TITLE: Covering spaces Hatcher question 6.
QUESTION [13 upvotes]: Let X be the shrinking wedge of circles. Which is the radius of circles $X \in R^2$ such that it's the union of $C_n$ circles centered at $(\frac{1}{n},0)$ with radius $\frac{1}{n}$ for n=1,2,3...

I'm having a bit of trouble with sheeted aspect. I'm thinking that the answer is two straight lines intersecting in a point. Does sheeted refer to covering it twice?
So like a straight line can cover that space I think.

REPLY [13 votes]: See this :

the composition is not a covering space because there is no open neighboorhood of the "connecting point" that pullbacks homeomorphically onto copies of itself.
For any neighboorhood you try, you will have to take a small circle in it, and this circle will be pulled apart in one of the pullbacks.


TITLE: Eigen values product
QUESTION [1 upvotes]: $ M= \left[ {
\begin{array}{ccccc}
 1 & 0 & 0 & 0 & 1\\
 0 & 1 & 1 & 1 & 0\\
 0 & 1 & 1 & 1 & 0\\
 0 & 1 & 1 & 1 & 0\\
1 & 0 & 0 & 0 & 1\\
 \end{array} } \right] $
What is the product of positive eigen values for the above matrix ?
I got the answer using normal method(by transforming to echelon form) to find the eigen values but it took me a lot of time.
Is there any shortcut or simple way to solve this ? After looking at the structure of the matrix, it seems that there is some shorter way to do it but I am not able to find it.
Can someone please help?

REPLY [2 votes]: Notably, the matrix $M$ has rank $2$, which is to say that it will have $2$ non-zero eigenvalues.  Both of these eigenvalues will be positive, since the matrix is positive semidefinite.
Let $\lambda_i$ denote the eigenvalues of $M$. We can note that
$$
\operatorname{tr}(M^2) = \sum_{i}\lambda_i^2\\
\operatorname{tr}(M)^2 = \left(\sum_{i}\lambda_i \right)^2 = \sum_{i, j} \lambda_i\lambda_j
$$
So that
$$
\operatorname{tr}(M)^2 - \operatorname{tr}(M^2) = \sum_{i \neq j} \lambda_i\lambda_j
$$
In other words, we can compute the product of the eigenvalues of this rank-2 positive definite matrix as
$$
\lambda_1 \lambda_2 = \frac 12\left[\operatorname{tr}(M)^2 - \operatorname{tr}(M^2) \right]
$$

On the other hand, we can find a permutation matrix $P$ such that
$$
PMP^T = \pmatrix{1&1\\1&1\\&&1&1&1\\&&1&1&1\\&&1&1&1}
$$
Since this matrix is block-diagonal, it isn't difficult to see that the non-zero eigenvalues of the individual blocks (and hence of the entire matrix) will be $2$ and $3$ (respectively).


text:
\begin{document}

\title{Characterization of $\ell_p$-like and $c_0$-like equivalence relations}
\author{Longyun Ding}
\address{School of Mathematical Sciences and LPMC, Nankai University, Tianjin, 300071, P.R.China}
\email{dinglongyun@gmail.com}
\thanks{Research partially supported by the National Natural Science Foundation of China (Grant No. 10701044).}

\subjclass[2000]{Primary 03E15, 54E35, 46A45}

\date{\today}

\begin{abstract}
Let $X$ be a Polish space, $d$ a pseudo-metric on $X$. If $\{(u,v):d(u,v)<\delta\}$ is ${\bf\Pi}^1_1$ for each $\delta>0$,
we show that either $(X,d)$ is separable or there are $\delta>0$ and a perfect set $C\subseteq X$
such that $d(u,v)\ge\delta$ for distinct $u,v\in C$.

Granting this dichotomy, we characterize the positions of $\ell_p$-like and $c_0$-like equivalence relations in the Borel reducibility hierarchy.
\end{abstract}
\maketitle

\section{Introduction}

Let $(X,d)$ be a pseudo-metric space. If $X$ is not separable, by Zorn's lemma, we can easily prove that, there are $\delta>0$ and a noncountable set $C\subseteq X$
such that $d(u,v)\ge\delta$ for distinct $u,v\in C$. However, if we do not assume CH, can we find such a $C$ whose cardinal is of $2^\Bbb N$? J. H. Silver \cite{silver}
answered a similar problem for equivalence relations under an extra assumption of coanalyticity.

A topological space is called a {\it Polish space} if it is separable and completely metrizable. As usual, We denote the Borel, analytic and coanalytic sets by
${\bf\Delta}^1_1,{\bf\Sigma}^1_1$ and ${\bf\Pi}^1_1$ respectively. For their effective analogues, the Kleene pointclasses and
the relativized Kleene pointclasses are denoted by $\Delta^1_1,\Sigma^1_1,\Pi^1_1,\Delta^1_1(\alpha),\Sigma^1_1(\alpha),\Pi^1_1(\alpha)$, etc.
For more details in descriptive set theory, one can see \cite{kechris} and \cite{MK}.

\begin{theorem}[Silver]
Let $E$ be a ${\bf\Pi}^1_1$ equivalence relation on a Polish space. Then $E$ has
either at most countably many or perfectly many equivalence classes.
\end{theorem}

In section 2, we use the Gandy-Harrington topology to establish the following dichotomy.

\begin{theorem}
Let $X$ be a Polish space, $d$ a pseudo-metric on $X$. If $\{(u,v):d(u,v)<\delta\}$ is ${\bf\Pi}^1_1$ for each $\delta>0$,
then either $(X,d)$ is separable or there are $\delta>0$ and a perfect set $C\subseteq X$
such that $d(u,v)\ge\delta$ for distinct $u,v\in C$.
\end{theorem}

Let $X,Y$ be Polish spaces and
$E,F$ equivalence relations on $X,Y$ respectively. A {\it Borel
reduction} of $E$ to $F$  is a Borel function $\theta:X\to Y$ such
that $(x,y)\in E$ iff $(\theta(x),\theta(y))\in F$, for all $x,y\in
X$. We say that $E$ is {\it Borel reducible} to $F$, denoted $E\le_B
F$, if there is a Borel reduction of $E$ to $F$. If $E\le_B F$ and
$F\le_B E$, we say that $E$ and $F$ are {\it Borel bireducible} and
denote $E\sim_B F$. We refer to \cite{gao} for
background on Borel reducibility.

In section 3, we will introduce notions of $\ell_p$-like and $c_0$-like equivalence relations.
Granting the dichotomy on pseudo-metric spaces, we answer that when is $E_1$ Borel reducible to an $\ell_p$-like or a $c_0$-like equivalence relation.
In the end, we compare $\ell_p$-like and $c_0$-like equivalence relations with some remarkable equivalence
relations $E_0,E_1,E_0^\omega$.
\begin{enumerate}
\item[(a)] For $x,y\in 2^\Bbb N$, $(x,y)\in E_0\Leftrightarrow\exists m\forall
n\ge m(x(n)=y(n)).$
\item[(b)] For $x,y\in 2^{\Bbb N\times\Bbb N}$, $(x,y)\in
E_1\Leftrightarrow\exists m\forall n\ge m\forall k(x(n,k)=y(n,k)).$
\item[(c)] For $x,y\in 2^{\Bbb N\times\Bbb N}$, $(x,y)\in
E_0^\omega\Leftrightarrow\forall k\exists m\forall n\ge
m(x(n,k)=y(n,k)).$
\end{enumerate}

The following dichotomies show us why these equivalence relations are so remarkable.

\begin{theorem}
Let $E$ be a Borel equivalence relation. Then
\begin{enumerate}
\item[(a)] {\rm (Harrington-Kechris-Louveau \cite{HKL})} either $E\le_B{\rm
id}(\Bbb R)$ or $E_0\le_B E$;
\item[(b)] {\rm (Kechris-Louveau \cite{KL})} if $E\le_B E_1$, then
$E\le_B E_0$ or $E\sim_B E_1$;
\item[(c)] {\rm (Hjorth-Kechris \cite{HK})} if $E\le_B E_0^\omega$,
then $E\le_B E_0$ or $E\sim_B E_0^\omega$.
\end{enumerate}
\end{theorem}

\section{Separable or not}

For a ${\bf\Pi}^1_1$ equivalence relation $E$ on $X$, let us consider the following
pseudo-metric on $X$:
$$d_E(u,v)=\left\{\begin{array}{ll}0, & (u,v)\in E,\cr 1, &(u,v)\notin E.\end{array}\right.$$
From Silver's theorem, we can see that either $d_E$ is separable or there is a perfect set $C\subseteq X$ such that $d_E(u,v)=1$ for distinct $u,v\in C$.

By the same spirit of the Silver dichotomy theorem, we define:

\begin{definition}
Let $X$ be a Polish space, $d$ a pseudo-metric on $X$. If $\{(u,v):d(u,v)<\delta\}$ is ${\bf\Pi}^1_1$ for each $\delta>0$, we say $d$ is {\it lower} ${\bf\Pi}^1_1$.
\end{definition}

For a pseudo-metric space $(X,d)$ and $\delta>0$, we say $(X,d)$ is {\it $\delta$-separable} if there is a countable set $S\subseteq X$ such that
$$\forall u\in X\exists s\in S(d(u,s)<\delta).$$
Hence $(X,d)$ is separable iff it is $\delta$-separable for arbitrary $\delta>0$.

\begin{theorem} \label{separable}
Let $X$ be a Polish space, $d$ a lower ${\bf\Pi}^1_1$ pseudo-metric. Then for $\delta>0$, either $(X,d)$ is $\delta$-separable or there is a perfect set $C\subseteq X$
such that $d(u,v)\ge\delta/2$ for distinct $u,v\in C$.
\end{theorem}

\begin{proof}
We denote $Q=\{(u,v):d(u,v)<\delta\}$ and $R=\{(u,v):d(u,v)<\delta/2\}$. We see that both $Q,R$ are ${\bf\Pi}^1_1$.

Then the theorem follows from the next lemma.
\end{proof}

\begin{lemma}
Let $X$ be a Polish space, $Q,R\subseteq X^2$. Assume that
\begin{enumerate}
\item[(i)] $Q$ is ${\bf\Pi}^1_1$ and $R$ is $\sigma({\bf\Sigma}^1_1)$ (the $\sigma$-algebra generated by the ${\bf\Sigma}^1_1$ sets);
\item[(ii)] $\Delta(X)=\{(u,u):u\in X\}$ contains in $Q$;
\item[(iii)] if there exists $v\in X$ such that $(v,u)\in R,(v,w)\in R$, then $(u,w)\in Q$.
\end{enumerate}
Then one of the following holds:
\begin{enumerate}
\item[(a)] there is a countable set $S\subseteq X$ such that $\forall u\in X\exists s\in S((u,s)\in Q)$;
\item[(b)] there is a perfect set $C\subseteq X$ such that $(u,v)\notin R$ for distinct $u,v\in C$.
\end{enumerate}
\end{lemma}

\begin{proof}
We follow the method as in Harrington's proof for Silver's theorem.

Without loss of generality we may assume $X=\Bbb N^\Bbb N$ and $Q\in\Pi^1_1$. The proof for $Q\in\Pi^1_1(\alpha)$ with $\alpha\in\Bbb N^\Bbb N$ is similar.
Let $\tau$ be the Gandy-Harrington topology (the topology generated by all $\Sigma^1_1$ sets) on $\Bbb N^\Bbb N$.

For $u\in X$ we denote $Q(u)=\{v\in X:(u,v)\in Q\}$. First we define
$$V=\{u\in X:\mbox{ there is no $\Delta^1_1$ set $U$ such that }u\in U\subseteq Q(u)\}.$$

If $V=\emptyset$, since there are only countably many $\Delta^1_1$ set, we can find a countable subset $S\subseteq X$ which meets every nonempty $\Delta^1_1$ set
at least one point. For each $u\in X$ there is a nonempty $\Delta^1_1$ set $U\subseteq Q(u)$. Let $s\in S\cap U$. Then $s\in Q(u)$, i.e. $(u,s)\in Q$.

For the rest of the proof we assume $V\ne\emptyset$. Note that

$$u\in V\iff\forall U\in\Delta^1_1(u\in U\to\exists v\in U(v\notin Q(u))).$$
With the coding of $\Delta^1_1$ sets (see \cite{gao} Theorem 1.7.4), there are $\Pi^1_1$ subsets $P^+,P^-\subseteq\Bbb N\times\Bbb N^\Bbb N$ and $D\subseteq\Bbb N$ such that
\begin{enumerate}
\item $\forall n\in D\forall u((n,u)\in P^+\Leftrightarrow(n,u)\notin P^-)$;
\item for any $\Delta^1_1$ set $A$ there is $n\in D$ such that $\forall u(u\in A\Leftrightarrow(n,u)\in P^+)$.
\end{enumerate}
Thus we have
$$u\in V\iff\forall n((n\in D,(n,u)\in P^+)\to\exists v((n,v)\notin P^-,(u,v)\notin Q)).$$
So $V$ is $\Sigma^1_1$.

By a theorem of Nikodym (see \cite{kechris} Corollary 29.14), the class of sets with the Baire property in any topological space is closed under the
Suslin operation. It is well known that all ${\bf\Sigma}^1_1$ sets are results of the Suslin operation applied on closed sets in the usual topology
(see \cite{kechris} Theorem 25.7). Note that all closed sets in usual topology are also closed in $\tau$, we see that every $\sigma({\bf\Sigma}^1_1)$
subset of $\Bbb N^\Bbb N$ (or $\Bbb N^\Bbb N\times\Bbb N^\Bbb N$) has Baire property in $\tau$ (or $\tau\times\tau$).

Toward a contradiction assume that for some $v\in V$, $R(v)$ is not $\tau$-meager in $V$. Since $R(v)$ has Baire property in $\tau$,
there is a nonempty $\Sigma^1_1$ set $U\subseteq V$
such that $R(v)$ is $\tau$-comeager in $U$. By Louveau's lemma (see \cite{MK} Lemma 9.3.2), $R(v)\times R(v)$ meets any nonempty $\Sigma^1_1$ set in $U\times U$.
We denote $\neg Q=(\Bbb N^\Bbb N\times\Bbb N^\Bbb N)\setminus Q$. If $\neg Q\cap(U\times U)\ne\emptyset$, since it is $\Sigma^1_1$, we have
$$(R(v)\times R(v))\cap\neg Q\cap(U\times U)\ne\emptyset,$$
which contradicts to clause (iii). Thus we have $U\times U\subseteq Q$. We define $W$ by
$$w\in W\iff\forall u(u\in U\to(u,w)\in Q).$$
Fix a $u_0\in U$. We can see that $W$ is $\Pi^1_1$ and $U\subseteq W\subseteq Q(u_0)$. By the separation property for $\Sigma^1_1$ sets there is $U_0\in\Delta^1_1$
such that $U\subseteq U_0\subseteq W$. Then we have $u_0\in U_0\subseteq Q(u_0)$, which contradicts $u_0\in U\subseteq V$. Therefore, $R(v)$ is $\tau$-meager in $V$ for
each $v\in V$.

Since $R$ has Baire property in $\tau\times\tau$, by the Kuratowski-Ulam theorem (see \cite{kechris} Theorem 8.41), $R$ is $\tau\times\tau$-meager in $V\times V$.
By the definition of $V$ and clause (ii),
we see that $V$ contains no $\Delta^1_1$ real, i.e. $V$ has no isolate point in $\tau$. Since the space $\Bbb N^\Bbb N$ with $\tau$ is strong Choquet
(see \cite{gao} Theorem 4.1.5), $V$ is a perfect Choquet space. From \cite{kechris} Exercise 19.5, we can find a perfect set $C\subseteq V$ such that
$(u,v)\notin R$ for distinct $u,v\in C$.
\end{proof}

\section{Characterization}

The notion of $\ell_p$-like equivalence relation was introduced in \cite{ding1}.

\begin{definition}
Let $(X_n,d_n),\,n\in\Bbb N$ be a sequence of pseudo-metric spaces, $p\ge 1$. We define an equivalence relation
$E((X_n,d_n)_{n\in\Bbb N};p)$ on $\prod_{n\in\Bbb N}X_n$ by
$$(x,y)\in E((X_n,d_n)_{n\in\Bbb N};p)\iff\sum_{n\in\Bbb
N}d_n(x(n),y(n))^p<+\infty$$ for $x,y\in\prod_{n\in\Bbb N}X_n$.
We call it an {\it $\ell_p$-like equivalence relation}. If $(X_n,d_n)=(X,d)$ for every $n\in\Bbb N$, we write
$E((X,d);p)=E((X_n,d_n)_{n\in\Bbb N};p)$ for the sake of brevity.
\end{definition}

If $X$ is a separable Banach space, we have $E(X;p)=X^\Bbb N/\ell_p(X)$ where $\ell_p(X)$ is the Banach space whose underlying space is
$\{x\in X^\Bbb N:\sum_{n\in\Bbb N}\|x(n)\|^p<+\infty\}$ with the norm $\|x\|=\left(\sum_{n\in\Bbb N}\|x(n)\|^p\right)^{\frac{1}{p}}$.
Then $E(X;p)$ is an orbit equivalence relation induced by a Polish group action, thus $E_1\not\le_B E(X;p)$ (see \cite{gao} Theorem 10.6.1).

Let $(X,d)$ be a pseudo-metric space, we denote
$$\delta(X)=\inf\{\delta:X\mbox{ is $\delta$-separable}\}.$$

\begin{theorem} \label{lp}
Let $X_n,\,n\in\Bbb N$ be a sequence of Polish spaces, $d_n$ a Borel pseudo-metric on $X_n$ for each $n$ and $p\ge 1$. Denote $E=E((X_n,d_n)_{n\in\Bbb N};p)$.
We have
\begin{enumerate}
\item[(i)] $\sum_{n\in\Bbb N}\delta(X_n)^p<+\infty\iff E\le_B E(c_0;p)$;
\item[(ii)] $\sum_{n\in\Bbb N}\delta(X_n)^p=+\infty\iff E_1\le_B E$.
\end{enumerate}
\end{theorem}

\begin{proof}
Because $E_1\not\le_B E(c_0;p)$, we only need to prove ($\Rightarrow$) for (i) and (ii).

(i) By the definition of $\delta(X_n)$, we see that $X_n$ is $(\delta(X_n)+2^{-n})$-separable, i.e. there is a countable set $S_n\subseteq X_n$ such that
$$\forall u\in X_n\exists s\in S_n(d_n(u,s)<\delta(X_n)+2^{-n}).$$
Let $S_n=\{s_m^n:m\in\Bbb N\}$. Without loss of generality, we assume that $d_n(s_k^n,s_l^n)>0$ for $k\ne l$, i.e. $d_n$ is a metric on $S_n$.
For $u\in X$ we denote $m(u)$ the least $m$ such that $d(u,s_m^n)<\delta(X_n)+2^{-n}$. Then we define $h_n:X_n\to S_n$ by
$h_n(u)=s_{m(u)}^n$ for $u\in X$. It is easy to see that $h_n$ is Borel. Define $\theta:\prod_{n\in\Bbb N}X_n\to\prod_{n\in\Bbb N}S_n$ by
$$\theta(x)(n)=h_n(x(n))$$
for $x\in\prod_{n\in\Bbb N}X_n$. Note that for each $x$ we have
$$\sum_{n\in\Bbb N}d_n(x(n),\theta(x)(n))^p<\sum_{n\in\Bbb N}(\delta(X_n)+2^{-n})^p\le2^{p-1}\sum_{n\in\Bbb N}(\delta(X_n)^p+2^{-np})<+\infty,$$
i.e. $(x,\theta(x))\in E$. It follows that $(x,y)\in E\Leftrightarrow(\theta(x),\theta(y))\in E$. Hence $E\le_B E((S_n,d_n)_{n\in\Bbb N};p)$.

Note that each $(S_n,d_n)$ is a separable metric space. From Aharoni's theorem \cite{aharoni}, there are $K>0$ and $T_n:S_n\to c_0$ satisfying
$$d_n(u,v)\le\|T_n(u)-T_n(v)\|_{c_0}\le Kd_n(u,v)$$
for every $u,v\in S_n$. Define $\theta_1:\prod_{n\in\Bbb N}S_n\to c_0^\Bbb N$ by
$$\theta_1(x)(n)=T_n(x(n))$$
for $x\in\prod_{n\in\Bbb N}S_n$. It is easy to check that $\theta_1$ is a Borel reduction of $E((S_n,d_n)_{n\in\Bbb N};p)$ to $E(c_0;p)$.

(ii) Without loss of generality, we may assume that $\delta(X_n)>0$ for each $n$. Select a sequence $0<\delta_n<\delta(X_n),\,n\in\Bbb N$ such that
$\sum_{n\in\Bbb N}\delta_n^p=+\infty$. Thus we can find a strictly increasing sequence of natural numbers $(n_j)_{j\in\Bbb N}$ such that $n_0=0$ and
$$\sum_{n=n_j}^{n_{j+1}-1}\delta_n^p\ge 2^p,\quad j=0,1,2,\cdots.$$

Since $(X_n,d_n)$ is not $\delta_n$-separable, from Theorem \ref{separable}, there is a Borel injection
$g_n:2^\Bbb N\to X_n$ such that $d_n(g_n(\alpha),g_n(\beta))\ge\delta_n/2$ for distinct $\alpha,\beta\in 2^\Bbb N$.
Define $\vartheta:(2^\Bbb N)^\Bbb N\to\prod_{n\in\Bbb N}X_n$ by
$$\vartheta(x)(n)=g_n(x(j)),\quad n_j\le n<n_{j+1}.$$
For each $x,y\in(2^\Bbb N)^\Bbb N$, if $x(j)\ne y(j)$ for some $j\in\Bbb N$, we have
$$\sum_{n=n_j}^{n_{j+1}-1}\!\!\!d_n(\vartheta(x)(n),\vartheta(y)(n))^p=\!\!\!\sum_{n=n_j}^{n_{j+1}-1}\!\!d_n(g_n(x(j)),g_n(y(j)))^p
\ge\!\!\!\sum_{n=n_j}^{n_{j+1}-1}\!\!\!(\delta_n/2)^p\ge 1.$$
Therefore
$$\begin{array}{ll}&(\vartheta(x),\vartheta(y))\in E\cr
\iff &\sum_{j\in\Bbb N}\sum_{n=n_j}^{n_{j+1}-1}d_n(\vartheta(x),\vartheta(y))^p<+\infty\cr
\iff &\exists k\forall j>k(x(j)=y(j))\cr
\iff &(x,y)\in E_1.\end{array}$$
Thus $\vartheta$ witnesses that $E_1\le_B E$.
\end{proof}

$c_0$-like equivalence relations were first studied by I. Farah \cite{farah}.

\begin{definition}
Let $(X_n,d_n),\,n\in\Bbb N$ be a sequence of pseudo-metric spaces. We define an equivalence relation
$E((X_n,d_n)_{n\in\Bbb N};0)$ on $\prod_{n\in\Bbb N}X_n$ by
$$(x,y)\in E((X_n,d_n)_{n\in\Bbb N};0)\iff\lim_{n\to\infty}d_n(x(n),y(n))=0$$ for $x,y\in\prod_{n\in\Bbb N}X_n$.
We call it a {\it $c_0$-like equivalence relation}. If $(X_n,d_n)=(X,d)$ for every $n\in\Bbb N$, we write
$E((X,d);0)=E((X_n,d_n)_{n\in\Bbb N};0)$ for the sake of brevity.
\end{definition}

Farah mainly investigated the case named $c_0$-equalities that all $(X_n,d_n)$'s are finite metric spaces and denoted it by ${\rm D}(\langle X_n,d_n\rangle)$.

\begin{theorem} \label{c0}
Let $X_n,\,n\in\Bbb N$ be a sequence of Polish spaces, $d_n$ a Borel pseudo-metric on $X_n$ for each $n$. Denote $E=E((X_n,d_n)_{n\in\Bbb N};0)$.
We have
\begin{enumerate}
\item[(i)] $\lim_{n\to\infty}\delta(X_n)=0\iff E\le_B\Bbb R^\Bbb N/c_0$;
\item[(ii)] $(\delta(X_n))_{n\in\Bbb N}$ does not converge to $0\iff E_1\le_B E$.
\end{enumerate}
\end{theorem}

\begin{proof}
We closely follows the proof of Theorem 3.2. Some conclusions will be made without proofs for brevity, since they follow by similar arguments.

(i) Note that for each $x$ we have
$$\lim_{n\to\infty}d_n(x(n),\theta(x)(n))=\lim_{n\to\infty}(\delta(X_n)+2^{-n})=0,$$
i.e. $(x,\theta(x))\in E$. It follows that $(x,y)\in E\Leftrightarrow(\theta(x),\theta(y))\in E$. Hence $E\le_B E((S_n,d_n)_{n\in\Bbb N};0)$.

Fix a bijection $\langle\cdot,\cdot\rangle:\Bbb N^2\to\Bbb N$. We define $\theta_2:\prod_{n\in\Bbb N}S_n\to\Bbb R^\Bbb N$ by
$$\theta_2(x)(\langle n,m\rangle)=T_n(x(n))(m)$$
for $x\in\prod_{n\in\Bbb N}S_n$ and $n,m\in\Bbb N$. It is easy to
see that $\theta_2$ is Borel.

Now we check that $\theta_2$ is a reduction.

For every $x,y\in\prod_{n\in\Bbb N}S_n$, if $(x,y)\in
E((S_n,d_n)_{n\in\Bbb N};0)$, then
$$\lim_{n\to\infty}d_n(x(n),y(n))\to 0.$$
So $\forall\varepsilon>0\exists N\forall
n>N(d_n(x(n),y(n))<\varepsilon)$. Since
$\|T_n(x(n))-T_n(y(n))\|_{c_0}\le Kd_n(x(n),y(n))<K\varepsilon$, we
have
$$\forall n>N\forall m(|T_n(x(n))(m)-T_n(y(n))(m)|<K\varepsilon).$$
For $n\le N$, since $T_n(x(n)),T_n(y(n))\in c_0$, we have
$$\lim_{m\to\infty}|T_n(x(n))(m)-T_n(y(n))(m)|=0.$$
Therefore, for all
but finitely many $(n,m)$'s, we have
$$|\theta_2(x)(\langle n,m\rangle)-\theta_2(y)(\langle
n,m\rangle)|=|T_n(x(n))(m)-T_n(y(n))(m)|<K\varepsilon.$$ Thus
$$\lim_{\langle n,m\rangle\to\infty}|\theta_2(x)(\langle n,m\rangle)-\theta_2(y)(\langle
n,m\rangle)|=0.$$ It follows that $\theta_2(x)-\theta_2(y)\in c_0$.

On the other hand, for every $x,y\in\prod_{n\in\Bbb N}S_n$, if
$\theta_2(x)-\theta_2(y)\in c_0$, then
$$\forall\varepsilon>0\exists N\forall n>N\forall m(|\theta_2(x)(\langle
n,m\rangle)-\theta_2(y)(\langle n,m\rangle)|<\varepsilon).$$
Therefore, for $n>N$ we have
$$\begin{array}{ll}d_n(x(n),y(n))&\le\|T_n(x(n))-T_n(y(n))\|_{c_0}\cr
&=\sup\limits_{m\in\Bbb N}|T_n(x(n))(m)-T_n(y(n))(m)|\cr
&=\sup\limits_{m\in\Bbb N}|\theta_2(x)(\langle
n,m\rangle)-\theta_2(y)(\langle
n,m\rangle)|\le\varepsilon.\end{array}$$ It follows that $(x,y)\in
E((S_n,d_n)_{n\in\Bbb N};0)$.

To sum up, we have $E\le_B E((S_n,d_n)_{n\in\Bbb N};0)\le_B\Bbb R^\Bbb N/c_0$.

(ii) Assume that $(\delta(X_n))_{n\in\Bbb N}$ does not converge to $0$. Then there are $c>0$ and a strictly increasing sequence of
natural numbers $(n_j)_{j\in\Bbb N}$ such that $\delta(X_{n_j})>c$ for each $j$. From Theorem \ref{separable}, there is a Borel injection
$g_j':2^\Bbb N\to X_{n_j}$ such that $d_{n_j}(g_j'(\alpha),g_j'(\beta))\ge c/2$ for distinct $\alpha,\beta\in 2^\Bbb N$.
Fix an element $a_n\in X_n$ for every $n\in\Bbb N$. Define $\vartheta':(2^\Bbb N)^\Bbb N\to\prod_{n\in\Bbb N}X_n$ by
$$\vartheta'(x)(n)=\left\{\begin{array}{ll}g_j'(x(j)), & n=n_j,\cr a_n, &\mbox{otherwise}.\end{array}\right.$$
Then $\vartheta'$ witnesses that $E_1\le_B E$.
\end{proof}

\section{Further remarks}

The following condition was introduced in \cite{ding1} to investigate the position of $\ell_p$-like equivalence relations.

$(\ell 1)$ {\it $\forall c>0\exists x,y\in\prod_{n\in\Bbb N}X_n$
such that $\forall n(d_n(x(n),y(n))^p<c)$ and
$$\sum_{n\in\Bbb N}d_n(x(n),y(n))^p=+\infty.$$}

Let $X_n,\,n\in\Bbb N$ be a sequence of Polish spaces, $d_n$ a Borel pseudo-metric on $X_n$ for each $n$ and $p\ge 1$. Denote $E=E((X_n,d_n)_{n\in\Bbb N};p)$.
It was proved in \cite{ding1} that
\begin{enumerate}
\item[(i)] if $(\ell 1)$ holds, then $\Bbb R^\Bbb N/\ell_1\le_B E$;
\item[(ii)] if $(\ell 1)$ fails, then either $E_1\le_B E,E\sim_B E_0$ or $E$ is trivial, i.e. all elements in $\prod_{n\in\Bbb N}X_n$ are equivalent.
\end{enumerate}

Thus we have a corollary of Theorem \ref{lp}.

\begin{corollary}
Denote $E=E((X_n,d_n)_{n\in\Bbb N};p)$. We have
\begin{enumerate}
\item[(a)] $\sum_{n\in\Bbb N}\delta(X_n)^p<+\infty$ and $(\ell 1)$ fails $\iff E\sim_B E_0$ or $E$ is trivial;
\item[(b)] $\sum_{n\in\Bbb N}\delta(X_n)^p<+\infty$ and $(\ell 1)$ holds $\iff\Bbb R^\Bbb N/\ell_1\le_B E\le_B E(c_0;p)$;
\item[(c)] $\sum_{n\in\Bbb N}\delta(X_n)^p=+\infty\iff E_1\le_B E$.
\end{enumerate}
\end{corollary}

Another condition was introduced by I. Farah \cite{farah} for investigating $c_0$-equalities.

$(*)$ {\it $\forall c>0\exists\varepsilon<c(\varepsilon>0$ and $\exists^\infty i\exists u_i,v_i\in X_i(\varepsilon<d_i(u_i,v_i)<c))$.}

It is easy to check that $(*)$ holds iff for arbitrary $c>0$, there exist $x,y\in\prod_{n\in\Bbb N}X_n$ such that $\forall n(d_n(x(n),y(n))<c)$ and
$(d_n(x(n),y(n)))_{n\in\Bbb N}$ does not converge to $0$.

With similar arguments, we get a corollary of Theorem \ref{c0}.

\begin{corollary}
Denote $E=E((X_n,d_n)_{n\in\Bbb N};0)$. We have
\begin{enumerate}
\item[(a)] $\lim_{n\to\infty}\delta(X_n)=0$ and $(*)$ fails $\iff E\sim_B E_0$ or $E$ is trivial;
\item[(b)] $\lim_{n\to\infty}\delta(X_n)=0$ and $(*)$ holds $\iff E_0^\omega\le_B E\le_B\Bbb R^\Bbb N/c_0$;
\item[(c)] $(\delta(X_n))_{n\in\Bbb N}$ does not converge to $0\iff E_1\le_B E$.
\end{enumerate}
\end{corollary}

config: arxiv


TITLE: Let $\sigma \in S_n$ be a product of $r$ disjoint cycles. Show that $\sigma$ can be written as $n-r$ transpositions.
QUESTION [2 upvotes]: Note that in this problem we are counting all $1-$cycles when computing $r$. For example, if we are in $S_4$ and we have the permutation $\sigma = (1 \ 2)$, $r$ in this case would be $3$ because $(1 \ 2)$ can be also written as $(1 \ 2)(3)(4)$. In this case, it is obvious that we can indeed write this as $4-3=1$ transpositions.
The trouble is that I'm not quite sure how to approach this question. I tried some things with induction (on $r$) and it doesn't seem to work for me. The base case when $r=1$ is quite simple but I cannot get the induction step to follow.
I also tried constructing an argument that would use the fact that each cycle of length $k$ can be written as exactly $k-1$ transpositions, but I'm not quite sure how to morph that into something that would produce a complete proof.
As always, any and all help is much appreciated.

REPLY [1 votes]: Fix the integer $m \ge 1$.
Statement $\quad \mathcal C(r): \;$ Any permutation $\sigma$ of a set $X$ with cardinality $n$ less than or equal to $m$ that can expressed as the product of $r$ disjoint cycles can be expressed as the product of $n-r$ transpositions.
Induction on $r$
Base Case: $r = 1$ - ✔.
Step Case:
Assume $\mathcal C(r)$ is true for $r \ge 1$ and let $\sigma$, a permutation on a set with $n$ elements where $n \le m$,  be decomposed into $r+1$ disjoint cycles ,
$\tag 1 \sigma = \bigr(\sigma_1 \circ \dots \circ \sigma_r \bigr) \circ (\sigma_{r+1})$
Let $k$ be the length of cycle $\sigma_{r+1}$ so that
$\quad \sigma_1 \circ \dots \circ \sigma_r$
represents a permutation on a set with $n - k$ elements as the product of $r$ cycles. Since $\mathcal C(r)$ is true and $n - k \le n \le m$, it can be expressed as $(n - k) -r$ transpositions.
Also, $\sigma_{r+1}$ can be expressed as $k - 1$ transposition (see base case).
Substituting these transposition back into $\text{(1)}$ we see that $\sigma$ can be written as
$\quad \bigr(n - k -r\bigr) + (k - 1) = n -(r+1)$
and the induction is complete.


TITLE: Why an unbounded operator is not-constructive relying on Hahn Banach theorem?
QUESTION [4 upvotes]: Let $T: \mathcal S(\mathbb R)\to L^2(\mathbb R)$ defined by $$Tf(x)=f'(x),$$
where $\mathcal S(\mathbb R)$ is the Schwarz space on $\mathbb R$. The question is : is there a continuous extension to $L^2(\mathbb R)$, and the answer is no, because $\frac{\|Tf_k\|}{\|f_k\|}\to \infty $ where $f_k(x)=e^{-k|x|},$ and thus $T$ is unbounded.
You'll see the example here
1) Why is this a counter-example since $e^{-k|x|}\notin \mathcal S(\mathbb R)$ ? and is even not derivable at $0$, so $f'(x)$ is not well defined on $\mathbb R$, is it ? Because on $L^2(\mathbb R)\setminus \mathcal S(\mathbb R)$, the operator $T$ can be different than $Tf(x)=f'(x)$ ? (for example the fourier transform is not the same on $\mathcal S(\mathbb R)$ than on $L^2(\mathbb R)$.
2) Just under the example in the previous link, they say that the existence of unbounded operators defined everywhere is non constructive relying on Hahn-Banach theorem. Could someone explain what this is true ? I guess that here unbounded mean not continuous ? (and not defined on a subspace of the domain).

REPLY [6 votes]: By a theorem of Solovay, it's consistent with ZF (the Zermelo-Frankel axioms, without Choice) + DC (Dependent Choice) that every subset of a complete separable metric space has the property of Baire.  From this it would follow that every everywhere-defined linear operator from a Hilbert space to itself is continuous.
So to get an unbounded operator you need Axiom of Choice (or something pretty close to it). See J.D.M. Wright, BAMS 79 (1973) 1247-1250.


TITLE: Is it possible to triangularize a matrix only by adding scalar multiples of rows to each other?
QUESTION [4 upvotes]: I am working on showing if $B$ is a $s \times s$ matrix, $D$ is a $t \times t$  matrix, $C$ is a $s \times t$ matrix, and $0$ is a $t \times s$ zero matrix, then $\det(A)=\det(B)\det(D)$, where $$A = \begin{bmatrix}B &C\\ 0& D\end{bmatrix}$$
I understand two more complicated proofs, but I figured out if I could triangularize a matrix by just adding scalar multiples of row i to row j (in that sense my determinant will not change), then I can make to proof much more easier.
However, is that generally true?
After put in some thought, I realized I can use all elementary row operations in my proof. It will not affect the proof. A scratch of my proof will be I am trying to triangularize B and D and A as well, then because the determinant is just the product of diagonal element I can show $\det(A)=\det(B)\det(D)$. If I used row swaps and multiply a row by a scalar, I will have shown $K\det(A)=K\det(B)\det(D)$, where K is some constant due to the row operations. This then gives the result I want.
So, the question no longer have anything to do with the proof now. But still, is it possible?

REPLY [0 votes]: The elementary matrix for adding a multiple of one row to another has determinant 1; thus, you can only possibly produce row operations that have determinant one. Thus, you can't perform all row operations. However, you have these building blocks:
How to swap two vectors $v$ and $w$ and negate one of them

$v \leftarrow v + w$
$w \leftarrow w - v$
$v \leftarrow v + w$

This converts $v,w$ into $w,-v$.
This is all you need to make the row reduction algorithm work, since it only uses swaps and adding one row to another. However, you can also do
How to multiply one vector by $c$ and divide another vector by $c$:

$v \leftarrow v + (1-c) w$
$w \leftarrow w - v$
$v \leftarrow v + \frac{c-1}{c} w$
$w \leftarrow w + cv$

This converts $v,w$ into $(1/c)v, cw$


TITLE: On finite $p$ -group of class two with cyclic center
QUESTION [3 upvotes]: Do there exist a finite $p$-group of class two such that

$Z(G)$ is not a subgroup of $\Phi(G)$, where $Z(G)$ is center of $G$ and $\Phi(G)$ is  frattini subgroup of $G$.
$\dfrac{G}{Z(G)}$  is generated by 2 elements.
$Z(G)$ is cyclic.

Explaintion:
For condition 1 and 2 we can say $D_{8}\times C_{2}.$
For condition 2 and 3 we can say $D_{8}$.
For condition 1 and 3 ?
For condition 1 and 2 and 3?
Thank you

REPLY [2 votes]: The central product of $D_8$ and $C_4$ is such an example. More generally, for arbitrary primes $p$, you can take the central product of an extraspecial group of order $p^3$ with $C_{p^2}$.


TITLE: Absorbing ramification and factoring finite flat maps
QUESTION [2 upvotes]: In his Algebraic surfaces book, Beauville gives a result allowing one to "absorb ramification" for certain maps (see below). There are also something similar one can do with number fields. I would really like to know how I should think about these results, (that is, are they specific to these two situations or are they vastly more general.) I'm really hoping for a reference (to EGA?) where such a version is stated, assuming it is true.
This question naturally lends itself to questions 2,3; which are hopefully covered in the same place as #1 (assuming again that they are true).
$\textbf{1.}\textrm{  Absorbing ramification}$
If $K|k$ is a finite extension of number fields, then there are infinitely many finite extensions $E|k$, so that $K\cap E = k$ and $EK|E$ is unramified. (You can get such extensions from the approximation lemma.) (Does this fact have a name?)
$\textbf{Setup:}$ Let $X \xrightarrow{\phi} Y$ be a finite, faithfully flat map of noetherian schemes. Feel free to assume conditions on $X, Y, \phi$, as you need.
Can we have something similar for faithfully flat maps $\phi$? More specifically, let $\phi, X,Y$ be as in the setup. Does there exist $Z \xrightarrow{\tau} Y$ a finite, faithfully flat map so that the base change along $\tau$ is etale. Is there a lemma (like the approximation lemma) showing that there are "many" such maps in good conditions, and allowing us to control the ramification of $\tau$.
-In certain cases, we can absorb ramification for a fibration of a smooth surface (Beauville, p73 Lemma VI.7):
http://books.google.com/books?id=KV1WiV7WmPIC&pg=PA73&dq=Complex+algebraic+surfaces+lemma+VI.7&hl=en&sa=X&ei=e_sCUd2MM4ac2QX8qoCAAg&ved=0CDAQ6AEwAA#v=onepage&q=Complex%20algebraic%20surfaces%20lemma%20VI.7&f=false

$\textbf{2.}\textrm{  Maximal etale subextensions}$
Let $\phi, X,Y$ be as in the setup.
Does there exist a maximal etale subextension? That is, can we factor $\phi$ as $X \rightarrow Y_{et} \xrightarrow{\psi} Y$ with $\psi$ etale, and so that $\psi$ is maximal amongst all such factorizations. That is, if $\phi$ also factors as  $$X \rightarrow Y' \xrightarrow{\psi'} Y$$ with $\psi'$ etale, then the map $Y_{et} \xrightarrow{\psi} Y$ factors as $Y_{et} \rightarrow Y' \xrightarrow{\psi'} Y$.
$\textbf{3.}\textrm{  Galois closure}$
The natural question is take $X, Y, \phi$ be as our initial setup, (assume that $\phi$ is separable, $X,Y$ are projective over a field, integral) and ask if there a galois closure $Z \rightarrow X$, so that $Z/Aut(Z/Y) \cong Y$ naturally. However, it's not even clear to me that the automorphism group would be finite (or more generally that the quotient exists as a scheme) - or there exists such $Z$ which is "minimal".
These questions bear a strong relationship to theorems true for number fields. One issue of course is that we use composition of number fields in number theory and in algebraic geometry we have base change along maps which are a little different operations.

REPLY [1 votes]: If $X$ and $Y$ are normal and separated then they each have a unique generic point, so the map betwen generic points gives a field extension, and $X$ is determined over $Y$ by its field extension.  Indeed, if $U$ is an affine open set of $Y$, and $U'$ is its inverse image in $X$, then $\Gamma(U',\mathcal O_X)$ is the normalization of $\Gamma(U,\mathcal O_Y)$ in the residue field of the generic point of $X$.  Moreover, this is an equivalence of categories - the morphisms are the same.
This, for instance, immediately tells us how to construct the Galois closure - just take the Galois closure of the corresponding field extension.
One could study maximal etale subextensions this way, but I think it's better to argue like this: If $X \to Y_1 \to Y$ and $X \to Y_2 \to Y$ are both etale, then $Y_1 \times_Y Y_2$ is etale and $X$ maps to it. If you multiply all the finite etale things that $X$ factors through like this, and take the image of $X$, you should get a maximal etale cover.


TITLE: Why do we call a vector space in terms of vector space over some field?
QUESTION [9 upvotes]: I am getting a bit confused with the terminology here. I understand that a field means some set of scalars like real numbers but why do we need a field for a vector space? Are not the numerical values used to define a vector are inherent properties of the vector space? Why do we term it like "vector space over real field"? If not numbers, what are the other fields possible for a vector space because you obviously numbers to define the value of magnitude and direction of the vectors?

REPLY [2 votes]: Are not the numerical values used to define a vector are inherent properties of the vector space?

Yes, they are. If you use other numerical values (like complex numbers, or $\mathbb F_2$), you get another vector space (for example $\mathbb F_2^3$ instead of $\mathbb R^3$). However you can also get a different vector spaces using the same field (for example, $\mathbb R^2$ and $\mathbb R^3$). Now it turns out that vector spaces over the same field are in some sense compatible with each other, in a way that vector spaces over different fields generally are not. For example, it's no problem to define a linear function $f:\mathbb R^2\to\mathbb R^3$. However there are no linear functions $f:\mathbb F_2^2\to\mathbb R^2$. Therefore it makes sense to always explicitly say on which field the vector space is based.


TITLE: Why the cone generated by a closed convex set containing the origin may not be closed?
QUESTION [3 upvotes]: I encounter this problem in the proof of Theorem $1.28$, page 21, Skiadas' Asset Pricing Theory.
$X$ is a constrained market which is defined to be a closed convex subset of $\Bbb R^{K+1}$, $C = \{kx : x \in X, k \in \Bbb R_+\}$, why it's possible that $C$ is not closed?
PS: If you take an interest in the proof of equivalence between non-arbitrage and the existence of present value function for a constrained market, you can find them here in the sample chapter.
Added: As David Mitra's example suggests, it's important to point out $X$ contains the origin, which is required in the definition of a constrained market.

REPLY [3 votes]: Take, in $\Bbb R^2$, $X$ to be the graph of $y=x^2$ together with the set of points "above it".   Then the cone $C$ of $X$ is the non-closed set $\bigl\{\,(x,y)\mid y>0\,\bigr\}\cup\{\,(0,0)\,\}$.
(I think any strictly convex closed body tangent to the origin will do.)


TITLE: Evaluating $\lim\limits_{x\to\infty} \frac{\ln(x^2+4)}{\sinh^{-1}x}$
QUESTION [2 upvotes]: $$\lim\limits_{x\to\infty} \frac{\ln(x^2+4)}{\sinh^{-1}x}$$

This is an exam practice question. BTW, I am refering to the inverse hyperbolic function above.
Since this is infinity/infinity, I used one application of L'Hopital's rule for this one, and then did a bit of algebraic manipulation to get $\frac{2}{1}$, i.e. $2$. Is this right?
Is there a quicker way to do this? It seems like Taylor Polynomials are often used, but they confuse me - Is this one, where Taylor Polynomials would have been easier?

REPLY [1 votes]: For the next solution we'll resort to sinh(x) that goes to $\infty$ when x tends to $\infty$. Just throw sinh(x) in the limit and then consider the dominant term on nominator. Then you get that:
$$\lim\limits_{x\to\infty} \frac{\ln(x^2+4)}{\sinh^{-1}x} = \lim\limits_{x\to\infty} \frac{\ln(\sinh^{2}x+4)}{x} = \lim\limits_{x\to\infty} \frac{\ln{e^{2x}}}{x} = \lim\limits_{x\to\infty} \frac{{2x}}{x}=2$$
The proof is complete.


TITLE: Weighted Avg for Billable Hour
QUESTION [0 upvotes]: I am trying to get to an avg reimbursement rate for an MD billable hour.  They have different types of appointments, each receiving a different reimbursement amount.  I ran an analysis which gave the historical mix of appointments across all the MDs.  Looks something like this:
Appt.  % of Time Appt Occurred.  Reimb Amt.
A      30%                       $50
B      50%                       $65
C      20%                       $90

Obviously from this I could get an weighted avg reimbursement amount of $65.50 per appointment.  The problem is that I need to get to an avg reimb amt per hour.  Each of these appointments have varying lengths.  The data set would be something like this:
Appt.  Appt Duration(Mins).  % of Time Appt Occurred.  Reimb Amt.
A      15                    30%                       $50
B      30                    50%                       $65
C      45                    20%                       $90

The added complexity of the time component has me stumped.  How can I get to an avg hourly reimbursement rate?
Thank you in advance!

REPLY [0 votes]: Find hourly reimbursement rate for each kind of appointment, then get the weighted average for the three kinds:
For the first kind: $\$50/(1/4) = \$200,$ etc.
Then $\$200(.3) + ?(.5) + ?(.2) = $ answer.


TITLE: How to expand integral of variance for continuous random variable?
QUESTION [0 upvotes]: Variance of continuous random variable: $$E[(X-\mu_X)^2]=\int_x (x-\mu_X)^2 f_X(x)dx $$
I should expand into two parts, but I don't know how to move from here and I doubt it's even correct:
$$\int_x (x-\mu_X)^2 f_X (x)dx = \int_x(x^2-2x\mu_x+\mu^2_X)f_X dx$$

REPLY [2 votes]: Well, yes, this is perfectly fine. If you want to continue computing, you could do something like this: $$\int x^2 f_X(x) dx - 2 \mu_X \int x f_X(x) dx + \mu_X^2 \int f_X(x) dx = \mathbb{E}[X^2] - 2 \mu_X \mathbb{E}[X] + \mu_X^2 = \mathbb{E}[X^2] - 2 (\mathbb{E}[X])^2 + (\mathbb{E}[X])^2 = \mathbb{E}[X^2] - (\mathbb{E}[X])^2,$$ where I only used the definitions of $\mu_X, \, f_X$ and the law of the unconscious statistician. I hope this is what you meant.


TITLE: For what $\alpha > 0$ do we have $\int_{D_{\alpha}} f \ d\lambda < + \infty$ ?
QUESTION [1 upvotes]: Problem: Define for $\alpha > 0$ the subset of $\mathbb{R}^2$ given as $$ D_{\alpha} = \left\{(x,y) \in \mathbb{R}^2 \mid 0 < y, 0 < x < y^{\alpha} < 1 \right\}. $$
Define the function $$f : (0, \infty) \times (0, \infty) \to [0, \infty): (x,y) \mapsto \frac{1}{(x+y)^2}. $$ For which values of $\alpha > 0$ is $$ \int_{D_{\alpha}} f \ d\lambda < + \infty $$ ?
Attempt: I first wanted to change coordinates to polar coordinates. But I don't know how to redefine the subset $D_{\alpha}$ in terms of polar coordinates. Since $f$ is a positive function, we can switch order of integration (Fubini). So I wrote (I'm not sure if the integration bounds are correct) $$ \int_{D_{\alpha}} f \ d \lambda = \int_{0}^1 \bigg( \int_{0}^{y^{\alpha}} \frac{1}{(x+y)^{2}} dx \bigg) dy = \int_{0}^{y^{\alpha}} \bigg( \int_0^1 \frac{1}{(x+y)^2} dy \bigg) dx = \int_{0}^{y^{\alpha}} \bigg( \frac{1}{x} - \frac{1}{x+1} \bigg) dx. $$ I'm not sure how to proceed. I can only use linearity if I know the two terms inside the brackets are integrable. But $$ \int_{0}^{y^{\alpha}} \frac{dx}{x} = \lim_{\epsilon \to 0} \ln \bigg( \frac{y^{\alpha}}{\epsilon} \bigg) $$ and I don't think I can get this finite for any $\alpha$.
Help/advice with this problem is appreciated.

REPLY [1 votes]: The integral equals
$$\tag 1 \int_0^1 \left (\frac{1}{y}-\frac{1}{y+y^\alpha}\right)\,dy.$$
as @Cuoredicervo found. Suppose $0<\alpha <1.$ Note we have
$$\frac{1}{y+y^\alpha} < \frac{1}{y^\alpha}.$$
Because $\int_0^1 (1/y^\alpha)\, dy$ converges, so does $\int_0^1 1/(y+y^\alpha)\, dy.$ Since $\int_0^1 (1/y)\, dy$ diverges, and divergent - convergent = divergent, $(1)$ diverges.
If $\alpha = 1,$ then $(1)= \int_0^1 1/(2y)\, dy = \infty.$
If $\alpha >1,$ then
$$\frac{1}{y}-\frac{1}{y+y^\alpha} = \frac{y^\alpha}{y(y+y^\alpha)}< \frac{y^\alpha}{y^2} = y^{\alpha -2}.$$
Since $\alpha -2 > -1,$ $\int_0^1  y^{\alpha -2}\, dy$ converges.
Conclusion: The integral diverges for $0<\alpha \le 1,$ and converges for $\alpha > 1.$


text:
{\bf Problem.} Given that $f(x) = x^k$ where $k < 0$, what is the range of $f(x)$ on the interval $[1, \infty)$?
{\bf Level.} Level 5
{\bf Type.} Algebra
{\bf Solution.} We are looking at the range of $f(x)$ when $x$ is in the interval $[1,\infty)$. Because $k < 0$, $f(x)$ is decreasing on the interval $[1, \infty)$. We see that $f(1) = 1^k = 1$, and as $x$ increases, $f(x) = x^k$ approaches 0, but never reaches it.  Hence, on the interval $[1,\infty)$, $f(x)$ takes on all values between 0 (exclusive) and 1 inclusive, which means the range of $f(x)$ is $\boxed{(0,1]}$.

set_name: MATH


TITLE: Always a prime between $x$ and $x+cf(x)$
QUESTION [5 upvotes]: What is the asymptotically slowest growing function $f(x)$, such that there exists constants $a$ and $b$, such that for all $x>a$, there is always a prime between $x$ and $x+bf(x)$?
$f(x)=x$ works, does $\sqrt{x}$ work, $\log(x)$, or $\log\log(x)$?

REPLY [10 votes]: What you are looking at is an upper bound for prime gaps.  Bertrands postulate states that there is always a prime between $x$ and $2x$, but this has been improved significantly.  The most recent result due to Baker Harman and Pintz states that $$\pi\left(x+x^{0.525}\right)
-\pi(x)\gg \frac{x^{0.525}}{\log x}.$$ This means that for sufficiently large $x$, there is always a prime between $x$ and $x+x^{0.525}$.  This implies for example that there is always a prime between consecutive cubes, that is there is always a prime in the interval $(x^3,(x+1)^3)$.
As for the expected result, the Wikipedia article covers this as well, see Conjectures about prime gaps.  In particular, the Riemann Hypothesis tells us that for any $\epsilon$, we will have a prime in the interval $(x,x+x^{\frac{1}{2}+\epsilon})$ for sufficiently large $x$.  Cramer made the much stronger conjecture that there is always be a prime between $x$ and $x+\log^2x$ for sufficiently large $x$.