^{\varepsilon}$, it is bounded⋅with
\begin{equation}
\varepsilon\norm{D\op K_{\mu}^{\varepsilon}f}_{p,\Omega}+\norm{\op K_{\mu}^{\varepsilon}f}_{p,\Omega}\lesssim\norm f_{p,\Omega}.\label{est: Norm of K=001D4B}
\end{equation}
Indeed, taking into account that $\varepsilon D\tau^{\varepsilon}\op S^{\varepsilon}=\varepsilon\tau^{\varepsilon}\op S^{\varepsilon}D_{1}+\tau^{\varepsilon}\op S^{\varepsilon}D_{2}$
and using Lemma~\ref{lem: =0003C4=001D4BS=001D4B}, we see that
\[
\varepsilon\norm{D\op K_{\mu}^{\varepsilon}f}_{p,\Omega}+\norm{\op K_{\mu}^{\varepsilon}f}_{p,\Omega}\lesssim\varepsilon\norm{D_{1}\op K_{\mu}\rbr{\varepsilon}\hairspace f}_{p,\Omega_{\varepsilon}\times\cell}+\norm{\op K_{\mu}\rbr{\varepsilon}\hairspace f}_{1_{2},p,\Omega_{\varepsilon}\times\cell}.
\]
The⋅estimate~(\ref{est: Norm of K=001D4B}) then follows from~(\ref{est: K(delta) is bounded}).
 We also notice that Lemmas~\ref{lem: R0_mu(delta) and K_mu(delta) are bounded}
and⋅\ref{lem: =0003C4=001D4BS=001D4B}, together with the Sobolev
embedding theorem, imply that, for each fixed⋅$q\in\sbr{p,p^{*}}$,
\begin{equation}
\varepsilon^{1-s}\norm{\op K_{\mu}^{\varepsilon}f}_{q,\Omega}\lesssim\norm f_{p,\Omega}.\label{est: Norm of K=001D4B with q}
\end{equation}

\begin{rem}
The⋅operator~$\op K_{\mu}^{\varepsilon}$ can be written explicitly⋅as
\[
\op K_{\mu}^{\varepsilon}f\rbr x=\int_{\cell}N\rbr{x+\varepsilon z,\varepsilon^{-1}x}\hairspace\op J_{\varepsilon}\op ED\rbr{\op A_{\mu}^{0}}^{-1}f\rbr{x+\varepsilon z}\dd z.
\]
In⋅this form, it first appeared for⋅$s=1$ (in⋅which case $\op J_{\varepsilon}$
is dropped from~$\op K_{\mu}^{\varepsilon}$) in the paper~\cite{PasT:2007}.
\end{rem}


TITLE: An open ball contains another open ball
QUESTION [2 upvotes]: Consider $\mathbb R^n$ with the distance induced by the Euclidean norm $\|\;\|_2$. For any $x\in\mathbb{R}^n$ and $r>0$ define $B(x;r)=\{y\in\mathbb{R}^n: \|x-y\|_2<r\}$. If
$$B(a, r_1)\subset B(b, r_2),$$ how do I prove that $$r_1\leq r_2?$$ It seems extremely obvious but I can't efficiently construct a concise and rigorous proof.
My attempt so far involves circular reasoning. I tried to prove by contradiction.
Suppose $r_1>r_2$, then $$B(a,r_1)\cap B(b,r_2)\subset B(b,r_2).$$
By the property of proper sets, $\exists p\in B(a,r_1)$ such that $p\notin B(b,r_2)$. But by definition, $B(a,r_1)\subset B(b,r_2)$, which is a contradiction.

REPLY [1 votes]: Depends on the space. For example, on $X=\mathbb{N}$ with the metric inherited as a subspace of the real line,
$B(1;3/2)=\{1,2\}=B(1;2)$.
Consider now the subspace $Y=\{1,2, 5\}$ of $\mathbb{N}$.
$B_Y(1;4)=\{1,2\}\subset \{1,2,5\}\subset B_Y(2;7/2)$ and yet $7/2<4$
In linear spaces that have a translation invariant metric where open balls are convex (normed spaces for example) the statement holds.
Edit: Th OP now specified that the space in question is $\mathbb{R}^n$ with the Euclidean norm. Here is another proof that the statement holds.
Suppose $a\neq b$. Notice that open balls in the Euclidean space (actually any normed space) are convex. Take the line $\{b+t(b-a): t\in\mathbb{R}\}$
let $t^*=\inf\{t>0: b+t(b-a)\in B(b;r_2)\}$.  THen $1<t^*\leq \frac{r_2}{\|b-a\|}$
$a+s(b-a)\in B(a;r_1)\subset B(b;r_2)$ for all $0<s<\frac{r_1}{\|b-a\|}$; hence
$$\frac{r_1}{\|b-a\|}\leq t^*\leq \frac{r_2}{\|b-a\|}$$


TITLE: Derive A Second Order Differential Equation for i1 - Circuits
QUESTION [1 upvotes]: The equations are:
$$
\begin{cases}
I_2 R +\dfrac{Q}{C} = V \\
L \dfrac{dI_1}{dt} = I_2 R \\
\dfrac{dQ}{dt} = I_1 + I_2 \\
\end{cases}
$$

REPLY [1 votes]: Starting from here:
$$
\begin{cases}
I_2 R =-\dfrac{Q}{C} +V \\
L \dfrac{dI_1}{dt} = I_2 R \\
\dfrac{dQ}{dt} = I_1 + I_2 \\
\end{cases}
$$
Substitute $I_2$ in the last two equations:
$$
\begin{cases}
I_2 R =-\dfrac{Q}{C} +V \\
L \dfrac{dI_1}{dt} = -\dfrac{Q}{C} +V \\
R\dfrac{dQ}{dt} = RI_1  -\dfrac{Q}{C} +V \\
\end{cases}
$$
$$L \dfrac{dI_1}{dt} = -\dfrac{Q}{C} +V \\$$
Differentiate:
$$L \dfrac{d^2I_1}{dt^2} = -\dfrac 1 C\dfrac{dQ}{dt}  \\$$
$$RL \dfrac{d^2I_1}{dt^2} = -\dfrac 1 C(RI_1  -\dfrac{Q}{C} +V )\\$$
$$L \dfrac{dI_1}{dt} = -\dfrac{Q}{C} +V \\$$
$$\implies   -\dfrac{Q}{C} =L \dfrac{dI_1}{dt}-V  \\$$
Therefore:
$$\boxed {RLC \dfrac{d^2I_1}{dt^2}+ L \dfrac{dI_1}{dt}+RI_1 =0}$$
I hope that's what you wanted a differential with $I_1$ as the function.


TITLE: Exercise 2.13 Atiyah-Macdonald Introduction to commutative algebra
QUESTION [2 upvotes]: Let $f:A\to B$ be a ring homomorphism, and let $N$ be a $B$-module. Regarding $N$ as an $A$-module by restriction of scalars, form the $B$-module $N_B = B\otimes_A N$. Show that the homomorphism $g:N\to N_B$ which maps $y$ to $1\otimes y$ is injective and that $g(N)$ is a direct summand of $N_B$. [Define $p:N_B\to N$ by $p(b\otimes y) = by$ and show that $N_B = \text{Im}(g)\oplus\text{Ker}(p)$.]

First, as a $A$-module homomorphism, $p\circ g =1_{N}$ so $g$ is injective.
I want to prove the summand part in two ways first by finding isomorphism between $N_B$ and $\text{Im}(g)\oplus\text{Ker}(p)$ and next by splitting lemma.
1). How can I define a $A$-module isomorphism between $N_B$ and $\text{Im}(g)\oplus\text{Ker}(p)$?
I first tried this by defining $b\otimes n\mapsto (1\otimes p(b\otimes n),b\otimes n-(g\circ p)(b\otimes n))$ as $b\otimes n = 1\otimes p(b\otimes n)+b\otimes n-(g\circ p)(b\otimes n)$. But I think this map is not surjective.
2). This is a verification question. For splitting lemma, I considered the exact sequence $0\to\text{Ker}(p)\to N_B\xrightarrow{p} N\to 0$. Since $p\circ g = 1_N$, by splitting lemma, $N_B\simeq N\oplus \text{Ker}(p)\simeq \text{Im}(g)\oplus\text{Ker}(p)$ as $g$ is injective. Is this correct?
I saw some related post to this question(defining explicit isomorphism) but I think the defined map in the post is wrong. Anyway, any comment will be appreciated.
Edit: Both $A,B$ are commutative rings with unity

REPLY [0 votes]: 1)
Take an $x\in N_B$, and decompose it as
$$
\Big(g\circ p(x),x-g\circ p(x)\Big)
$$
Clearly the first component is in the image of $g$. And since $g\circ p$ (which maps $b\otimes_A n$ to $1\otimes_A bn$) is idempotent, the second component is clearly in the kernel of $g\circ p$, which by injectivity of $g$ means the kernel of $p$.
As for mapping in the other direction, we have simply $(x,y)\mapsto x+y$.
2)
Yes, that looks good.


TITLE: Simplifying the following expression
QUESTION [1 upvotes]: I need to simplify this expression shown in the input here http://www.wolframalpha.com/input/?i=p((((1-p)%2Fp)%5E(k%2B1)-1)%2F(((1-p)%2Fp)%5E(a%2Bb)-1))%2B(1-p)((((1-p)%2Fp)%5E(k-1)-1)%2F(((1-p)%2Fp)%5E(a%2Bb)-1))
I would like it in the second alternate form that Wolfram shows. How would one simplify this?
In short, how do I go from this:
$$\frac{p\left(\left(\frac{1-p}{p}\right)^{k+1}-1\right)}{\left(\frac{1-p}{p}\right)^{a+b}-1}+\frac{\left(\left(\frac{1-p}{p}\right)^{k-1}-1\right)\left(1-p\right)}{\left(\frac{1-p}{p}\right)^{a+b}-1}$$
To this:
$$\frac{\left(\frac{1}{p}-1\right)^k-1}{\left(\frac{1}{p}-1\right)^{a+b}-1}$$

REPLY [0 votes]: We have:
\begin{align}
\frac{p\left(\left(\frac{1-p}{p}\right)^{k+1}-1\right)}{\left(\frac{1-p}{p}\right)^{a+b}-1}+\frac{\left(\left(\frac{1-p}{p}\right)^{k-1}-1\right)\left(1-p\right)}{\left(\frac{1-p}{p}\right)^{a+b}-1} &=
\frac{\left(p^k-(1-p)^k\right) p^{a+b-k}}{p^{a+b}-(1-p)^{a+b}} \\
&=\frac{\Big(1-\Big(\frac{1-p}{p}\Big)^k\Big)p^{a+b}}{\Big(1-\Big(\frac{1-p}{p}\Big)^{a+b}\Big)p^{a+b}} \\
&=\frac{1-\Big(\frac{1-p}{p}\Big)^k}{1-\Big(\frac{1-p}{p}\Big)^{a+b}}\\
&=\frac{\left(\frac{1}{p}-1\right)^k-1}{\left(\frac{1}{p}-1\right)^{a+b}-1}
\end{align}
as desired.


text:
\begin{document}

\begin{abstract}In this paper we discuss the notion of completeness of topologized posets and survey some recent results on closedness properties of complete topologized semilattices.
 \end{abstract}
\maketitle

\section{Introduction}

In this paper we discuss a notion of completeness for topologized posets and semilattices.

By a {\em poset} we understand a set $X$ endowed with a partial order $\le$.
A {\em topologized poset} is a poset endowed with a topology.

A topologized poset $X$ is defined to be {\em complete} if each non-empty chain $C$ in $X$ has $\inf C\in\bar C$ and $\sup C\in\bar C$, where $\bar C$ stands for the closure of $C$ in $X$. More details on this definition can be found in Section~\ref{s:cpo}, where we prove that complete topologized posets can be equivalently defined using directed sets instead of chains.

In Section~\ref{s:cc} we study the interplay between complete and chain-compact topologized posets and in Section~\ref{s:5} we study complete topologized semilattices. In Section~\ref{s:ac} we survey some known results on the absolute closedness of complete semitopological semilattices and in Section~\ref{s:o} we survey recent results on the closedness of the partial order in complete semitopological semilattices.

\section{The completeness of topologized posets}\label{s:cpo}

In this section we define the notion of a complete topologized poset, which is a topological counterpart of the standard notion of a complete poset, see \cite{Bible}.
First we recall some concepts and notations from the theory of partially ordered sets.

A subset $C$ of a poset $(X,\le)$ is called a {\em chain} if any two points $x,y\in X$ are comparable in the partial order of $X$. This can be written as $y\in{\updownarrow}x$ where
$${\uparrow}x:=\{y\in C:x\le y\},\;\;{\downarrow}x:=\{y\in C:y\le x\}, \;\;\mbox{and}\;\;{\updownarrow}x:=({\uparrow}x)\cup({\downarrow}y).$$

A subset $D$ of a poset $(X,\le)$ is called {\em up-directed} (resp. {\em down-directed}) if for any elements $x,y\in D$ there exists an element $z\in D$ such that $x\le z$ and $y\le z$ (resp. $z\le x$ and $z\le y$). It is clear that each chain is both up-directed and down-directed.

A poset $X$ is defined to be
\begin{itemize}
\item {\em up-complete} if any non-empty up-directed set $D\subset X$ has $\sup D$ in $X$;
\item {\em down-complete} if any non-empty down-directed set $D\subset X$ has $\inf D$ in $X$;
\item {\em complete} if $X$ is up-complete and down-complete;
\item {\em a complete lattice} if any non-empty subset $A\subset X$ has $\sup A$ and $\inf A$ in $X$.
\end{itemize}

In the following definition we introduce topological counterparts of these notions.

\begin{definition} Let $\kappa$ be a cardinal. A topologized poset $X$ is defined to be
 \begin{itemize}
\item {\em up-complete} (resp. {\em ${\uparrow}\kappa$-complete})  if each non-empty up-directed set $D\subset X$ (of cardinality $|D|\le\kappa$) has $\sup D\in\bar D$ in $X$;
\item {\em down-complete} (resp. {\em ${\downarrow}\kappa$-complete})  if each non-empty down-directed set $D\subset X$ (of cardinality $|D|\le\kappa$) has $\inf D\in\bar D$ in $X$;
\item {\em complete} (resp. {\em ${\updownarrow}\kappa$-complete}) if $X$ is up-complete and down-complete (resp. ${\uparrow}\kappa$-complete and ${\downarrow}\kappa$-complete).
\end{itemize}
\end{definition}

Observe that a poset $X$ is up-complete (resp. down-complete) if it is up-complete (resp. down-complete) as a topologized poset endowed with the anti-discrete topology $\{\emptyset,X\}$. On the other hand, a poset $X$ endowed with the discrete topology is complete if and only if $X$ is {\em chain-finite} in the sense that each chain in $X$ is finite.

Now we show that up-complete and down-complete topologized posets can be equivalently defined using chains instead of up-directed and down-directed subsets.
The following lemma is a topologized version of Iwamura's Lemma \cite{Iwamura}  (cf. \cite{Bruns}, \cite{Mark}).

\begin{lemma}\label{l:up-p} Let $\kappa$ be an infinite cardinal. A topologized poset $X$ is ${\uparrow}\kappa$-complete if and only if any non-empty chain $C\subset X$ of cardinality $|C|\le\kappa$ has $\sup C\in\bar C$.
\end{lemma}

\begin{proof} The ``only if'' part is trivial as each chain is an up-directed subset in $X$. The ``if'' part will be proved by transfinite induction.

First we prove the lemma for the cardinal $\w$. Assume that each countable chain $C\subset X$ has $\sup C\in \bar C$. To prove that $X$ is ${\uparrow}\w$-complete,  take any non-empty countable up-directed subset $D=\{x_n\}_{n\in\w}$ in $X$. Put $y_0:=x_0$ an for every $n\in\IN$ choose an element $y_n\in D$ such that $y_n\ge x_n$ and $y_n\ge y_{n-1}$ (such an element $y_n$ exists as $D$ is up-directed).

By our assumption, the chain $C:=\{y_n\}_{n\in\w}$ has $\sup C\in \bar C\subset\bar D$. We claim that $\sup C$ is the least upper bound for the set $D$. Indeed, for any $n\in\w$ we get $x_n\le y_n\le\sup D$ and hence $\sup C$ is an upper bound for the set $D$. On the other hand, each upper bound $b$ for $D$ is an upper bound for $C$ and hence $\sup C\le b$. Therefore $\sup D=\sup C\in\bar C\subset\bar D$.
\smallskip

Now assume that for some uncountable cardinal $\kappa$  we have proved that each up-directed set $D$ of cardinality $|D|<\kappa$ in $X$ has $\sup D\in\bar D$ if each non-empty chain $C\subset X$ of cardinality $|C|<\kappa$ has $\sup C\in\bar C$. Assume that each chain $C\subset X$ of cardinality $|C|\le\kappa$ has $\sup C\in\bar C$. To prove that the topologized poset $X$ is a ${\uparrow}\kappa$-complete, fix any up-directed subset $D\subset X$ of cardinality $|D|\le\kappa$.

Since $D$ is up-directed, there exists a function $f:D\times D\to D$ assigning to any pair $(x,y)\in X\times X$ a point $f(x,y)\in D$ such that $x\le f(x,y)$ and $y\le f(x,y)$. Given any subset $A\subset D$, consider the increasing sequence $(A_n)_{n\in\w}$ of subsets of $A$, defined by the recursive formula $A_0:=A$ and $A_{n+1}:=A_n\cup f(A_n\times A_n)$ for $n\in\w$.
Finally, let $\langle A\rangle:=\bigcup_{n\in\w}A_n$ and observe that $\langle A\rangle$ is an up-directed subset of $D$. By induction it can be proved that $|A_n|\le \max\{\w,|A|\}$ for every $n\in\w$, and hence $|\langle A\rangle|\le\max\{\w,|A|\}$. Moreover, it can be also shown that for any subsets $A\subset B$ of $D$ we have $\langle A\rangle\subset \langle B\rangle$.

Write the set $D$ as $D=\{x_\alpha\}_{\alpha\in\kappa}$. For every $\beta\in\kappa$ let $D_\beta:=\langle\{x_\alpha\}_{\alpha\le\beta}\rangle$ and observe that $D_\beta\subset D$ is an up-directed set of cardinality $|D_\beta|\le\max\{\w,|\beta|\}<\kappa$. By the inductive assumption,  the up-directed set $D_\beta$ has $\sup D_\beta\in\bar D_\beta\subset\bar D$. For any ordinals $\alpha<\beta$ in $\kappa$ the inclusion $D_\alpha\subset D_\beta$ implies that $\sup D_\alpha\le \sup D_\beta$. By our assumption, the chain $C:=\{\sup D_\alpha:\alpha\in\kappa\}\subset \bar D$ has $\sup C\in \bar C\subset \bar D$. It remains to observe that $\sup C=\sup D$.
\end{proof}

\begin{corollary}\label{c:up-p}  A topologized poset $X$ is up-complete if and only if any non-empty chain $C\subset X$ has $\sup C\in\bar C$ in $X$.
\end{corollary}

Since a poset is up-complete if and only if it is up-complete as a topologized poset with the anti-discrete topology, Corollary~\ref{c:up-p} implies the following known characterization of up-complete posets, see \cite{Iwamura}, \cite{Bruns}, \cite{Mark}.

\begin{corollary}\label{c:up-poset}  A poset $X$ is up-complete if and only if any non-empty chain $C\subset X$ has $\sup C$ in $X$.
\end{corollary}

 We shall say that an up-directed set $D$ in a topologized poset $X$ {\em up-converges} to a point $x\in X$ if for each neighborhood $U_x\subset X$ of $x$ there exists $d\in D$ such that $D\cap{\uparrow}d\subset U_x$.

\begin{lemma}\label{l:up-conv} Let $\kappa$ be an infinite cardinal. If a topologized poset $X$ is ${\uparrow}\kappa$-complete, then each up-directed set $D\subset X$ of cardinality $|D|\le\kappa$ up-converges to its $\sup D$.
\end{lemma}

\begin{proof} To derive a contradiction, assume that some up-directed set $D\subset X$ of cardinality $|D|\le\kappa$ does not converge to $\sup D$. Then there exists an open neighborhood $U\subset X$ of $\sup D$ such that for every $d\in D$ the set $D\cap {\uparrow}d\setminus U$ is not empty. Then the set $E:=D\setminus U$ is directed and cofinal in $D$. To see that $E$ is directed, take any points $x,y\in E\subset D$ and find a point $z\in D$ with $x\le z$ and $y\le z$. Since $E$ is cofinal in $D$, there exists $e\in E$ such that $z\le e$. Then $x\le e$ and $y\le e$ by the transitivity of the partial order. By the up-completeness of $X$, the directed set $E$ has $\sup E\in \bar E$. The inclusion $E\subset D$ implies that $\sup E\le \sup D$ and the cofinality of $E$ in $D$ that $\sup E=\sup D$. Taking into account that $E\cap U=\emptyset$, we obtain a contradiction with $\sup D=\sup E\in\bar E$.
This contradiction completes the proof.
\end{proof}

We shall say that a down-directed set $D$ in a topologized poset $X$ {\em down-converges} to a point $x\in X$ if for each neighborhood $U_x\subset X$ of $x$ there exists $d\in D$ such that $D\cap{\downarrow}d\subset U_x$.

 Applying Lemmas~\ref{l:up-p}, \ref{l:up-conv} to the opposite partial order on a topologized poset, we obtain the following dual versions of these lemmas.

\begin{lemma}\label{l:down-p} Let $\kappa$ be an infinite cardinal. A topologized poset $X$ is ${\downarrow}\kappa$-complete if and only if any non-empty chain $C\subset X$ of cardinality $|C|\le\kappa$ has $\inf C\in\bar C$.
\end{lemma}

\begin{lemma}\label{down-conv} Let $\kappa$ be an infinite cardinal. If a topologized poset $X$ is ${\downarrow}\kappa$-complete, then each down-directed set $D\subset X$ of cardinality $|D|\le\kappa$ down-converges to its $\inf D$.
\end{lemma}

Lemma~\ref{l:down-p} implies the following two characterizations.

\begin{corollary}\label{c:down-p}  A topologized poset $X$ is down-complete if and only if any non-empty chain $C\subset X$ has $\inf C\in\bar C$ in $X$.
\end{corollary}

\begin{corollary}\label{c:up-poset}  A poset $X$ is down-complete if and only if any non-empty chain $C\subset X$ has $\inf C$ in $X$.
\end{corollary}

Unifying Corollaries~\ref{c:up-p} and \ref{c:down-p} we obtain the following useful characterization of completeness of topologized posets.

\begin{theorem}\label{t:completeT} A topologized poset $X$ is complete if and only if each non-empty chain $C\subset X$ has $\sup C\in\bar C$ and $\inf C\in\bar C$.
\end{theorem}

Since a poset is complete if and only if it is complete as a topologized poset with the anti-discrete topology, Corollary~\ref{t:completeT} implies the following (known) characterization of complete posets.

\begin{theorem}\label{t:complete} A poset $X$ is complete if and only if each non-empty chain $C\subset X$ has $\sup C$ and $\inf C$ in $X$.
\end{theorem}

\section{Preserving the completeness by Tychonoff products of topologized posets}

Now we prove that the completeness of topologized posets is preserved by Tychonoff products. On the Tychonoff product $\prod_{\alpha\in A}X_\alpha$ of topologized posets $(X_\alpha,\le_\alpha)$ we consider the pointwise partial order $\le$ defined by $(x_\alpha)_{\alpha\in A}\le(y_\alpha)_{\alpha\in A}$ iff $x_\alpha\le_\alpha y_\alpha$ for each $\alpha\in A$.

\begin{theorem}\label{t:up-prod} Let $\kappa$ be an infinite cardinal. The Tychonoff product $X:=\prod_{\alpha\in A}X_\alpha$ of ${\uparrow}\kappa$-complete topologized posets $X_\alpha$, $\alpha\in A$, is an ${\uparrow}\kappa$-complete topologized poset.
\end{theorem}

\begin{proof}
By Lemma~\ref{l:up-p}, the ${\uparrow}\kappa$-completeness of $X$ will follow as soon as we prove that each non-empty chain $C\subset X$ of cardinality $|C|\le\kappa$ has $\sup C\in \bar C$. For every $\alpha\in A$ let $\pr_\alpha:X\to X_\alpha$ denote the coordinate projection. Since the projection $\pr_\alpha$ is monotone, the image $C_\alpha:=\pr_\alpha(C)$ of the chain $C$ is a chain in $X_\alpha$ of cardinality $|C_\alpha|\le|C|\le\kappa$. By the ${\uparrow}\kappa$-completeness of the topologized poset $X_\alpha$, the chain $C_\alpha$ has $\sup C_\alpha\in \bar C_\alpha$.

Consider the elements $c:=(\sup C_\alpha)_{\alpha\in A}\in X$. It is clear that $c$ is an upper bound for $C$. We claim that $c=\sup C$. Indeed, given any other upper bound $b=(b_\alpha)_{\alpha\in A}\in X$ of $C$, we have $\sup C_\alpha\le b_\alpha$ for all $\alpha\in A$ and hence $c\le b$. So, $c=\sup C$.

It remains to show that $c\in\bar C$. Assuming that $c\notin\bar C$, we could find an open neighborhood $O_c\subset X$ such that $O_c\cap C=\emptyset$. Replacing $O_c$ by a smaller neighborhood, we can assume that $O_c$ is of the basic form $O_c=\prod_{\alpha\in A}U_\alpha$, where the set $F=\{\alpha\in A:U_\alpha\ne X_\alpha\}$ is finite. For every $\alpha\in F$ the set $U_\alpha$ is a neighborhood of $\sup C_\alpha$. Applying Lemma~\ref{l:up-conv}, find a point $u_\alpha\in C_\alpha$ such that $C_\alpha\cap{\uparrow}u_\alpha\subset U_\alpha$. Since $u_\alpha\in C_\alpha=\pr_\alpha(C)$, there exists an element $v_\alpha\in C$ such that $\pr_\alpha(v_\alpha)=u_\alpha$. Since $C$ is a chain, the finite set  $\{v_\alpha:\alpha\in F\}$ has a largest element $v$. For this element we have $\pr_\alpha(v)\in C_\alpha\cap{\uparrow}u_\alpha\subset U_\alpha$ for all $\alpha\in F$. Consequently, $v\in C\cap O_c$, which contradicts the choice of the neighborhood $O_c$.
\end{proof}

Applying Theorem~\ref{t:up-prod} to the opposite order on a topologized poset, we get the following dual version of Theorem~\ref{t:up-prod}.

\begin{theorem}\label{t:down-prod} Let $\kappa$ be an infinite cardinal. The Tychonoff product $X:=\prod_{\alpha\in A}X_\alpha$ of ${\downarrow}\kappa$-complete topologized posets $X_\alpha$, $\alpha\in A$, is a ${\downarrow}\kappa$-complete topologized poset.
\end{theorem}

Theorems~\ref{t:up-prod} and \ref{t:down-prod} imply:

\begin{theorem}\label{t:updown-prod} Let $\kappa$ be an infinite cardinal. The Tychonoff product $X:=\prod_{\alpha\in A}X_\alpha$ of ${\updownarrow}\kappa$-complete topologized posets $X_\alpha$, $\alpha\in A$, is an ${\updownarrow}\kappa$-complete topologized poset.
\end{theorem}

Let us also note the following obvious preservation property of complete topologized posets.

\begin{proposition}\label{p:sub} If a topologized poset $X$ is complete (resp. up-complete, down-complete), then so is each closed topologized subposet in $X$.
\end{proposition}

\section{Interplay between completeness and chain-compactness}\label{s:cc}

In this section we establish the relation between the completeness and chain-compactness of topologized posets.

A topologized poset is defined to be {\em chain-compact} if each closed chain in $X$ is compact.

\begin{lemma}\label{l:c=>cc} Each complete topologized poset $X$ is chain-compact.
\end{lemma}

\begin{proof} Given a non-empty closed chain $C$ in a complete topologized poset $X$, we shall prove that $C$ is compact. Given any open cover $\U$ of $C$, we should find a finite subfamily $\U'\subset \U$ such that $C\subset\bigcup\U'$.  By the down-completeness of $X$, the closed chain $C$ has $\inf C\in\bar C=C$, which means that $c:=\inf C$ is the smallest element of $C$. Consider the set $A\subset C$ consisting of points $a\in C$ such that the closed interval $[c,a]:=\{x\in C:c\le x\le a\}$ can be covered by a finite subfamily of the cover $\U$.
The set $A$ contains the point $c$ and hence is not empty. By the up-completeness of $X$, the set $A$ has $\sup A\in\bar A\subset\bar C$.

We claim that the point $b:=\sup A$ belongs to $A$. To derive a contradiction, assume that $b\notin A$. Choose any open set $U_b\in \U$ containing the point $b:=\sup A\subset \bar C$.
By Lemma~\ref{l:up-conv}, the chain $A$ contains a point $a\in A$ such that $A\cap{\uparrow}a\subset U_b$. Taking into account that for any $x\in A$ the interval $[c,x]\subset \bar C$ is contained in $A$, we see that $A\cap{\uparrow}a=[a,b]\setminus\{b\}$. Then $[a,b]=(A\cap{\uparrow}a)\cup\{b\}\subset U_b$.
Now the definition of the set $A\ni a$ yields a finite subfamily $\V\subset\U$ such that $[c,a]\subset\bigcup\V$. Then for the finite  subfamily $\U'=\V\cup\{U_b\}$ of $\U$ we have $[c,b]=[c,a]\cup[a,b]\subset\bigcup \U'$, which means that $b\in A$.

To complete the proof, it suffices to show that $C\subset \bigcup\U'$. Assuming that the (closed) set $E:=C\setminus\bigcup\U'$ is not empty, we can apply the completeness of $X$ and find $\inf E\in \bar E=E$. Choose any open set $U_e\in\U$ containing the point $e:=\inf E$ and observe that $\U'\cup\{U_e\}$ is a finite subfamily covering the set $[c,e]$, which means that $e\in A$. On the other hand, the (non)inclusion $[c,b]\subset\bigcup\U'\not\ni e$ implies that $b<e$, which contradicts the equality $b=\sup A$. \end{proof}

Lemma~\ref{l:c=>cc} can be reversed for ${\uparrow}{\downarrow}$-closed topologized posets. We define a topologized poset $X$ to be
\begin{itemize}
\item {\em ${\uparrow}$-closed} if the upper set ${\uparrow}x$ of any point $x\in X$ is closed in $X$;
\item {\em ${\downarrow}$-closed} if the lower set ${\downarrow}x$ of any point $x\in X$ is closed in $X$;
\item {\em ${\downarrow}{\uparrow}$-closed} if it is ${\uparrow}$-closed and ${\downarrow}$-closed;
\item {\em ${\updownarrow}$-closed} if for any point $x\in X$ the set ${\updownarrow}x:={\uparrow}x\cup{\downarrow}x$ is closed in $X$;
\item a {\em pospace} if the partial order $\le$ is a closed subset of $X\times X$;
\item {\em chain-closed} if the closure of each chain in $X$ is a chain.
\end{itemize}
For topologized posets we have the following implications:
$$\mbox{pospace} \Ra \mbox{${\uparrow}{\downarrow}$-closed} \Ra \mbox{${\updownarrow}$-closed} \Ra \mbox{chain-closed}.$$
The last implication is not entirely trivial and is proved in the following lemma.

\begin{lemma}\label{l:cc1} Each ${\updownarrow}$-closed topologized poset is chain-closed.
\end{lemma}

\begin{proof}  Given a chain $C\subset X$, we should prove that its closure $\bar C$ in $X$ is a chain. Assuming that $\bar C$ contains two incomparable elements $x$  and $y$, observe that $V_x:=X\setminus {\updownarrow}y$ is an open neighborhood of $x$. Since $x\in\bar C$, there exists an element $z\in C\cap V_x$. It follows from $z\notin{\updownarrow}y$ that $V_y=X\setminus {\updownarrow}z \supset X\setminus C$ is an open neighborhood of $y$, disjoint with $C$, which is not possible as $y\in\bar C$.
\end{proof}

\begin{theorem}\label{t:cc<=>kc} An ${\uparrow}{\downarrow}$-closed topologized poset $X$ is complete if and only if it is chain-compact.
\end{theorem}

\begin{proof} The ``only if'' part follows from Lemma~\ref{l:c=>cc}. To prove the ``if'' part, assume that an ${\uparrow}{\downarrow}$-closed topologized poset $X$ in chain-compact. To prove that $X$ is complete, take any non-empty chain $C$. By Lemma~\ref{l:cc1}, the closure $\bar C$ of $C$ is a chain in $X$. By the chain-compactness of $X$, the closed chain $\bar C$ is compact. By the compactness of $\bar C$, the centered family $\F=\{\bar C\cap{\downarrow}x:x\in \bar C\}$ of closed subsets of $\bar C$ has non-empty intersection, which is a singleton, containing the smallest element $s$ of the compact chain $\bar C$. It is clear that $s$ is a lower bound for the set $C$. On the other hand, for any other lower bound $b$ for the set $C$, we get $C\subset{\uparrow}b$ and hence $s\in\bar C\subset\overline{{\uparrow}b}={\uparrow}b$ and finally, $b\le s$. So, $s=\inf C$.

By analogy we can prove that the compact chain $\bar C$ has the largest element which coincides with $\sup C$.
\end{proof}

\section{Complete topologized semilattices}\label{s:5}
In this section we study the notion of completeness in the framework of topologized semilattices.

By a {\em semilattice} we understand a commutative semigroup $X$ of idempotents (the latter means that $xx=x$ for all $x\in X$). Each semilattice $X$ carries a natural partial order $\le$ defined by $x\le y$ iff $xy=x$. So, we can consider a semilattice $X$ as a poset such that each non-empty finite subset $A=\{a_1,\dots,a_n\}\subset X$ has $\inf A=a_1\cdots a_n$.

By a {\em topologized semilattice} we understand a semilattice endowed with a topology. A topologized semilattice $X$ is called a
({\em semi\/}){\em topological semilattice} if the semilattice operation $X\times X\to X$, $(x,y)\mapsto xy=\inf\{x,y\}$, is (separately) continuous.

A topologized semilattice is {\em complete} (resp. {\em up-complete, down-complete}) if it is complete (resp. up-complete, down-complete) as a topologized poset endowed with the natural order, induced by the semilattice operation. By Theorem~\ref{t:completeT}, a topologized semilattice $X$ is {\em complete} if and only if each non-empty chain $C\subset X$ has $\inf C\in\bar C$ and $\sup C\in\bar C$. Observe that a discrete topological semilattice is complete if and only if it is {\em chain-finite} in the sense that each chain in $X$ is finite.

The completeness of topologized semilattices is preserved by many operations. The following two propositions are partial cases of Proposition~\ref{p:sub} and Theorem~\ref{t:updown-prod}.

\begin{proposition} Let $X$ be a closed subsemilattice of a topologized semilattice $Y$. If $Y$ is complete (resp. up-complete, down-complete), then so is the topologized semilattice $X$.
\end{proposition}

\begin{proposition}  If topologized semilattices $X_\alpha$, $\alpha\in A$, are complete (resp. up-complete, down-complete), then so is their Tychonoff product $\prod_{\alpha\in A}X_\alpha$.
\end{proposition}

A topologized poset $X$ is defined to be {\em weakly ${\uparrow}$-closed} if for every $x\in X$ we have $\overline{\{x\}}\subset{\uparrow}x$. It is easy to see that a topologized poset is weakly ${\uparrow}$-closed if it is ${\uparrow}$-closed or satisfies the separation axiom $T_1$.

\begin{lemma}\label{l:homo} Let $\kappa$ be an infinite cardinal. Let $h:X\to Y$ be a continuous surjective homomorphism from a topologized semilattice $X$ to a weakly ${\uparrow}$-closed topologized semilattice $Y$. If the topologized semilattice $X$ is down-complete (and ${\uparrow}\kappa$-complete), then so is the topologized semilattice $Y$.
\end{lemma}

\begin{proof} Assume that the topologized semilattice $X$ is down-complete (and ${\uparrow}\kappa$-complete). By Corollary~\ref{c:down-p} (and Lemma~\ref{l:up-p}), the down-completeness (and ${\uparrow}\kappa$-completeness) of $X$ will follow as soon as we show that each non-empty chain $C\subset X$ (of cardinality $|C|\le\kappa$) has $\inf C\in\bar C$ (and $\sup C\in\bar C$).

Observe that for every $c\in C$ the preimage $h^{-1}(c)$ is a subsemilattice in $X$ and hence is a down-directed set in $X$. By the down-completeness of $X$, it has $\inf h^{-1}(c)\in\overline{h^{-1}(c)}$. Let $b_c:=\inf h^{-1}(c)$.
By the continuity of $h$ and the weak $\uparrow$-closedness of $Y$, $$h(b_c)\in h(\overline{h^{-1}(c)})\subset\overline{h(h^{-1}(c))}=\overline{\{c\}}\subset {\uparrow}c$$ and hence $c\le h(b_c)$. On the other hand, for any $x\in h^{-1}(c)$, we get $b_c\le x$ and hence $h(b_c)\le h(x)=c$ and finally $$h(b_c)=c.$$ Let us show that $b_c=\inf h^{-1}({\uparrow}c)$. Indeed, for any $x\in h^{-1}({\uparrow}c)$ we get $h(x\cdot b_c)=h(x)\cdot h(b_c)=h(x)\cdot c=c$ and hence $b_c\le x\cdot b_c\le x$. So, $b_c$ is a lower bound for the set $h^{-1}({\uparrow}c)$. On the other hand, any lower bound $b$ of the set $h^{-1}({\uparrow}c)$ is a lower bound of the set $h^{-1}(c)\subset h^{-1}({\uparrow}c)$ and hence $b\le b_c$, which means that $b_c=\inf h^{-1}({\uparrow}c)$.

For every elements $x\le y$ in $C$ the inclusion $h^{-1}({\uparrow}y)\subset h^{-1}({\uparrow}x)$ implies $b_x=\inf h^{-1}({\uparrow}x)\le\inf h^{-1}({\uparrow}y)=b_y$. So the set $D=\{b_c:c\in C\}$ is a chain (of cardinality $\le\kappa$).
By the down-completeness (and ${\uparrow}\kappa$-completeness) of $X$ the chain $D$ has $\inf D\in\bar D$ (and $\sup D\in\bar D$). The continuity of $h$ implies that $f(\inf D)\in f(\bar D)\subset\overline{f(D)}=\bar C$ (and
$f(\sup D)\in f(\bar D)\subset\overline{f(D)}=\bar C$).

It remains to prove that $f(\inf D)=\inf C$ (and $f(\sup D)=\sup C$). The monotonicity of $f$ implies that $f(\inf D)$ is a lower bound (and $f(\sup D)$ is an upper bound) for $f(D)=C$. For any lower bound $\lambda$ of $C$, we get $C\subset {\uparrow}\lambda$ and $D\subset h^{-1}(C)\subset h^{-1}({\uparrow}\lambda)$. Then $b_\lambda=\inf h^{-1}({\uparrow}\lambda)\le \inf D$ and $\lambda=h(b_\lambda)\le h(\inf D)$, which means that $h(\inf D)=\inf C$. (For any upper bound $u$ of $C$ and any $c\in C$ the inequality $c\le u$ implies $b_c\le b_u$. So, $b_u$ is an upper bound of the chain $D=\{b_c:c\in C\}$. Then $\sup D\le b_u$ and $h(\sup D)\le h(b_u)=u$ by the monotonicity of $h$, witnessing that $h(\sup D)=\sup C$).
\end{proof}

Finally, we discuss the relation of the completeness of a topologized semilattice $X$ to the compactness of the the {\em weak$^\bullet$ topology} $\Zar_X$, which is generated by the subbase consisting of complements to closed subsemilattices in $X$. A topologized semilattice $X$ is called {\em $\Zar$-compact} if its weak$^\bullet$ topology $\Zar_X$ is compact. The weak$^\bullet$-topology was introduced and studied in \cite{BBw}. According to Lemmas 5.4, 5.5 of \cite{BBw}, for any topologized semilattice we have the implications:
$$\mbox{complete} \Ra \mbox{$\Zar$-compact} \Ra \mbox{chain-compact}.
$$These implications combined with Theorem~\ref{t:cc<=>kc} yield the following characterization.

\begin{theorem} For an ${\uparrow}{\downarrow}$-closed topologized semilattice $X$ the following conditions are equivalent:
\begin{enumerate}
\item $X$ is complete;
\item$X$ is $\Zar_X$-compact;
\item $X$ is chain-compact.
\end{enumerate}
\end{theorem}

\section{Absolute closedness of complete topologized semilattices}\label{s:ac}

Quite often the notion of completeness is connected with the absolute closedness, understood in an appropriate sense. For example, a metric space $X$ is complete if and only if it is closed in each metric space $Y$, containing $X$ as a metric subspace. A uniform space $X$ is complete if and only if $X$ is closed in each uniform space, containing $X$ as a uniform subspace. A topological group $X$ is complete in its two-sided uniformity if and only if $X$ is closed in any topological group, containing $X$ as a topological subgroup. A similar phenomenon happens in the category of (semi)topological semilattices.

Historically the first result in this direction belongs to J.W.~Stepp \cite{Stepp1969}, \cite{Stepp1975} who
 proved that any chain-finite semilattice $X$ is closed in each Hausdorff topological semilattice, containing $X$ as a subsemilattice. This result of Stepp was improved to the following characterizations, which can be found in \cite{BBm}.

\begin{theorem} A semilattice $X$ is chain-finite if and only if $X$ is closed in any Hausdorff topological semilattice $Y$ containing $Z$ as a subsemilattice.
\end{theorem}

\begin{theorem} A Hausdorff topological semilattice $X$ is complete if and only if each closed subsemilattice $Z$ of $X$ is closed in any Hausdorff topological semilattice containing $Z$ as a topological subsemilattice.
\end{theorem}

The ``only if'' parts of these characterizations are partial cases of the following theorems on closedness of (chain-finite) complete topologized semilattices under continuous homomorphisms.

\begin{theorem}\label{t:cf} For any  homomorphism $h:X\to Y$ from a chain-finite semilattice $X$ to a Hausdorff  semitopological semilattice $Y$, the image $h(X)$ is closed in $Y$.
\end{theorem}

\begin{theorem}\label{t:ct} For any continuous homomorphism $h:X\to Y$ from a complete topologized semilattice $X$ to a Hausdorff topological semilattice $Y$, the image $h(X)$ is closed in $Y$.
\end{theorem}

In fact, Theorems~\ref{t:cf} and \ref{t:ct} are corollaries of more general results related to upper semicontinuous $T_i$-multimorphisms between topologized semilattices.

A {\em multi-valued map} $\Phi:X\setmap Y$ between sets $X,Y$ is a function assigning to each point $x\in X$ a subset $\Phi(x)$ of $Y$. For a subset $A\subset X$ we put $\Phi(A):=\bigcup_{x\in A}\Phi(x)$. A multi-valued map $\Phi:X\setmap Y$ between semigroups is called a {\em multimorphism} if $\Phi(x)\cdot\Phi(y)\subset\Phi(xy)$ for any $x,y\in X$. Here $\Phi(x)\cdot\Phi(y):=\{ab:a\in\Phi(x),\;b\in\Phi(y)\}$.

A multi-valued map $\Phi:X\setmap Y$ between topological spaces is called
{\em upper semicontinuous} if for any closed subset $F\subset Y$ the preimage $\Phi^{-1}(F):=\{x\in X:\Phi(x)\cap F\ne\emptyset\}$ is closed in $X$.

A subset $F$ of a topological space $X$ is called {\em $T_1$-closed} (resp.
{\em $T_2$-closed}) in $X$ if each point $x\in X\setminus F$ has a (closed) neighborhood, disjoint with $F$.

A multimorphism $\Phi:X\setmap Y$ is called a {\em $T_i$-multimorphism} for $i\in\{1,2\}$ if for any $x\in X$ the set $\Phi(x)$ is $T_i$-closed in $Y$.

The following two theorems (implying Theorems~\ref{t:cf} and \ref{t:ct}) are proved in \cite{BBm}.

\begin{theorem} For any $T_1$-multimorphism $\Phi:X\to Y$ from a chain-finite semilattice $X$ to a topological semilattice $Y$, the image $\Phi(X)$ is closed in $Y$.
\end{theorem}

\begin{theorem}\label{t:multi} For any upper semi-continuous $T_2$-multimorphism $\Phi:X\setmap Y$ from a complete topologized semilattice $X$ to a topological semilattice $Y$, the image $\Phi(X)$ is closed in $Y$.
\end{theorem}

Looking at Theorems~\ref{t:cf} and \ref{t:ct}, one can ask the following problem.

\begin{problem}\label{prob:abscl} Let $h:X\to Y$ be a continuous homomorphism from a complete topologized semilattice $X$ to a Hausdorff semitopological semilattice $Y$. Is the set $h(X)$ closed in $Y$?
\end{problem}

Problem~\ref{prob:abscl} has affirmative answer for homomorphisms to sequential  semitopological semilattices.

We recall that a topological space $X$ is {\em sequential} if each sequentially closed subset in $X$ is closed. A subset $A$ of a topological space $X$ is called {\em sequentially closed} if $A$ contains the limit points of all sequences $\{a_n\}_{n\in\w}\subset A$ that converge in $X$.

A topological space $X$ is {\em countably tight} if for any subset $A\subset X$ and point $a\in\bar A$ there exists a countable subset $B\subset A$ such that $a\in\bar B$. It is well-known \cite[1.7.13(c)]{Engelking1989} that each subspace of a sequential topological space has countable tightness. The following (non-trivial) result was proved in \cite{BBs}.

\begin{theorem}\label{t:seq} For any continuous homomorphism $h:X\to Y$ from a countably tight complete topologized semilattice $X$ to a Hausdorff semitopological semilattice $Y$, the image $h(X)$ is sequentially closed in $Y$.

\end{theorem}

\begin{corollary}\label{c:seq} For any continuous homomorphism $h:X\to Y$ from a complete topologized semilattice $X$ to a sequential Hausdorff semitopological semilattice $Y$, the image $h(X)$ is closed in $Y$.

\end{corollary}

Also Problem~\ref{prob:abscl} has an affirmative answer for semitopological semilattices satisfying the separation axiom $\vv{T}_{2\delta}$ (which is stronger than the Hausdorff axiom $T_2$ but weaker than axiom $\vv{T}_{3\frac12}$ of the functional Hausdorffness).

Let us recall that a topological space $X$ satisfies the separation axiom
\begin{itemize}
\item $T_1$ if for any distinct points $x,y\in X$ there exists an open set $U\subset X$ such that $x\in U\subset X\setminus\{y\}$;
\item $T_2$ if for any distinct points $x,y\in X$ there exists an open set $U\subset X$ such that $x\in U\subset\bar U\subset X\setminus\{y\}$;
\item $T_3$ if $X$ is a $T_1$-space and for any open set $V\subset X$ and point $x\in V$ there exists an open set $U\subset X$ such that $x\in U\subset\bar U\subset V$;
\item $T_{3\frac12}$ if $X$ is a $T_1$-space and for any open set $V\subset X$ and point $x\in V$ there exists a continuous function $f:X\to[0,1]$ such that $x\in f^{-1}([0,1))\subset V$;
\item $T_{2\delta}$ if $X$ is a $T_1$-space and for any open set $V\subset X$ and point $x\in V$ there exists a countable family $\U$ of closed neighborhoods of $x$ in $X$ such that $\bigcap\U\subset V$;
\item $\vv{T}_i$ for $i\in\{1,2,2\delta,3,3\frac12\}$ if $X$ admits a bijective continuous map $X\to Y$ to a $T_i$-space $Y$.
\end{itemize}

Topological spaces satisfying a separation axiom $T_i$ are called {\em $T_i$-spaces}. The separation axioms $T_{2\delta}$ and $\vv{T}_{2\delta}$ were introduced in \cite{BBR}.

The following diagram describes the implications between the separation axioms $T_i$ and $\vv{T}_i$ for $i\in\{1,2,2\delta,3,3\frac12\}$.
$$\xymatrix{
T_{3\frac12}\ar@{=>}[r]\ar@{=>}[d]&T_3\ar@{=>}[r]\ar@{=>}[d]&T_{2\delta}\ar@{=>}[r]\ar@{=>}[d]&T_2\ar@{<=>}[d]\ar@{=>}[r]&T_1\ar@{<=>}[d]\\
\vv{T}_{3\frac12}\ar@{=>}[r]&\vv{T}_3\ar@{=>}[r]&\vv{T}_{2\delta}\ar@{=>}[r]&\vv{T}_2\ar@{=>}[r]&T_1
}
$$
Observe that a topological space $X$ satisfies the separation axiom $\vv{T}_{3\frac12}$ if and only if it is {\em functionally Hausdorff\/} in the sense that for any distinct points $x,y\in X$ there exists a continuous function $f:X\to\mathbb R$ with $f(x)\ne f(y)$. Therefore, each functionally Hausdorff space is a $\vv{T}_{2\delta}$-space.

The following (non-trivial) result was proved in \cite{BBR}.

\begin{theorem}\label{t:Tdelta} For any continuous homomorphism $h:X\to Y$ from a complete topologized semilattice to a semitopological semilattice $Y$ satisfying the separation axiom $\vv{T}_{2\delta}$, the image $h(X)$ is closed in $Y$.
\end{theorem}

Next, we give a partial answer to Problem~\ref{prob:abscl} for $\bar G_\delta$-separated semitopological semilattices. A topologized semilattice $X$ is defined to be {\em $\bar G_\delta$-separated} if for any distinct points $x,y$ in $X$ there exists a countable family $\U$ of closed neighborhoods of $x$ in $X$ such that $\bigcap\U$ is a closed subsemilattice of $X$ which does not contain $y$. It is easy to see that each $\bar G_\delta$-separated topologized semilattice is Hausdorff.

\begin{proposition} Each Hausdorff topological semilattice $X$ is $\bar G_\delta$-separated.
\end{proposition}

\begin{proof} Given two distinct points $x,y\in X$, choose a decreasing sequence $(U_n)_{n\in\w}$ of open neighborhoods of $x$ such that $y\notin\bar U_0$ and $\bar U_{n}\cdot \bar U_n\subset \bar U_{n-1}$ for all $n\in\IN$. The choice of $U_0$ is possible by the Hausdorff property of $X$ and the choice of the neighborhoods $U_n$ is possible by the continuity of the semilattice operation at $(x,x)$. It follows that the intersection $\bigcap_{n\in\w}\bar U_n$ is a closed subsemilattice of $X$ containing $x$ but not $y$. This witnesses that the topological semilattice $X$ is $\bar G_\delta$-separated.
\end{proof}

\begin{theorem}\label{t:Gd} For any continuous homomorphism $h:X\to Y$ from a complete topologized semilattice $X$ to a $\bar G_\delta$-separated semitopological semilattice $Y$, the image $h(X)$ is closed in $Y$.
\end{theorem}

\begin{proof} The semitopological semilattice $Y$, being $\bar G_\delta$-separated, is Hausdorff and hence ${\uparrow}{\downarrow}$-closed. By Lemma~\ref{l:homo}, the completeness of the topologized semilattice $X$ implies the completeness of the semitopological semilattice $Z=h(X)\subset Y$. To derive a contradiction, assume that the subsemilattice $h(X)=Z$ is not closed in $Y$ and find a point $y\in\overline{Z}\setminus Z$.

Let $\mathfrak U$ be the set of all countable families $\U$ of closed neighborhoods of $y$ such that $\bigcap\U$ is a subsemilattice of $Y$.

\begin{claim}\label{cl:ne} For any $\U\in\mathfrak U$ the intersection $Z\cap\bigcap\U$ is not empty.
\end{claim}

\begin{proof} Let $\U=\{U_n\}_{n\in\w}$ be an enumeration of the countable family $\U$. For every $n\in\w$ let $U_n^\circ$ be the interior of $U_n$ in $Y$. By induction we shall construct a sequence of points $(z_n)_{n\in\w}$ in $Z$ such that for every $k\le n$ the following condition is satisfied for every $k\le n$:
\begin{itemize}
\item[$(*_{k,n})$] $z_k\cdots z_n,z_k\cdots z_ny\in\bigcap_{i\le k}U_i^\circ$.
\end{itemize}
 To start the inductive construction, choose an open neighborhood $V_0\subset Y$ of $y$ such that $V_0\cup V_0y\subset U_0^\circ$. Next, choose any point $z_0\in Z\cap V_0$ and observe that the condition $(*_{0,0})$ is satisfied.

Now assume that for some $n\in\w$ we have chosen points $z_0,\dots,z_{n}\in Z$ such that for every $k\le n$ the condition $(*_{k,n})$ is satisfied.  So, for every $k\le n$ the point $z_k\cdots z_ny=z_k\cdots z_nyy$ belongs to $\bigcap_{i\le k}U_i^\circ$ and by the separate continuity of the semilattice operation, we can find a neighborhood $V_k\subset Y$ of $y$ such that the sets  $z_k\cdots z_{n}V_k$ and $z_k\cdots z_nV_ky$ are contained in $\bigcap_{i\le k}U_i^\circ$. Now choose any point $z_{n+1}\in Z\cap\bigcap_{k=0}^nV_k$ and observe that for every $k\le n+1$ the condition $(*_{k,n+1})$ is satisfied. This completes the inductive construction.

By the down-completeness of $Z$, for every $k\in\w$ the chain $C_k:=\{z_k\cdots z_n:n\ge k\}$ has $\inf C_k\in \bar C_k\cap Z$. Observe that for every $k<n$ we have $\inf C_k\le z_kz_{k+1}\cdots z_n\le z_{k+1}\cdots z_n$, which means that $\inf C_k$ is a lower bound of the chain $C_{k+1}$ and hence $\inf C_k\le\inf C_{k+1}$. By the up-completeness of $Z$, the chain $C=\{\inf C_k\}_{k\in\w}$ has $\sup C\in \bar C\cap Z$.

We claim that $\sup C\in \bigcap\U$. Observe that for every $k\le n$ we have $z_k\cdots z_n\in \bigcap_{i\le k}U_i$. Then $C_k\subset \bigcap_{i\le k}U_i$ and $\inf C_k\in Z\cap\bar C_k\subset Z\cap\bigcap_{i\le k}U_i$ for every $k\in\w$. Finally, $\sup C=\sup_{k\in\w}\inf C_k\in Z\cap\bigcap_{k\in\w}\bigcap_{i\le k}U_i=Z\cap\bigcap\U$. So $Z\cap\bigcap\U$ is not empty.
\end{proof}

By Claim~\ref{cl:ne}, for every $\U\in\mathfrak U$ the closed subsemilattice $Z\cap\bigcap\U$ of $Z$ is not empty and hence has $\inf (Z\cap \bigcap\U)\in Z\cap\bigcap\U$ (by the down-completeness of $Z$). Next, observe that the set $D:=\{\inf(Z\cap\bigcap\U):\U\in\mathfrak U\}$ is up-directed. By the up-completeness of $Z$, the set $D$ has $\sup D\in\bar D\cap Z$.

 Since the semitopological semilattice $Y$ is $\bar G_\delta$-separated, there exists a countable family $\V\in\mathfrak U$ such that $\sup D\notin \bigcap\V$.  Consider the set $\mathfrak U'=\{\U\in\mathfrak U:\V\subset\U\}$ and observe that the set $D'=\{\inf(Z\cap\bigcap\U):\U\in\mathfrak U'\}$ is cofinal in $D$. Indeed, for any $\U\in\mathfrak \U$ the family $\U':=\U\cup\V$ belongs to the family $\mathfrak U'$
 and has $\inf (Z\cap\bigcap\U')\ge\inf (Z\cap\U)$ as $\bigcap\U'\subset\bigcap\U$.

By the cofinality of $D'$ in $D$, we have $\sup D=\sup D'\in \bar D'\subset Z\cap\bigcap\V$, which contradicts the choice of the family $\V$. This contradiction completes the proof of the closedness of $h(X)$ in $Y$.
\end{proof}

Corollary~\ref{c:seq} and Theorems~\ref{t:Tdelta}, \ref{t:Gd} imply the following partial answer to Problem~\ref{prob:abscl}.

\begin{corollary}\label{c:final} For a continuous homomorphism $h:X\to Y$ from a complete topologized semilattice $X$ to a Hausdorff semitopological semilattice $Y$, the image $h(X)$ is closed in $Y$ if one of the following conditions is satisfied:
\begin{enumerate}
\item the topologized semilattice $Y$ is $\bar G_\delta$-separated;
\item $Y$ is a topological semilattice;
\item the topological space $Y$ is sequential;
\item the topological space $Y$ satisfies the separation axiom $\vv{T}_{2\delta}$;
\item the topological space $Y$ is functionally Hausdorff.
\end{enumerate}
\end{corollary}

Comparing Theorems~\ref{t:Tdelta} and \ref{t:multi} we can ask the following problem.

\begin{problem} Let $\Phi:X\setmap Y$ be an upper semi-continuous $T_2$-multimorphism from a complete topologized semilattice $X$ to a semitopological semilattice $Y$ satisfying the separation axiom $\vv{T}_{2\delta}$. Is the image $\Phi(X)$ closed in $Y$?
\end{problem}

\section{The closedness of the partial order in Hausdorff semitopological semilattices}\label{s:o}

Observing that the partial order $\{(x,y)\in X\times X:xy=x\}$ of a Haudorff topological semilattice $X$ is a closed subset of $X\times X$ we can ask the following problem, considered also in \cite{BBs} and \cite{BBR}.

\begin{problem}\label{prob2} Let $X$ be a complete Hausdorff semitopological semilattice. Is the partial order $\{(x,y)\in X\times X:xy=x\}$ closed in $X\times X$?
\end{problem}

In this section we survey some results giving partial answers to Problem~\ref{prob2}. The following theorem is proved in \cite{BBs}.

\begin{theorem}\label{t:seq2} Let $X$ be a countable tight Hausdorff semitopological semilattice. If $X$ is ${\downarrow}\w$-complete and ${\uparrow}\w_1$-complete, then the partial order $\{(x,y)\in X\times X:xy=x\}$ is sequentially closed in $X\times X$.
\end{theorem}

This theorem implies the following partial answer to Problem~\ref{prob2}.

\begin{corollary} Let $X$ be a Hausdorff semitopological semilattice whose square $X\times X$ is sequential. If $X$ is ${\downarrow}\w$-complete and ${\uparrow}\w_1$-complete, then the partial order $\{(x,y)\in X\times X:xy=x\}$ is closed in $X\times X$.
\end{corollary}

Another partial answer to Problem~\ref{prob2} was given in \cite{BBR}.

\begin{theorem} For any complete semitopological semilattice $X$ satisfying the separation axiom $\vv{T}_{2\delta}$ the partial order $\{(x,y)\in X\times X:xy=x\}$ is closed in $X\times X$.
\end{theorem}

A topologized semilattice $X$ is called a {\em $V$-semilattice} if for any point $x\in X$ and $y\in X\setminus{\downarrow}x$ there exists a point $z\in X\setminus {\downarrow}x$ whose upper cone ${\uparrow}z$ contains the point $y$ in its interior.

\begin{proposition} For any $\uparrow$-closed topologized $V$-semilattice $X$, the partial order $P=\{(x,y)\in X\times X:x\le y\}$ is closed in $X\times X$.
\end{proposition}

\begin{proof} Given any pair $(x,y)\notin P$, we conclude that $x\not\le y$ and hence $x\notin{\downarrow}y$. Since $X$ is a $V$-semilattice, there exists a point $z\in X\setminus{\downarrow}y$ such that ${\uparrow}z$ contains $x$ in its interior $({\uparrow}z)^\circ$.
Then $U_x:=({\uparrow}z)^\circ$ and $U_y=X\setminus{\uparrow}z$ are two open sets in $X$ such that $U_x\times U_y$ is disjoint with $P$. So $P$ is closed in $X\times Y$.
\end{proof}

A semilattice $X$ is called {\em $\uparrow$-finite} if for every $x\in X$ the upper set ${\uparrow}x$ is finite. The following  example is constructed in \cite{BBR2}.

\begin{example} There exists an ${\uparrow}$-finite metrizable semitopological semilattice $X$ whose partial order $P=\{(x,y)\in X\times X:x\le y\}$ is a dense non-closed subset of $X\times X$.
\end{example}

A topologized semilattice is called {\em Lawson} if its topology has a base consisting of open subsemilattices.

\begin{problem} Is there a Lawson Hausdorff semitopological semilattice with non-closed partial order?
\end{problem}

config: arxiv


TITLE: Find the slope of the line such that the area of triangle formed inside the circle is maximum
QUESTION [4 upvotes]: Here is the question. It states:

A circle of radius 1 unit touches positive x  -axis and positive y  -axis at A  and B  respectively. A  variable line passing through origin intersects the circle in two points D  and E . If the area of the triangle DEB is maximum when the slope of the line is $m$ , then the value of $\frac{1}{m^2}$  is

When I tried plotting its graph it came out to be something like this:

Now When I thought, I Felt like the triangle with maximum area should be an equilateral triangle.
So from the center of the circle I joined a line to B which makes 90 with the x axis. This line should also bisect the angle of the equilateral triangle so formed. Hence If I assume point D to be the point intersecting the circle on left. Then the line DB should have a slope $\tan(120) = -\sqrt{3}$. This line shall make and angle 60 with the line DE, so I calculated that and the slope comes out to be $m = 0$
However the answer to $\frac{1}{m^2}$ is $3$ and I am not able to understand how. Any hint/suggestions are appreciated. Thanks for your time

REPLY [2 votes]: First a bit of Euclidean geometry:
$$S_{\triangle{DEB}} = S_{\triangle{OEB}} - S_{\triangle{ODB}} = \frac12 \times OB \times (EH - DK) = \frac12 \times 1 \times (x_E - x_D)$$
Next, we use analytic geometry to find the line that maximizes $(x_E - x_D)$.
The line $y=mx$ meets the circle $(x-1)^2+(y-1)^2=1$ at two points ($E , D$). We can plug $mx$ for $y$ in the equation of the circle to find $x_E$ and $x_D$:
$$(x-1)^2+(mx-1)^2=1$$
Or
$$x^2 - 2\frac{1+m}{1+m^2}x + \frac{1}{1+m^2} = 0$$
So
$$x =  - \frac{1+m}{1+m^2} \pm \sqrt{\frac{(1+m)^2}{(1+m^2)^2} - \frac{1}{1+m^2}}$$
The difference between the two roots (which is what we are looking for) is twice the square root term.
$$x_E - x_D = 2\sqrt{\frac{(1+m)^2}{(1+m^2)^2} - \frac{1}{1+m^2}} = \frac{2\sqrt{2m}}{1+m^2}$$
What $m$ maximizes $\frac{\sqrt{m}}{1+m^2}$? We'll use the derivative to find out:
$$\frac{d}{dm}(\frac{\sqrt{m}}{1+m^2}) = \frac{\frac12 \frac{1}{\sqrt m}(1+m^2) - 2m \sqrt m}{(1+m^2)^2} = 0 $$
$$\therefore \frac12 \frac{1}{\sqrt m}(1+m^2) = 2m \sqrt m$$
$$\frac12 (1+m^2) = 2m^2 $$
which simplifies to $m^2 = \frac13$ .
Did we just solve the problem?
Not yet! We still don't know whether this slope $m$ corresponds to a maximum or a minimum. An analytical way to answer this important question is to calculate the second derivative and analyze its sign, or somehow analyze the variations in the sign of the first derivative. If interested readers would like to do that as an exercise then I salut them. But here we will take a different approach:
Let's look back at the figure. At two extreme situations, the line passes through points $A$ and $B$. In these two extremes, the points $D$ and $E$ fall over each other and the difference $x_E - x_D$ is zero. In the in-between situations, such as shown in the figure, $x_E - x_D$ is greater than zero. Evidently at some point this difference has to be maximum. Et voila! Now we can say for certain that our calculated $m^2 = \frac13$ corresponds to a maximum value for $x_E - x_D$ and therefore maximum $S_{\triangle{DEB}}$ .
------
Update: This is not an alternative solution but just an observation, I am not yet sure if it leads to anywhere. Triangles $OBD$ and $OEB$ are similar (why?). It follows that $\frac{OD}{OB} = \frac{OB}{OE}$ , and because $OB=1$ we have $OD \times OE=1$ . We also have the ratio of areas of these two triangles:
$$\frac{S_{\triangle OBD}}{S_{\triangle OEB}} = (\frac{OB}{OE})^2 = \frac{1}{OE^2} = \frac{1}{x_E^2 (1+m^2)}$$
I will post an update if I find a geometric solution based on these.


TITLE: How to proof $z^{n}-1= \prod_{k=1}^{n}({z- e^{ \frac{2 \pi i k}{n} } )}$
QUESTION [0 upvotes]: i tried to rewrite as
$z^{n-1}+ \dots + 1 =\prod_{k=1}^{n-1}({z- e^{ \frac{2 \pi i k}{n} } )}$
and could not do much. Also, tried to use periodicity identity yet it did not fit in.

REPLY [0 votes]: Hint
$$z^n=1\iff z=e^{\frac{2 ik\pi}{n}}$$
with  $k=0,...,n-1$.

REPLY [0 votes]: For any integer $r,$ $$\left(e^{2r\pi i/n}\right)^n=e^{2r\pi i}=1$$
But if $f(z)=z^n-1$
$f'(z)=nz^{n-1}$
Hence $f(z)$ can not have repeated roots
Again if $e^{2a\pi i/n}=e^{2b\pi i/n}\iff e^{2(a-b)\pi i/n}=1$
$\iff n|(a-b)\iff a\equiv b\pmod n$
So, the roots of $z^n=1$ are $e^{2r\pi i/n}, r\equiv0,1,\cdots,n-2,n-1\pmod n$


TITLE: What is the significance of the integral of the Hessian determinant?
QUESTION [5 upvotes]: The integral of a function over some region measures the total value of the function in that region:
$$T(u)=\int u\thinspace\mathrm{d}V$$
The integral of the squared norm of the gradient of the function measures its variability over some region and is known as the Dirichlet energy:
$$E(u)=\int \lvert\nabla u\rvert^2\thinspace\mathrm{d}V$$
Is there a geometric significance to the integral of the determinant of the Hessian matrix (the square matrix of second-order partial derivatives) of a function over some region?
$$\int\det H(u)\thinspace\mathrm{d}V$$
What other integral functionals of a similar form provide interesting information about the behavior of a function over some region?

REPLY [4 votes]: The  Hessian determinant allows a considerable flexibility to the function, more than the Dirichlet integral. For example, all functions of the form $u(x,y) = \epsilon x^2+\epsilon^{-1}y^2$ have the same Hessian determinant $4$, while taking extremely large values on the unit disk if $\epsilon$ is very large or very small.
The Hessian determinant is the Jacobian determinant of the gradient map $\nabla u$. The geometric significance of the integral of Jacobian is that it's the volume of the image, accounting for multiplicity and orientation. This is also not exactly geometric unless the map happens to be orientation preserving and (more or less) injective. An important case when $\nabla u$ has this property is when $u$ is convex.
For a convex function, the integral $\int_U \det D^2u$ is the volume of the image of $U$ under $\nabla u$. This can be interpreted as some measure of the spread of directions of tangent plane to the graph of $u$ on the boundary of $U$.
In these brief notes Caffarelli explains how this simple fact can be used: for a convex function the supremum of $\nabla u$ is controlled by the oscillation of $u$ on a larger domain $V\Supset U$. Hence, $\int_U \det D^2u\le C (\operatorname{osc}_V u)^n$, which is neat: second derivatives controlled by the function itself.


text:
\begin{document}

\title{Nonsingular Ricci flow on a noncompact manifold in dimension three}

\author{ Li Ma, Anqiang Zhu}

\address{Li Ma, Department of Mathematical Sciences, Tsinghua University,
 Peking 100084, P. R. China}

\email{lma@math.tsinghua.edu.cn}

\begin{abstract} We
consider the Ricci flow $\frac{\partial}{\partial t}g=-2Ric$ on the
3-dimensional complete noncompact manifold $(M,g(0))$ with
non-negative curvature operator, i.e., $Rm\geq 0, |Rm(p)|\rightarrow
0, ~as ~d(o,p)\rightarrow 0.$ We prove that the Ricci flow on such a
manifold is nonsingular in any finite time.
\end{abstract}

\keywords{Ricci flow, non-negative curvature operator, nonsingular}

\subjclass[msc2000]{53Cxx,35Jxx}

\maketitle

\section{introduction}
The aim of this paper is to get a global existence of Ricci flow
with bounded non-negative curvature operator in three dimensions.
This kind of question was asked by Hamilton \cite{HS}. We remark
that the local existence of the flow was obtained by Shi \cite{S}.
So we only need to show that the curvature is bounded in finite
time. Our research is based on previous important results obtained
by Hamilton and Perelman (\cite{HS}, \cite{KL}, and \cite{MT}),
which will be recalled in next section.

The Ricci flow
$$
\frac{\partial}{\partial t}g_{ij}=-2R_{ij}
$$
was first introduced by Richard Hamilton \cite{H82}. Using it,
Hamilton has obtained some interesting theorem, such as \cite{H82}.
Hamilton's program is to prove Poincar$\acute{e}$ conjecture and
Thurston's geometrization conjecture by Ricci flow. In three
remarkable papers \cite{P1},\cite{P2},\cite{P3}, Perelman
significantly advanced the theory of the Ricci flow. Perelman
introduced canonical neighborhood, and analyze the high curvature
region. Perelman also analyzed one of the special solution to the
Ricci flow, $\kappa$ solution, which is usually the limit solution
of the blow up sequence. Before Perelman, Hamilton\cite{HS} had
defined asymptotic volume and obtained that the asymptotic volume is
constant under Ricci flow. By a induction argument, Perelman
obtained that the asymptotic volume is zero when the solution is
$\kappa$ solution. In order to analyze the high curvature region,
Hamilton obtained some compactness of Ricci flow \cite{HC}. But in
order to apply this compactness, one has to check the non-collapse
and curvature bound assumption. We can use these ideas to prove the
existence of a non-singular Ricci flow on a 3-manifold.

Our main result is the following
\begin{thm}\label{thm4}
Assume that $(M,g(t)), t\in [0,T)$ is a Ricci flow on 3-dimensional
complete noncompact Riemannian manifold. Suppose $Rm(0)\geq 0,
|Rm(p,0)|\rightarrow 0, d(o,p)\rightarrow \infty.$ Then $T=\infty,$
i.e Ricci flow is nonsingular in finite time on such a manifold.
\end{thm}
We now give a interesting example.

\begin{exam}
Consider the paraboloid of revolution
$x_{4}=x_{1}^{2}+x_{2}^{2}+x_{3}^{2},$ $~(x_{1},\cdots ,x_{4})\in
R^{4}.$ We know the curvature $Rm(x)\rightarrow 0, ~x\rightarrow
\infty.$ and $Rm>0$. So Ricci flow on it will not blow up in finite
time. We also know that the asymptotic volume is $0$, so the
paraboloid can't converge uniformly to flat $R^{3}$. We also note
that when the infinite of the manifold is asymptotical to a cone,
the Ricci flow doesn't blow up in finite time. But this time the
asymptotic volume is between $0$ and $\omega,$ where $\omega$ is the
volume of standard ball $B(0,1)\subset R^{3}$.
\end{exam}

We remark that in the radial symmetrical case, a similar result was
obtained in \cite{OW}, where another assumption such as the
asymptotic flatness was used. One may also see our previous work
\cite{M} for more results in this direction.

\section{Some famous results}
In this section, we recall the following results of Hamilton and
Perelman, which will be used in our proof of Theorem \ref{thm4}.

By monotonicity of $\mathscr{W}$ functional and reduced volume,
Perelman proved twice the non-collapse of Ricci flow. Furthermore
Perelman obtained the following convergence theorems about Ricci
flow, which we will use many times (see \cite{P1}, \cite{KL}, and
\cite{MT}).

\begin{thm}\label{thm1}
Fix canonical neighborhood constance $(C,\epsilon),$ and a
non-collapsing constance $r>0,\kappa>0.$ Let
$(\mathscr{M}_{n},G_{n},x_{n})$ be a sequence of based generalized
3-dimensional Ricci flows. We set $t_{n}=t(x_{n})$ and
$Q_{n}=R(x_{n}).$ We denote by $M_{n}$ the time $t_{n}$ time slice
of $\mathscr{M}_{n}.$ We suppose that:
\begin{enumerate}
  \item Each $(\mathscr{M}_{n},G_{n})$ has time interval of
  definition contained in $[0,\infty)$ and has curvature pinched
  toward positive.
  \item Every point $y_{n}\in (\mathscr{M}_{n},G_{n})$ with $t(y_{n})\leq
  t_{n}$ and $R(y_{n})\geq 4R(x_{n})$ has a strong $(C,\epsilon)$
  canonical neighborhood.
  \item $lim_{n\rightarrow \infty} Q_{n}=\infty.$
  \item For each $A<\infty$ the following holds for all n
  sufficiently large. The Ball
  $B(x_{n},t_{n},AQ_{n}^{-\frac{1}{2}})$ has compact closure in
  $M_{n}$ and the flow is $\kappa$ non-collapsed on scales $\leq r$
  at each point of $B(x_{n},t_{n},AQ_{n}^{-\frac{1}{2}}).$
  \item There is $\mu>0$ such that for every $A<\infty$ the
  following holds for all n sufficiently large. For $y_{n}\in
  B(x_{n},t_{n},AQ^{-\frac{1}{2}})$ the maximum flow line through
  $y_{n}$ extends backwards for a time at least $\mu (max(Q_{n},R(y_{n})))^{-1}.$

  Then after passing to a subsequence and shifting the times of each
  the Ricci flow so that $t_{n}=0$ for every n, there is a geometry
  limit $(M_{\infty},g_{\infty},x_{\infty})$ of the sequence of
  based Riemannian manifolds $(M_{n},Q_{n}G_{n}(0),x_{n}).$ The
  limit is a complete 3-dimensional Riemannian manifold of bounded,
  non-negative curvature. Furthermore, for some $t_{0}>0$ which
  depends on the curvature bound for $(M_{\infty},g_{\infty}),$
  there is a geometric limiting Ricci flow defined
  on $(M_{\infty},g_{\infty}(t)), -t_0\leq t\leq
  0,$ with $g_{\infty}(0)=g_{\infty.}$
\end{enumerate}
\end{thm}
\begin{thm}\label{thm2}
Suppose that $\{\mathscr{M}_{n},G_{n},x_{n}\}_{n=1}^{\infty}$ is a
sequence of 3-dimensional Ricci flow satisfying all the hypothesis
of above theorem. Suppose in addition that there is $T_{0}$ with
$0<T_{0}\leq \infty$ such that the following holds. For any
$T<T_{0},$ for each $A<\infty,$ and all n sufficiently large, there
is an embedding $B(x_{n},t_{n},AQ_{n}^{-\frac{1}{2}})\times
(t_{n}-TQ_{n}^{-1},t_{n}]$ into $\mathscr{M}_{n}$ compatible with
time and with the vector field and at every point of the image the
flow is $\kappa$ non-collapsed on scales $\leq r.$ Then, after
shifting the times of the generalized flows so that $t_{n}=0$ for
all n and passing to a subsequence there is a geometric limiting
Ricci flow
$$
(M_{\infty},g_{\infty}(t),x_{\infty}), -T_{0}<t\leq 0,
$$
for the rescaled flows $(\mathscr{M}_{n},Q_{n}G_{n},x_{n}).$ This
limiting flow is complete and of nonnegative curvature. Furthermore,
the curvature is locally bounded n time. If in addition
$T_{0}=\infty,$ then it is a $\kappa$ solution.
\end{thm}

In this paper, we also need the following result of Hamilton on
Ricci flow. By using barrier functions, Hamilton has proved the
asymptotically property is preserved under Ricci flow (see
\cite{HS}).

\begin{thm}\label{thm3}
Assume that we have a solution to the Ricci flow on a complete
noncompact manifold with bounded curvature. If $|Rm(p,0)|\rightarrow
0, d_{g(0)}(o,p)\rightarrow \infty,$ this remains true for $t\geq
0.$
\end{thm}

\begin{rem} One may see \cite{M} for an improvement of this result.
\end{rem}

\section{proof of Theorem \ref{thm4}}
In the following, we will consider the 3-dimensional Ricci flow on
$(M,g(0)),$ with $Rm\geq 0,~|Rm(p)|\rightarrow 0, ~as
~d(o,p)\rightarrow \infty,$ where $o$ is a fixed point. In order to
applied the above convergence theorem, we first show two lemmas. The
method is from Perelman's famous papers (see \cite{P1}, \cite{P2}).
But the condition is different from us here.

\begin{lem}\label{lem1}
For sufficiently small $r>0,$ there is $\kappa>0$ such that the
Ricci flow on this noncompact manifold is $\kappa$ non-collapse on
scales $\leq r.$
\end{lem}

\begin{proof} Fix $(x,t_{0})\in M\times [0,T).$ Since $Rm\geq 0$, Shi has
proved that it's preserved by Ricci flow \cite{S}. By using a well
known result of Gromoll and Meyer \cite{CE}, we have an injectivity
radius estimate
$$
inj(M^{n},g(t))\geq \frac{\pi}{\sqrt{R_{max}(t)}}.
$$
Since
$$
R(x,t)<C_{1}, ~(x,t)\in M\times [0,\frac{1}{2}t_{0}],
$$
by the above injective estimate, we have
$$
Vol B(x,t,r)\geq V^{'}r^{3}, ~(x,t)\in M\times [0,\frac{1}{2}t_{0}].
$$
where $V^{'}$ is a constant. By the inequality of reduced length,
$$
\frac{\partial l_{x}}{\partial \tau}(q,\tau)+\Delta
l_{x}(q,\tau)\leq \frac{(\frac{n}{2})-l_{x}(q,\tau)}{\tau},
$$
we know there is a point $(\tilde{q},\tilde{t}),
~\tilde{t}=\frac{1}{4}t_{0},$ such that
$l_{x}(\tilde{q},\tilde{\tau})\leq \frac{3}{2},$ where
$\tilde{\tau}=t_{0}-\tilde{t}$. By the inequality (see \cite{P1} and
\cite{MT})
$$
|\nabla l_{x}(q,\tau)|^{2}\leq |\nabla
l_{x}(q,\tau)|^{2}+R(q,\tau)\leq \frac{(1+2n)l_{x}(q,\tau)}{\tau},
$$
we have that
$$
l_{x}(q,\tilde{\tau})\leq
(\frac{\sqrt{2n+1}d_{g(t_{0}-\tilde{\tau})}(q,\tilde{q})}{2}+\sqrt{\frac{n}{2}})^{2},
$$
So, for any $A<\infty,$ we have $l_{x}(q,\tilde{\tau})<C(A),$ when
$(q,t_{0}-\tilde{\tau})\in B(\tilde{q},t_{0}-\tilde{\tau},A),$ where
$C(A)$ is a constant depend on $r.$ By Perelman's non-collapse
theorem, we know that if
$$
|Rm(p,t)|<r^{-2}, ~(p,t)\in B(x,t_{0},r)\times [t_{0}-r^{2},t_{0}]
$$
then
$$
Vol B(x,t_{0},r)\geq \kappa r^{n}.
$$
\end{proof}

\begin{lem}\label{lem2}
Fix $0<\epsilon<1.$ Then there is $r>0$ such that for any point
$(x_{0}, t_{0})$ in the flow with $R(x_{0},t_{0})\geq r^{-2}$ the
following hold. $(x_{0},t_{0})$ has a strong canonical
$(C(\epsilon),\epsilon)$ neighborhood.
\end{lem}
\begin{proof} By Hamilton's theorem \ref{thm3}, we know that for any $t^{'}_{n}<T,
~t^{'}_{n}\rightarrow T,$ the curvature is bounded
$$
|Rm(x,t)|<C(t^{'}_{n}), ~(x,t)\in M\times [0,t^{'}_{n}].
$$
Set
$$
A_{n}=\{(x,t)\in M\times [0,t^{'}_{n}]|~(x,t) ~doesn't ~have
~canonical~ neighborhood\}.
$$
Then there is also a up bound to curvature of point in $A_{n},$
$$
R(x,t)<\tilde{C}(t^{'}_{n}), ~(x,t)\in A_{n}.
$$
We can pick point $(x_{n},t_{n})\in A_{n},$ such that
$$
R(x_{n},t_{n})>\frac{1}{2}\tilde{C}(t^{'}_{n}), ~t_{n}\leq t^{'}_{n}
$$ We need to prove
that
$$
\overline{lim}_{n\rightarrow \infty}R(x_{n},t_{n})<\infty.
$$
Suppose not. Then we may assume that
$Q_{n}=R(x_{n},t_{n})\rightarrow \infty.$ From the construction, we
know that $\forall (x,t)\in M\times [0,t_{n}],$ if
$R(x,t)>4R(x_{n},t_{n}),$ $(x,t)$ have a canonical neighborhood.
Since it's a 3-dimensional Ricci flow, the curvature is pinching
toward positive. By Lemma \ref{lem1}, we have the non-collapse
assumption. Since $Q_{n}\rightarrow \infty, ~t_{n}\rightarrow T,$
for any fix $T>0,$ we have $(t_{n}-TQ_{n}^{-1},t_{n}]\subset
[0,t_{n}]$ for sufficiently large n. That is the addition assumption
of Theorem \ref{thm2} is satisfied.

By Theorem \ref{thm2}, we have that
$(M,Q_{n}g(t_{n}+\frac{t}{Q_{n}}),(x_{n},t_{n}))$ converges to a
limit flow $(M_{\infty}, g_{\infty}(t),(x_{\infty},0)),$ which is a
$\kappa$ solution. So for sufficiently large n, $(x_{n},t_{n})$ has
a canonical neighborhood. This contradicts our assumption that none
of the point $(x_{n},t_{n})$ has a canonical neighborhood.
\end{proof}

\emph{ Proof of Theorem \ref{thm4}}: Suppose that the Ricci flow
blows up at time $T.$ By Theorem \ref{thm3}, we know there is a
limit metric $g(T)$ at infinity such that $g(t)|_{M-K}\rightarrow
g(T)|_{M-K},$ where $K$ is a suitable compact set of $M.$ Since the
Ricci flow blows up at the time $t=T,$ we have $sup_{x\in
M}Rm(x,t)\rightarrow \infty, ~t\rightarrow T.$ Otherwise, we can
extend the flow keeping the curvature bounded (That contradicts the
maximum of existence time). So we have that there is a point $p\in
K,$ such that the scalar curvature blows up at $T$, that is
$R(p,t)\rightarrow \infty, ~t\rightarrow T.$ We pick a sequence
$t_{n}\rightarrow T,$ such that $Q_{n}=R(p,t_{n})\rightarrow
\infty.$ By Lemma \ref{lem2}, we have the assumptions of Theorem
\ref{thm2} is satisfied. So
$(M,Q_{n}g(t_{n}+\frac{t}{Q_{n}}),(p,t_{n}))$ converges to a
$\kappa$ solution $(M_{\infty},g_{\infty}(t),(x_{\infty},0)),$

we know that the asymptotic volume of $\kappa$ solution is zero,
that is,
$$lim_{r\rightarrow \infty}\frac{vol
B(x_{\infty},r)}{r^{3}}=0.
$$
Fixing any $\epsilon>0,$ there is a sufficient large $r$ such that
$$
\frac{vol B(x_{\infty},r)}{r^{3}}\leq \frac{\epsilon}{2}.
$$
Since $(M,Q_{n}g(t_{n}+\frac{t}{Q_{n}}),(p,t_{n}))$ converges to
$(M_{\infty},g_{\infty}(t),(x_{\infty},0)),$  for large n, we have
$$
\frac{volB_{Q_{n}g(t_{n})}((p,t_{n}),r)}{r^{3}}\leq \epsilon.
$$
That is
$$
\frac{volB_{g(t_{n})}((p,t_{n}),\frac{r}{Q_{n}^{\frac{1}{2}}})}{(\frac{r}{Q_{n}^{\frac{1}{2}}})^{3}}\leq
\epsilon.
$$
On the other hand, there is a compact region $\Omega\subset M-K,$
such that $g(t)|_{\Omega}\rightarrow g(T)|_{\Omega}.$ Since $R>0,$
$~\frac{d}{dt}\int_{\Omega}d\mu=-\int_{\Omega}Rd\mu\leq 0,$ we have
$$
Vol_{g(t)} \Omega\geq \delta,~t\in [0,T].
$$
Since $\Omega$ is compact, we can find $\tilde{r}>0,$ such that
$$
\Omega \subset B_{g(0)}(p,\tilde{r}).
$$
Since $Ric\geq 0,$ the distance is decreasing. We have $\Omega
\subset B_{g(t)}(p,\tilde{r}), ~t\in [0,T)$. So
$$
Vol B_{g(t_{n})}(p,\tilde{r})\geq \delta.
$$
Since $Q_{n}\rightarrow \infty,$ for sufficiently large $n$,
$\tilde{r}>\frac{r}{Q_{n}}$. We now choose
$\epsilon<\frac{\delta}{\tilde{r}^{3}}.$ Using $Ric\geq 0,$ we have
by the volume comparison theorem that
\begin{eqnarray*}
\frac{\delta}{\tilde{r}^{3}}>\epsilon
>\frac{VolB_{g(t_{n})}((x_{n},t_{n}),\frac{r}{Q_{n}^{\frac{1}{2}}})}{(\frac{r}{Q_{n}^{\frac{1}{2}}})^{3}}
>\frac{Vol B_{g(t_{n})}(x_{n},\tilde{r})}{\tilde{r}^{3}}\geq
\frac{\delta}{\tilde{r}^{3}},
\end{eqnarray*}
which is a contradiction. So the Ricci flow on $(M,g(0))$ is
nonsingular at finite time. That is, $T=\infty.$

\begin{rem}
X. Dai and L. Ma proved that the Ricci flow on the asymptotically
flat manifold can not converge uniformly to flat manifold by ADM
mass (see \cite{M}).
\end{rem}

{\em Acknowledgement:} The research is partially supported by the
National Natural Science Foundation of China 10631020 and SRFDP
20060003002. Part of the work is done when the first name author is
visiting Chern Institute of Mathematics in Nankai University,
Tianjin, which the first named author would like to thank the
director Prof.Y.M.Long for invitation.


TITLE: Let $\gamma$ be a piece wise smooth closed curve .show that $∫z^mdz=0$
QUESTION [1 upvotes]: Let  $\gamma$ be a piece wise smooth closed curve .show that $∫z^mdz=0$ where $m=0,1,2...$
since  $\gamma$ is closed curve and $z^m$  is analytic so by cauchy theorem
$∫z^mdz=0$ is i am right?

REPLY [0 votes]: The equality is only correct if $m\neq -1.$ In that case $F(z)=\frac{z^{m+1}}{m+1}$ has $f$ as its derivative. So then
\begin{equation}
\int_\gamma f(z) dz=\int_0^1F'(\gamma(t))\gamma'(t) dt=F(\gamma(1))-F(\gamma(0))=0,
\end{equation} where the last inequality holds because $\gamma$ is closed.


TITLE: Is the subset of a set and its negation equivalent to the empty set?
QUESTION [0 upvotes]: ie. is $\subseteq A \cap \bar A$ equivalent to $\emptyset$ ?
Just thinking about it logically, this seems to make sense to me, but I couldn't find any similar questions when I searched so I'm still unsure. I apologize if this is a poor question but I'm a beginner when it comes to set theory.

REPLY [0 votes]: Your question essentially asks whether the statement $$B\subseteq A\cap \bar A$$ implies $$B=\Phi $$
The answer is yes because the empty set is a subset of every set and nothing belongs to both the set and its complement.