TITLE: "Fat" Cantor Set
QUESTION [16 upvotes]: So the standard Cantor set has an outer measure equal to $0$, but how can you construct a "fat" Cantor set with a positive outer measure? I was told that it is even possible to produce one with an outer measure of $1$. I don't see how changing the size of the "chuck" taken out will change the value of the outer measure. Regardless of the size, I feel like it will inevitably reach a value of $0$ as well...
Are there other constraints that need to be made in order to accomplish this?

REPLY [0 votes]: The measure of a Cantor set cannot be 1, but certainly can be arbitrary close to it.  Let the total measure of the pieces removed at step $n$ be equal $m_n$, so we remove an open intervalof length $m_1$ in the first step, then remove two open intervals of total length  $m_2$ from the two pieces remaining after step 1.  Then remove four open intervals of total length $m_3$ from the remaining four pieces left after step 2.  Continuing this way, we remove open intervals with total measure $M=m_1+m_2+...$. The measure of the compact set $C$ left is $1-M$.  If $M=1$, the measure of $C$ is $0$, if $M<1$ then the measure of $C$ is positive---it is fat.


TITLE: Linear operators and change of basis in a vector space
QUESTION [1 upvotes]: Suppose we have a vector space $V$ over a scalar field $\mathbb{F}$ and two different bases $\mathcal{B}=\lbrace\mathbf{v}_{i}\rbrace_{i=1,\ldots , n}$ and $\mathcal{E}=\lbrace\mathbf{w}_{i}\rbrace_{i=1,\ldots , n}$. A given vector $\mathbf{x} \in V$ can be described in terms of either of these bases, by the (unique) linear combinations $$\mathbf{x}= \sum_{i=1}^{n}a_{i}\mathbf{v}_{i} \quad \text{or}\quad  \mathbf{x}= \sum_{i=1}^{n}b_{i}\mathbf{w}_{i}$$ where $a_{i} \neq b_{i}$.
Now, suppose that we have a linear operator $\mathcal{S}$ which maps the basis $\mathcal{E}$ to the basis $\mathcal{B}$, defined in the following manner, $$\mathcal{S}\left(\mathbf{w}_{j}\right)= \mathbf{v}_{j}$$ As $\mathbf{v}_{j} \in V$ it can itself be described in terms of a (unique) linear combination of the basis vectors $\mathbf{w}_{i}$ $\left(i=1,\ldots , n\right)$ and as such, we have that $$ \mathbf{v}_{j}=\sum_{i=1}^{n}\left[\mathcal{S}\left(\mathbf{w}_{j}\right)\right]_{i}\mathbf{w}_{i}=\sum_{i=1}^{n}S_{ij}\mathbf{w}_{i}$$ where we have defined $\left[\mathcal{S}\left(\mathbf{w}_{j}\right)\right]_{i} \equiv S_{ij}$, which is the $i^{th}$ component of the $j^{th}$ basis vector $\mathbf{v}_{j} \in \mathcal{B}$ with respect to the basis $\mathcal{E}$.
Is this a correct description of how to change between two different bases for the same vector space? (The equation $\mathbf{v}_{j}=\sum_{i=1}^{n}S_{ij}\mathbf{w}_{i}$ is given in a text book that I've been reading as a formula for switching between two different basis sets for a given vector space and I'm trying to justify its form).

REPLY [0 votes]: To see the pattern it helps to write out specific cases.  Such as $$v_1=a_1 w_1 + a_2 w_2 + a_3 w_3 + ...$$ and $$v_2=b_1 w_1 + b_2 w_2 + b_3 w_3 + ...$$ and $$v_3 = c_1 w_1 + c_3 w_3 + ...$$ Now you're really tired of the alphabet game and aching for something more concise, so you think maybe $a$ can be denoted by $S_1$ and $b$ by $S_2$ and ... etc.  And you're into double subscripts. Note that the subscript of $S$ matches up with the subscript of $v$ and the subscript that increments through each sum matches up with $w$.  Then you've understand the equation your author used.Now look back at what you originally wrote:$$\sum_{i=1}^{n}\left[\mathcal{S}\left(\mathbf{w}_{j}\right)\right]_{i}\mathbf{w}_{i}$$That glob with the brackets is some kind of private notation of yours unknown to the rest of us. In my alphabet-infested equations, if I replace $a_1$ by $S$ with 1,1 subscript and $a_2$ by $S$ with 1,2 subscript and $a_3$ by $S$ with 1,3 subscript, and if I use $i$ for the varying subscript in the sum, then I should use $j$ for the subscript of the $v$'s on the left hand side of my equations.  If you rewrite my alphabet-infested equations with that convention, I think you'll get what your author wrote, or the equivalent. (Which subscript you decide to designate by $i$ and which by $j$ obviously doesn't matter.  If that's not obvious you'll have to ponder it and fool around with concrete examples of double-subscripting until it is.)


TITLE: Any linear maps between two isomorphic sets
QUESTION [0 upvotes]: There is a similar discussion here:
Is any homomorphism between two isomorphic fields an isomorphism?
What I want to assume further is:
Suppose there are two isomorphic groups $\mathcal{M}$, $\mathcal{N}$ with operator multiplication, i.e., there exists an isomorphism between two sets, $\phi: \mathcal{M} \mapsto \mathcal{N}$.
Now suppose that I have another homomorphism $\psi: \mathcal{M} \mapsto \mathcal{N}$ and I know $\psi$ is onto does this guarantee that $\psi$ is an isomorphism?

REPLY [1 votes]: As sets, $\mathbb{R}$ is isomorphic to $\mathbb{R}^2 = \mathbb{C}$, but as groups we have $\varphi : \mathbb{C} \to \mathbb{R}$ given by $\varphi(x,y)=x$ is an homomorphism, is onto but not injective.
Edit: The question was changed, but the example still works since $\mathbb{R}$ and $\mathbb{C}$ are isomorphic as groups too.


TITLE: Cohen-Macaulayness of the direct image of the canonical sheaf
QUESTION [2 upvotes]: Let $Y$ be a normal projective variety and let $f:X\to Y$ be a desingularization. Define $\mathcal K_X=f_*\omega_X$, the Grauert--Riemenschneider canonical sheaf of $X$. It is independent of the resolution. Does this have some good properties?
In particular, are there mild assumptions that ensure it is Cohen--Macaulay?

REPLY [3 votes]: Let me expand Donu's answer a bit.  Say that $Y$ is normal for simplicity.  Then we always have:
$$f_* \omega_X \subseteq \omega_Y.$$
The sheaf $\omega_Y$ is Cohen-Macaulay if $Y$ is Cohen-Macaulay, but in general, it is not.
On the other hand, $\omega_Y$ is always S2 (see for instance Kollár-Mori).  Furthermore, if $Y$ is normal, then $f_* \omega_X \subseteq \omega_Y$ is an isomorphism outside a set of codimension 2.  Hence
Observation For normal $Y$, $f_* \omega_X$ will be S2 if and only if $f_* \omega_X = \omega_Y$.  This equality is equivalent to rational singularities if $Y$ is Cohen-Macaulay.
In other words, the only reasonable way to guarantee that $f_* \omega_X$ is Cohen-Macaulay is to require that $Y$ has rational singularities (at least if $Y$ is normal).


TITLE: Simplification of an infinite sum consisting of Legendre polynomials
QUESTION [1 upvotes]: In an article about Legendre Polynomials, I encountered the following simplification.
\begin{align}
 (something)\dots&=\int_{-1}^{1} \int_{-1}^{1} \left[\sum_{i=n+1}^{\infty} \sum_{j=n+1}^{\infty} a_{ij}\, p_{i}(x) \,p_{j}(y) \right]^2\, dx\, dy \qquad (1)\\
&=\int_{-1}^{1} \int_{-1}^{1} \sum_{i=n+1}^{\infty} \sum_{j=n+1}^{\infty} \left[a_{ij}^2\, p_{i}^2(x) \,p_{j}^2(y) \right]\, dx\, dy \, \qquad (2)
\end{align}
where $p_i(x)$ and $p_j(y)$ are Legendre polynomials.
Could you please explain how we can simplify equation $(1)$ to equation $(2)$? As you know infinite sums do not have many properties that finite sums have. By the way, the article can be found here. The above equations are on page 88.

REPLY [1 votes]: The use of $$ \int_{-1}^{1} P_{n}(x) \, P_{m}(x) \, dx = \frac{2}{2 n + 1} \, \delta_{n,m}$$ will be made as follows:
\begin{align}
I &= \int_{-1}^{1} \int_{-1}^{1} \left[\sum_{i=n+1}^{\infty} \sum_{j=n+1}^{\infty} b_{i,j} \, P_{i}(x) \, P_{j}(y) \right]^2 \, dx\, dy  \\
&= \sum_{i} \sum_{j} \sum_{k} \sum_{m} b_{i,j} \, b_{k,m} \, \int_{-1}^{1} \int_{-1}^{1} P_{i}(x) \, P_{k}(x) \, P_{j}(y) \, P_{m}(y) \, dx \, dy \\
&= \sum_{i} \sum_{j} \sum_{k} \sum_{m} b_{i,j} \, b_{k,m} \, \int_{-1}^{1} P_{i}(x) \, P_{k}(x) \, dx \times \, \int_{-1}^{1} P_{j}(y) \, P_{m}(y) \, dy \\
&= \sum_{i} \sum_{j} \sum_{k} \sum_{m} \frac{4 \, b_{i,j} \, b_{k,m}}{(2 i+1)(2 j +1)} \, \delta_{i,k} \, \delta_{j,m} \\
&= \sum_{i} \sum_{j} \frac{4 \, b_{i,j}^{2}}{(2 i+1)(2 j+1)}
\end{align}
Now, since $$ \int_{-1}^{1} P_{n}^{2}(x) \, dx = \frac{2}{2 n + 1} $$ then
\begin{align}
I &= \sum_{i} \sum_{j} \frac{4 \, b_{i,j}^{2}}{(2 i+1)(2 j+1)} \\
&= \sum_{i} \sum_{j} b_{i,j}^{2} \, \int_{-1}^{1} \int_{-1}^{1} P_{i}^{2}(x) \, P_{j}^{2}(x) \, dx \\
&= \int_{-1}^{1} \int_{-1}^{1} \sum_{i=n+1}^{\infty} \sum_{j=n+1}^{\infty} \left( b_{i,j}^2 \, P_{i}^2(x) \, P_{j}^2(y) \right) \, dx\, dy.
\end{align}
This gives $$ \int_{-1}^{1} \int_{-1}^{1} \left[\sum_{i=n+1}^{\infty} \sum_{j=n+1}^{\infty} b_{i,j} \, P_{i}(x) \, P_{j}(y) \right]^2 \, dx\, dy = \int_{-1}^{1} \int_{-1}^{1} \sum_{i=n+1}^{\infty} \sum_{j=n+1}^{\infty} \left( b_{i,j}^2 \, P_{i}^2(x) \, P_{j}^2(y) \right) \, dx\, dy $$


TITLE: Local path connected and connected implies path connected
QUESTION [1 upvotes]: I am very close to proving that if a metric space $X$ is locally path-connected and connected, then it is path connected. I am stuck on the last step.

Let $X$ be connected and locally path connected. Let $x \in X$, and define $A$ to be the set of points path connected to $x$. $x$ is connected to itself, so $A$ is nonempty. Because $X$ is connected, if $A$ is open and closed, it must equal $X$ (since it is not $\emptyset$). We show that $A$ is open. Let $a \in A$. By local path connectedness, there exists an open ball $U \ni a$ that is path connected. Let $u \in U$, so $u$ is path connected to $a$. But $a$ is path connected to $x$, so $u$ is path connected to $a$. So $u \in A$, so $U \subset A$, so we've found an open ball around $a$ that's contained in $A$, so $A$ is open.

My difficulty is in showing $A$ is closed. It's enough to show that $A = \overline{A}$.

To show $A$ is closed, we prove $A = \overline{A}$. Clearly $A \subset \overline{A}$. For the reverse inclusion, let $y \in \overline{A}$. As $\overline{A} \subset X$ and $X$ is path-connected, there exists a open ball $U \ni y$ that is path=connected. Then $U \cap A$ is nonempty. Let $t \in U \cap A$. Since $t \in A$, $t$ is path-connected to $x$. Since $t \in U$ and $y \in U$, $t$ is path-connected to $y$. Therefore, $y$ is path-connected to $x$, so $y \in A$. So $\overline{A} \subset A$, so $A = \overline{A}$.

I think every step of this proof is fine, but to talk about such a $t$, I have to argue that $U \cap A$ is nonempty, but I can't think of an argument for this.

REPLY [1 votes]: The fact that $U \cap A \neq \emptyset$ follows immediately from the fact that $U$ is a neighbourhood of $y \in \overline{A}$. One of the many equivalent conditions for membership in the closure of $A$ is that every neighbourhood of the given point intersects $A$.
If you prefer it done sequentially, we know that $y \in \overline{A}$ if and only if there exists a sequence $a_n \to y$, meaning that, for all $\varepsilon > 0$, there exists some $N$ such that
$$n \ge N \implies d(a_n, y) < \varepsilon.$$
Taking $\varepsilon$ to be the radius of the open ball $U$, we get
$$n \ge N \implies a_n \in U \cap A.$$
Thus, $U \cap A \neq \emptyset$.


text:
{\bf Problem.} Let $a,$ $b,$ $c,$ $d$ be nonzero integers such that
\[\begin{pmatrix} a & b \\ c & d \end{pmatrix}^2 = \begin{pmatrix} 7 & 0 \\ 0 & 7 \end{pmatrix}.\]Find the smallest possible value of $|a| + |b| + |c| + |d|.$
{\bf Level.} Level 3
{\bf Type.} Precalculus
{\bf Solution.} We have that
\[\begin{pmatrix} a & b \\ c & d \end{pmatrix}^2 = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{pmatrix} a^2 + bc & ab + bd \\ ac + cd & bc + d^2 \end{pmatrix},\]so $a^2 + bc = bc + d^2 = 7$ and $ab + bd = ac + cd = 0.$  Then $b(a + d) = c(a + d) = 0.$  Since $b$ and $c$ are nonzero, $a + d = 0.$

If $|a| = |d| = 1,$ then
\[bc = 7 - a^2 = 6.\]To minimize $|a| + |b| + |c| + |d| = |b| + |c| + 2,$ we take $b = 2$ and $c = 3,$ so $|a| + |b| + |c| + |d| = 7.$

If $|a| = |d| = 2,$ then
\[bc = 7 - a^2 = 3.\]Then $|b|$ and $|c|$ must be equal to 1 and 3 in some order, so $|a| + |b| + |c| + |d| = 8.$

If $|a| = |d| \ge 3,$ then $|a| + |b| + |c| + |d| \ge 8.$

Therefore, the minimum value of $|a| + |b| + |c| + |d|$ is $\boxed{7}.$

set_name: MATH


TITLE: Can I say $x>0 \implies x\ge0$?
QUESTION [2 upvotes]: By the logic, I can understand that if $P$ is true, $P \lor Q$ is also true whatever $Q$ is. Since $x\ge 0 \iff (x>0 \lor x＝0)$, if I know $x>0$, can I conclude $x\ge 0$?
Sorry for some sort of stupid question.

REPLY [0 votes]: Yes.  $P \implies P$ or $Q$.
so if $P$ is $x > 0$ and $Q$ is $x=0$ then if $x> 0$ is true, then $x > 0$ or $x =0$ is true.  Which is the same thing as $x \ge 0$ is true.
I can see why this seems weird: We know something specific ($x > 0$) and we are suggesting something less specific and even implying something we know to be false ($x \ne 0$) to possibly be true ("$x \ge 0$??? so that means... $x$ could be $0$???").
But consider this.  Suppose you knew $x = 3$.  Can we conclude that $x \ge 0$?  Well, of course we can!  $3 \ge 0$ and as $x = 3$ we know $x \ge 0$.  But $x \ge 0$ implies that $x$ could be $15$. Are we saying that $x$ could be $15$?  Are we saying that $3$ could be $15$?!?!?!
Well, no.  The statement that $x \ge 0$ isn't necessarily a statement of all potential possible things[1] and the only thing we know for certain is that $x < 0$ is false.  The statement that $x \ge 0$ is literally just a statement of fact; it is the case that $x$ is greater than or equal to $0$.  We may or may not know any more than that but that much is true.[2]
Another analogy is "Penguins are animals".  Well, "animals" can be many things including camels or elephants or eagles.  But "Penguins are animals" doesn't mean that all those possibilities ("Penguins might be camels"?  No Way!) are open.
[1] although it could be.
[2] actually $x$ must be something.  Its not possible that is both $x = 0$ and $x > 0$.  So it is never a case that all possibilities are open.  $x$ must be something and if its that, it can't be anything else. We might not know what it is, but it can't be more than one unknown option.  But that's okay.  Saying $x$ is one of several things isn't implying that all are equally valid.


text:
\section{Conclusion} \label{S:concl}

Parametric mappings $r:\C{M}\to\C{U}$ have been analysed with in a  variety of
settings via the associated linear map $R:\C{U}\to\C{Q}\subseteq\D{R}^{\C{M}}$,
enabling the linear analysis.  The RKHS setting allows a first representation, and
essentially reproduces everything in $\C{U}$ in the function space $\C{R}$.
The choice of another inner product and corresponding Hilbert space $\C{Q}$
leads to measures of importance in  $\C{M}$, or, more precisely, in $\D{R}^{\C{M}}$.

It is shown that each separated representation defines an associated linear map,
and that conversely under some more restrictive conditions, the normally more general
notion of an associated linear map defines a representation.

Several refinements are presented to represent some additional structure in the
linear map.  One such structure is the information of dealing with a coupled system.
This can be reflected in the structure of the associated linear map.

\ignore{
\subsection{Test of fonts} \label{SS:testf}

\paragraph{Slanted:} --- serifs
\[ a, \alpha, A, \Phi, \phi, \vphi, \qquad
     \vek{a}, \vek{\alpha}, \vek{A}, \vek{\Phi}, \vek{\phi}, \vek{\vphi} \]

\paragraph{Slanted:} --- sans serif
\[ \tns{a}, \tns{\alpha}, \tns{A}, \tns{\Phi}, \tns{\phi}, \tns{\vphi}, \qquad
  \tnb{a}, \tnb{\alpha}, \tnb{A}, \tnb{\Phi}, \tnb{\phi}, \tnb{\vphi} \]

\paragraph{Other:}  --- no small letters
\[ \C{E}, \C{Q}, \C{R}; \D{E}, \D{Q}, \D{R}; \E{E}, \E{Q}, \E{R}\]

\paragraph{Fraktur:}
\[ \F{e}, \F{q}, \F{r}; \F{E}, \F{Q}, \F{R} \]

\paragraph{Upright:} --- serifs, no small greek letters
\[ \mrm{a}, \mrm{A}, \mrm{\Phi},\qquad \mat{a}, \mat{A}, \mat{\Phi} \]

\paragraph{Upright:} --- sans serif, no small greek letters
\[ \ops{a}, \ops{A}, \ops{\Phi},\qquad \opb{a}, \opb{A}, \opb{\Phi} \]
}

config: arxiv


TITLE: How to describe a summation of $\frac{1}{2^x3^y}$ and evaluate.
QUESTION [3 upvotes]: I want too calculate the value of this sum:
$$\sum  \frac{1}{2^x3^y}$$
Where we sum up all permutations of terms involving a nonnegative integer $x$ and a nonnegative integer $y$.
How can I notationally describe this sum and how can I calculate it (assuming it does converge).

REPLY [7 votes]: Due to the Fubini/Tonelli Theorem, you can just sum over one index first, and then over the other. That is,
$$\sum_{n = 0}^\infty \sum_{m = 0}^\infty \frac{1}{2^n 3^m} = \sum_{n = 0}^\infty \frac 1 {2^n}\left(\sum_{m = 0}^\infty \frac 1{3^y} \right) = \sum_{n = 0}^\infty \frac 1 {2^n} \left(\frac{1}{1-(1/3)}\right) = \frac 3 2 \cdot\frac{1}{1-(1/2)} = 3.$$ It'd be the exact same if you summed over $n$ first, then $m$.


TITLE: Coproduct of two modules
QUESTION [0 upvotes]: Suppose that $M$ is an $A$-module, and $N$ is a $B$-module. The coproduct of $A$ and $B$ is $A\otimes_{\mathbb{Z}}B$, and the coproduct of $M$ and $N$ is $M\oplus N$. I was wondering if $M\oplus N$ could be a module over $A\otimes_{\mathbb{Z}}B$, so that it is a "coproduct of two modules." But it appears that there is no $A\otimes_{\mathbb{Z}}B$-module structure on $M\oplus N$. Is it true that "the category of modules" doesn't have coproducts?
Edit: All rings are assumed to be commutative. I'm envisaging a "category of modules," where  objects are pairs $(A,M)$, where $A$ is a ring and $M$ is an abelian group, together with $A$-scalar multiplication structure. A morphism $(A,M)→(B,N)$ is a pair $(f,ϕ)$, where $f:A→B$ is a ring homomorphism, $ϕ:M→N$ is an abelian group homomorphism, such that $ϕ(am)=f(a)ϕ(m)$.
When I said "coproduct of $M$ and $N$", I'm treating them as abelian groups.
The question is whether this category has coproducts. My initial guess was that the coproduct of $(A,M)$ and $(B,N)$ is $(A\otimes_{\mathbb{Z}}B,M\oplus N)$. I think this is wrong, but I think a more complicated coproduct exists for any family of modules $(A_\alpha,M_\alpha)_\alpha$. My current guess looks something like $(\varinjlim\otimes A_\alpha,\varinjlim\otimes_{\beta\neq\alpha}A_\beta\otimes M_\alpha)$.

REPLY [2 votes]: $\newcommand{\dlim}{\varinjlim}
\newcommand{\sst}{\subset}
\newcommand{\sbe}{_\beta}
\newcommand{\sal}{_\alpha}
\newcommand{\iso}{isomorphism}
\newcommand{\al}{\alpha}
\newcommand{\be}{\beta}
\newcommand{\fall}{\ \ \forall}
\newcommand{\dra}{\dashrightarrow}
\newcommand{\mt}{\mapsto}
\newcommand{\ra}{\rightarrow}
\newcommand{\cc}{\circ}
\newcommand{\id}{\operatorname{id}}
\newcommand{\itk}[1]{{\it #1}}
\newcommand{\ras}{\overset{\sim}{\ra}}
\newcommand{\amod}{$A$-module}
\newcommand{\Hom}{\operatorname{Hom}}
\newcommand{\ten}{\otimes}
\newcommand{\opl}{\oplus}
\newcommand{\rf}[1]{\overset{#1}{\rightarrow}}
\newcommand{\ifof}{if and only if }
\newcommand{\la}{\leftarrow}
\newcommand{\Ra}{\Rightarrow}
\newcommand{\set}[1]{\left\{ #1 \right\}}
\newcommand{\setst}[2]{\left\{ #1 \ | \ #2 \right\}}
\newcommand{\lf}[1]{\overset{#1}{\leftarrow}}
\newcommand{\lff}[1]{\overset{#1}{\longleftarrow}}
\newcommand{\im}{\operatorname{im}}
\newcommand{\sur}{\twoheadrightarrow}
\newcommand{\krn}{\operatorname{ker}}
\newcommand{\hra}{\hookrightarrow}
\newcommand{\las}{\overset{\sim}{\la}}
\newcommand{\Z}{\mathbb{Z}}
\newcommand{\bb}{\backslash}
\newcommand{\os}[2]{\overset{#1}{#2}}
\newcommand{\us}[2]{\underset{#1}{#2}}
\newcommand{\amods}{$A$-modules}
\newcommand{\sm}{\smallskip}
\newcommand{\beq}[1]{\begin{equation}#1\end{equation}}
\newcommand{\ffall}{\qquad\forall}
\newcommand{\del}{\delta}
\newcommand{\isos}{isomorphisms}
\newcommand{\Tcd}{There is a commutative diagram}
\newcommand{\cd}{commutative diagram}
\newcommand{\tcd}{there is a \cd}
\newcommand{\tis}{there is an \iso}
\newcommand{\Tis}{There is an \iso}
\newcommand{\gam}{\gamma}
\newcommand{\hmms}{homomorphisms}
\newcommand{\Del}{\Delta}
\newcommand{\fgas}{f.g.~\amods}
\newcommand{\lam}{\lambda}
\newcommand{\Tesq}{There is an exact sequence}
\newcommand{\isoc}{isomorphic}
\newcommand{\bopl}{\bigoplus}
\newcommand{\wrt}{with respect to}
\newcommand{\esqs}{exact sequences}
\newcommand{\qqq}{\qquad\quad}
\newcommand{\ds}{_*}
\newcommand{\dla}{\dashleftarrow}
\newcommand{\sab}{_{\al\be}}
\newcommand{\sx}{_x}
\newcommand{\Calg}{\mathsf{Alg}}
\newcommand{\Cab}{\mathsf{Ab}}
\newcommand{\ang}[1]{\langle #1\rangle}
\newcommand{\hpsi}{\hat{\psi}}
\newcommand{\hphi}{\hat{\phi}}
\newcommand{\resp}{respectively}
\newcommand{\cds}{commutative diagrams}
\newcommand{\qq}{\qquad}
\newcommand{\mf}{\mapsfrom}
\newcommand{\aalg}{$A$-algebra}
\newcommand{\aalgs}{$A$-algebras}
\newcommand{\Cring}{\mathsf{Ring}}
\newcommand{\Cmod}{\mathsf{Mod}}
\newcommand{\alg}{algebra}
\newcommand{\algs}{algebras}
\newcommand{\hmm}{homomorphism}$
Proposition. The category $\Cmod$ has coproducts.
Proof. Let $(A\sal,M\sal)_{\al\in\Lambda}$ be a nonempty family in $\Cmod$.  Let $A'$ be the direct limit of finite tensor products of $(A\sal)\sal$, i.e., $A'=\dlim\ten_{\al\in I}A\sal$, where $I\sst\Lambda$ is a finite subset. Given $\al\in\Lambda$, let $A'\sal:=\dlim\ten_{\be\in I\sal}A\sbe$, where $I\sal\sst\Lambda\bb\set{\al}$ is a finite subset. We claim that $A'\simeq A\sal\ten A'\sal\fall\al\in\Lambda$. To see this, first note that for every $\al\in\Lambda$, there is an obvious ring \hmm\ $\tau'\sal:A'\sal\dra A'$.
This induces a ring \hmm
$$\phi:A\sal\ten A'\sal\dra A',\qqq a\sal\ten x\mt\tau_{\set{\al}}(a\sal)\tau'\sal(x),$$
where $(\tau_I)_{I\sst\Lambda}$ is the direct limit for $A'$.
On the other hand, let $(\sigma_{I\sal})_{I\sal}$ be the direct limit for $A'\sal$.
Given a finite subset $I\sst\Lambda$, define a  ring \hmm
$$\psi_I:A\sal\ten A'\sal\la \ten_{\al\in I}A\sal$$
as follows: if $\al\not\in I$, then $\psi_I(x):=1_{A\sal}\ten\sigma_I(x)$. If $\al\in I$, then
$$\psi_I(\ten_{\be\in I}a\sbe)=a\sal\ten\sigma_{I'\sal}(\ten_{\be\in I'\sal}a\sbe),$$
where $I'\sal:=I\bb\set{\al}$. It is easy to see that $\psi_I$ is compatible with inclusions of the finite subsets $I$. Hence, there is an induced ring \hmm
$$\hpsi:A\sal\ten A'\sal\dla A'.$$
It is easy to see that $\phi$ and $\hpsi$ are inverses of each other.
Now let $M'\sal:=M\sal\ten A'\sal$. It is then an $A'$-module, and we will show that $(A',\opl\sal M'\sal)$ is the coproduct of $(A\sal,M\sal)\sal$.
There is an abelian group \hmm
$$j\sal:M\sal\ra M'\sal,\qqq m\sal\mt m\sal\ten1'\sal.$$
Let $(i\sal)\sal$ be the coproduct for $\opl\sal M'\sal$. It is easy to see that
$$(\tau_{\set{\al}},i\sal\cc j\sal):(A\sal,M\sal)\dra(A',\opl\sal M'\sal)$$
is a morphism in $\Cmod$.
Let $(\phi\sal,\psi\sal)\sal$ be a family of morphisms in $\Cmod$, where
$$(\phi\sal,\psi\sal):(A\sal,M\sal)\ra(B,N).$$
Then there are induced ring \hmms\ $\hphi:A'\dra B$, $\hphi\sal:A'\sal\dra B$, and a unique abelian group \hmm
$$\hpsi\sal:M'\sal\dra N$$
\st $m\sal\ten x\mt\hphi\sal(x)\cdot\psi\sal(m\sal)$. Note that $\hpsi\sal\cc j\sal=\psi\sal$.
It is easy to see that
$$(\hphi,\hpsi\sal):(A',M'\sal)\ra(B,N)$$
is a morphism of modules, so there is an induced morphism of modules
$$(\hphi,\hpsi):(A',\opl\sal M'\sal)\dra(B,N).$$
It is easy to see that $\hpsi\sal$ is the unique abelian group \hmm\ \st for every $\al\in\Lambda$, $\hpsi\sal\cc j\sal=\psi\sal$ and $(\hphi,\hpsi\sal)$ is a morphism of modules. Hence, the uniqueness of $(\hphi,\hpsi)$ follows.


TITLE: $\Bbb Q\cap\Bbb Z[\frac{1+\sqrt n}2]=\Bbb Z$ when $n=4k+1$ is square-free
QUESTION [1 upvotes]: I have a question concerning how to prove $\Bbb Q\cap\Bbb Z[\frac{1+\sqrt n}2]=\Bbb Z$ when $n=4k+1$ (maybe not necessary, I am not sure) is square-free .
The nontrivial containment is "$\subset$". A general element in $Z[\frac{1+\sqrt n}2]$ has the form of $\sum_i a_i(\frac{1+\sqrt n}2)^i$. I wonder what trick I should use to simplify it.

REPLY [1 votes]: Note that
$$\left(\frac{1+\sqrt n}2\right)^2=\frac{2k+1+\sqrt n}2
=k+\frac{1+\sqrt n}2.$$
It follows by induction that
$$\left(\frac{1+\sqrt n}2\right)^r=a_r+b_r\left(\frac{1+\sqrt n}2\right)$$
where $a_r$ and $b_r$ are integers, ans so
$$\Bbb Z\left[\frac{1+\sqrt n}2\right]
=\left\{a+b\frac{1+\sqrt n}2:a,b\in\Bbb Z\right\}.$$
As $\frac12(1+\sqrt n )$ is irrational the only rationals in this set
are the $a+\frac b2(1+\sqrt n)$ with $b=0$, that is the integers.


TITLE: Multivariate Non-Differentiability
QUESTION [2 upvotes]: This example says that "continuous partial derivatives imply differentiability but not vice-versa". Based on transposition logic, I would then assume that if a multivariate function has discontinuous partials at a certain point, then the function is not differentiable at that point.  Is this the right interpretation?
Any guidance would be appreciated -- thanks.

REPLY [1 votes]: Discontinuous partial derivatives do not imply that the function is not differentiable. See this link for a counter example: http://mathinsight.org/differentiable_function_discontinuous_partial_derivatives


TITLE: Diagram hasse in natural.
QUESTION [0 upvotes]: Is there a natural to other than 0 such that in the diagram of Hasse of natural with divisibility there is a natural between a and 0?

REPLY [1 votes]: Let $a$ be any natural number, we want to show that there exists a natural number $b$ such that $b$ is between $a$ and $0$ in the Hasse diagram.
Since the relationship is divisibility, we want either $a | b$ and $b | 0$ or $0 | b$ and $b | a$. It turns out that every natural number divides into $0$, so we can just pick any multiple of $a$, say $2a$, then
$$a | 2a \text{ and } 2a | 0.$$
Notice that I made the claim that every natural number $x$ divides into $0$, i.e. $x|0$. To show this is true, it suffices to show that there is an integer $k$ such that $xk = 0$; we can simply pick $k=0$ to make this equation true thus $x|0$.


TITLE: Faraday-Henry: proof
QUESTION [3 upvotes]: I know Faraday-Newmann-Lenz's law is...(easy form)
$$\mathcal E_{\text{inducted}}=-\frac{d\Phi_S(\mathbf{B})}{dt} \tag 1$$
The flow through a Gaussian surface is given by a surface integral through $S$,
$$\Phi_S(\mathbf{B})\equiv\int_S \mathbf{B} \cdot d\mathbf{a}$$
Why the relation
$$\boxed{\frac{dB(t)}{dt}=-kI(t)} \tag 2$$
derives from Faraday-Henry's law that given the induced electromotive force $\mathcal E_{\text{inducted}} $ in a spiral of surface $S$? Note that $k$ is a positive dimensional constant.
For the $(2)$ I have not a flux for $B$! What is the proof of $(2)$: I don't remember doing something like this when I was in my University. That was 23 years ago.

I'm adding the image just to show users the exercise (which doesn't care if someone executes it) where there is the green expression that I have called the (2).

REPLY [2 votes]: I've actually never seen this form of the law before, so if you have a reference, please let me know. I just worked out how to derive it from the "standard" form of Faraday's Law you've given at the start of your question, but I'm going to make a couple of assumptions, let me know if they aren't justified:

The magnetic field is constant in space, it only varies in time,
The area of the spiral or coil is fixed and unchanging,
The constant $k$ depends on the parameters like the resistance of the wire $R$ and the area of the loop $S$.

(If anyone can figure out how to do it without these assumptions, I'd be very interested to know.)
Given these assumptions, the flux simply reduces to the component of the magnetic field along the surface, since $\vec{B}$ is constant over $S$ and so $\Phi_B = \oint \vec{B}\cdot\text{d}\vec{S} = \vec{B}\cdot\vec{S} = B S \cos{\theta}$, where $\theta$ is the angle between the magnetic field and the area $S$.
Using this in Faraday's Law, $$\mathcal{E}_\text{induced} = -\frac{\text{d}}{\text{d}t}\left( B S \cos{\theta}\right) = -\frac{\text{d}B}{\text{d}t}\times \left(S \cos{\theta}\right).$$
Now, this is the induced electromotive force (emf) in the loop, the actual current that flows through it will depend on its resistance. The current flowing through the loop will thus be (from Ohm's Law) $$I(t) = \frac{\mathcal{E}_\text{induced}}{R}.$$
Therefore,
\begin{equation*}
\begin{aligned}
I(t) = \frac{\mathcal{E}_\text{induced}}{R} = -\frac{\text{d}B}{\text{d}t}\times \left(\frac{S \cos{\theta}}{R}\right)
\end{aligned}
\end{equation*}
Rearranging, I get: $$\frac{\text{d}B(t)}{\text{d}t} = - \left(\frac{R}{S\cos{\theta}}\right) I(t).$$
I suppose the constant $k$ should then be $$k = \frac{R}{S\cos{\theta}},$$
which is indeed positive, as $0<\theta<\pi/2$.


TITLE: A new approach to find value of $x^2+\frac{1}{x^2}$
QUESTION [7 upvotes]: When I was teaching in a college class, I write this question on board.
If we now $\bf{x+\dfrac{1}{x}=4}$ show the value of  $\bf{x^2+\dfrac{1}{x^2}=14}$
Some student asks me for a multi idea to show or prove that.
I tried these : $$ $$1 :$$\left(x+\frac{1}{x}\right)^2=4^2\\x^2+\frac{1}{x^2}+2x\times \frac{1}{x}=16\\x^2+\frac{1}{x^2}=16-2 $$
 2:solving quadratic equation ,and putting one of roots$$x+\frac{1}{x}=4\\\frac{x^2+1}{x}=4\\x^2+1=4x\\x^2-4x+1=0\\x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}=\frac{-(-4)\pm \sqrt{(-4)^2-4(1)(+1)}}{2}=\\x=\frac{4\pm \sqrt{12}}{2}=2\pm \sqrt{3}\\x=2+\sqrt3 \to x^2=4+4\sqrt3+3=7+4\sqrt3\\x^2+\frac{1}{x^2}=7+4\sqrt3+\frac{1}{7+4\sqrt3}=\\7+4\sqrt3+\frac{1}{7+4\sqrt3}\cdot\frac{7-4\sqrt3}{7-4\sqrt3}=\\7+4\sqrt3+\frac{7-4\sqrt3}{49-16\cdot 3}=\\7+4\sqrt3+\frac{7-4\sqrt3}{1}=14$$
3: visual approach .assume side length of a square is $x+\frac{1}{x}=4$
now I am looking for new idea to proof.Any hint will be appreciated.(more visual proof - geometrical - trigonometrical - using complex numbers ...)

REPLY [1 votes]: Here's a trigonometric approach:
Let $x=\tan\theta=\sin\theta/\cos\theta$.  Then
$$4=x+{1\over x}={\sin\theta\over\cos\theta}+{\cos\theta\over\sin\theta}={1\over\sin\theta\cos\theta}={2\over\sin2\theta}$$
so $\sin2\theta=1/2$.  Now
$$x^2+{1\over x^2}={\sin^2\theta\over\cos^2\theta}+{\cos^2\theta\over\sin^2\theta}={1\over\cos^2\theta}-1+{1\over\sin^2\theta}-1={1\over\sin^2\theta\cos^2\theta}-2\\
={4\over\sin^22\theta}-2=16-2=14$$


TITLE: Easy way to find distribution of transformation?
QUESTION [0 upvotes]: Suppose you have $X$ is $\operatorname{Normal}(\mu,\sigma^{2})$ and you want to find the pdf and cdf  of Y= $X^{a}$. What's an easy way to do this?

REPLY [0 votes]: If $a$ is an odd positive integer you can say
$$
F_{X^a}(t)=P[X^a \leq t] = P[X \leq t^{1/a}] = F_X(t^{1/a})
$$
Then differentiating
$$
f_{X^a}(t) = f_X(t^{1/a})\cdot \frac{1}{a}t^{1/a-1}.
$$
If $a$ is an even positive integer you can say for $t>0$
$$
F_{X^a}(t)=P[X^a \leq t] = P[|X| \leq t^{1/a}] =P[-t^{1/a} \leq X \leq t^{1/a}] = F_X(t^{1/a})-F_X(-t^{1/a})
$$
and $F_{X^a}(t)=0$ for all $t \leq 0$.
Differentiating then gives
$$
f_{X^a}(t) = 1_{t\geq 0}\cdot (f_X(t^{1/a})+f_X(-t^{1/a}))\frac{1}{a}t^{1/a-1}
$$
$$
= 1_{t\geq 0}\cdot f_X(t^{1/a})\frac{2}{a}t^{1/a-1}.
$$
You can plug in and simplify more using the specific form of $F_X$ and $f_X$ if you like.


TITLE: Verify that $ y(x) = Ae^{2x} + Be^{-2x} - 2\cos(x) - 5x\sin(x) $ is a solution of the ODE for any value of the constants A and B.
QUESTION [0 upvotes]: Verify that
$ y(x) = Ae^{2x} + Be^{-2x} - 2\cos(x) - 5x\sin(x) $
is a solution of the ordinary differential equation
$ \dfrac{d^2y}{dx^2} - 4y = 25x\sin(x) $
for any value of the constants A and B. $\\$
I seem to have differentiated correctly, so I'm unsure of why my reasoning is incorrect. Please explain why my reasoning is incorrect and how I can reason about this correctly to get the correct solution. Thank you.
My reasoning is as follows:
We must first find the second derivative of $y(x)$. We can then substitute this into the ODE to check if it is a solution. $\\$
$ \dfrac{dy}{dx} = 2Ae^{2x} - 2Be^{-2x} + 2\sin(x) - 5\sin(x) - 5x\cos(x) $ $\\[5pt]$
$ \Rightarrow \dfrac{dy}{dx} = 2Ae^{2x} - 2Be^{-2x} - 3\sin(x) - 5x\cos(x) $ $\\[5pt]$
$ \dfrac{d^2y}{dx^2} = 4Ae^{2x} + 4Be^{-2x} - 3\cos(x) - 5\cos(x) + 5x\sin(x) $ $\\[5pt]$
$ \Rightarrow \dfrac{d^2y}{dx^2} = 4Ae^{2x} + 4Be^{-2x} - 8\cos(x) + 5x\sin(x) $ $\\[5pt]$
We now substitute the second derivative into the ODE. $\\[5pt]$
$ \dfrac{d^2y}{dx^2} - 4y = 25x\sin(x) $ $\\[5pt]$
$ 4Ae^{2x} + 4Be^{-2x} - 8\cos(x) + 5x\sin(x) - 4y = 25x\sin(x) $ $\\[5pt]$
$ \Rightarrow 4Ae^{2x} + 4Be^{-2x} - 8\cos(x) - 4y = 20x\sin(x) $ $\\[5pt]$
We can now verify if this is a solution. Let $A = 2$, $B = 3$, $ x = 1 $, $ y = 4 $. $\\$
$ 4(2)e^{2(1)} + 4(3)e^{-2(1)} - 8\cos(1) - 4(4) = 20(1)\sin(1) $ $\\[5pt]$
$ \Rightarrow 8e^{2} + 12e^{-2} - 8\cos(1) - 16 = 20(1)\sin(1) $ $\\[5pt]$
Which is false.
My error was in not substituting $y$. My corrected solution is as follows.
$ \Rightarrow 4Ae^{2x} + 4Be^{-2x} - 8\cos(x) -  4Ae^{2x} - 4Be^{-2x} + 8cos(x) + 20x\sin(x) = 20x\sin(x) $ $\\[5pt]$
$ \Rightarrow 20x\sin(x) = 20x\sin(x) $ $\\[5pt]$

REPLY [1 votes]: You must substitute $Ae^{2x}+Be^{-2x}-2\cos x-5x\sin x$ instead of $y$ in the equation just before 'We can now verify if this is a solution'.


TITLE: Galois group of a characteristic polynomial
QUESTION [3 upvotes]: Quick question.
I have a 3x3 matrix with integer entries, say $J$. $J$ has rank 2 and determinant equal to zero. According to GAP the matrix has characteristic polynomial $f = X^3-9X^2+8X$. Moreover GAP tells me $J$ has splitting field the "rationals", $\mathbb Q$. How I do compute the Galois group of $f$? Is it non trivial?

REPLY [8 votes]: We have that $f=X^3-9X^2+8X=X(X-1)(X-8)$. So the splitting field of $f$ is $\mathbb{Q}$ itself. So the Galois group of $f$ over $\mathbb{Q}$ is trivial.


TITLE: Silly question: Why is $\sqrt{(9x^2)} $ not $3x$?
QUESTION [4 upvotes]: I had to find the derivative of $f(x) = \sqrt{(9x^2)}$. I applied chain rule with the following steps.
Let $f(x)$ be $\sqrt{x}$ and $g(x)$ be $9x^2$
$$ \begin{align} &f'(g(x)) \times g'(x)  \\
 & = \frac{1}{2\sqrt{(9x^2)}} \times 18x \\
 & = \frac{18x}{2\sqrt{9x^2}} \\
 & = \frac{9x}{3\sqrt{x^2}}\\
 & = \frac{9x}{3\sqrt{x^2}} \\
 & = \frac{3x}{\sqrt{x^2}} \end{align}$$
I got the answer but I don't understand why the last bit doesn't simplify to $3$ because $\sqrt{x^2}$ is $x$ and if it does then why does the back of my textbook and W|A say that it is not?
EDIT:
Okay, so from what I understand, it should be actually $3|x|$. If for example, I had $\sqrt{4x^2}$, I will have $2|x|$, if my understanding is correct.

REPLY [5 votes]: $\sqrt{x^2}=x$ if and only if $x\ge 0$; if $x<0$, then $\sqrt{x^2}=-x$. The correct general formula is $\sqrt{x^2}=|x|$.

REPLY [4 votes]: Hint: We always know that: $$\sqrt{a^2x^2}=|ax|$$

REPLY [4 votes]: Remember that $\sqrt{x^2}$ is the positive number whose square is $x^2$ (this is the definition of the square root sign). If $x$ is positive, then that number is $x$, otherwise, it is $-x$. For example, $\sqrt{(-3)^2} = \sqrt{9} = 3$, not $-3$. You can always say that $\sqrt{x^2} = \left| x \right|$.


TITLE: Number of undirected forests with $n$ vertices
QUESTION [0 upvotes]: $S(n) = $ Number of undirected forests with $n$ vertices.
I need to find $S(n)$ recursively.
The solution is:Remove the tree with the vertex $n$ along with all other vertices in it.
If there are $k$ vertices in the tree then order them by Cayley's formula.The remained vertices are a forest with $n-k$ vertices. Hence:
$$S(n)=\sum_{k=1}^n \binom{n-1}{k-1}k^{k-2}S(n-k)$$
I don't understand where $\binom{n-1}{k-1}$ came from. I would love some more detailed explanation over the whole solution if possible as well.

REPLY [2 votes]: The tree containing vertex $n$ contains $k-1$ other vertices, and they can be any $k-1$ of the $n-1$ vertices $1,2,\ldots,n-1$. Thus, for each $k$ there are $\binom{n-1}{k-1}$ ways to choose the other $k-1$ vertices of the tree containing vertex $n$, $k^{k-2}$ ways to build that tree, and $S(n-k)$ forests on the remaining $n-k$ vertices.


text:
\begin{document}
\maketitle
\begin{abstract}
Let $G$ be a model of Presburger arithmetic. Let $\mathcal{L}$ be an expansion of the language of Presburger $\mathcal{L}_{Pres}$. In this paper we prove that the $\mathcal{L}$-theory of $G$ is $\mathcal{L}_{Pres}$-minimal iff it has the exchange property and any bounded definable set has a maximum.
\end{abstract}

\section{Introduction}
Let $G$ be a model of Presburger arithmetic i.e., $G$ is a group elementary equivalent to $\Z$ (as ordered group i.e. in the language $\mathcal{L}_{Pres}=(+,-,0,1,<,\equiv_n (n\in \N))$ - where $\equiv_n$ is the congruence relation).
We fix $\mathcal{L}$ an expansion of $\mathcal{L}_{Pres}$ and an interpretation of the elements of $\mathcal{L}$ in $G$. We will consider the theory of $G$ as a $\mathcal{L}$-structure.
\begin{definition} We say that $(G,\mathcal{L})$ is $\mathcal{L}_{Pres}$-minimal if for all $X\subset G$ definable, $X$ is $\mathcal{L}_{Pres}$-definable.  $Th(G,\mathcal{L})$ is $\mathcal{L}_{Pres}$-minimal if for all $\mathcal{M}\equiv G$ (as $\mathcal{L}$-structures), $(\mathcal{M},\mathcal{L})$ is $\mathcal{L}_{Pres}$-minimal.
\end{definition}
\begin{Remark}
All definable sets are definable with parameters.
\end{Remark}
\par It is known that $\mathcal{L}_{Pres}$-minimal theories have many good properties. For instance in \cite{Cluckers}, R. Cluckers proves a cell decomposition theorem for models of such theory. It follows that it admits a good notion of dimension and the exchange property.
\begin{definition} Let $\mathcal{M}=(M,\cdots)$ be a $\mathcal{L}$-structure. Let $A\subset M$. Then $dcl(A)$ is the set of $a\in M$ such that there is $\Phi(x,\overline{y})$ a $\mathcal{L}$-formula and $\overline{b}\subset A$ such that $\mathcal{M}\vDash \Phi(a,\overline{b})\wedge \forall y \Phi(y,\overline{b})\rightarrow y=a$.
\end{definition}
\begin{definition} A theory $T$ satisfies EP (exchange property) if for all $\mathcal{M}\vDash T$, for all $A\subset M$, for all $a,b\in M$,
$$a\in dcl(A\cup \{b\})\setminus dcl(A)\mbox{ implies } b\in dcl(A\cup\{a\})$$
\end{definition}
\par If $G=\Z$, then the theory $Th(G,\mathcal{L})$ satisfies EP if and only if it is $\mathcal{L}_{Pres}$-minimal. This follows easily from results of C. Michaux and R. Villemaire \cite{M-V}. In their paper the authors prove that  (1) $Th(\Z,\mathcal{L})$ is $\mathcal{L}_{Pres}$-minimal iff $(\Z,\mathcal{L})$ is $\mathcal{L}_{Pres}$-minimal and (2) that $(\Z,\mathcal{L})$ is $\mathcal{L}_{Pres}$-minimal iff there is no definable expanding set in $\Z$.
\begin{definition} Let $X\subset G$ definable. We say that $X$ is expanding if for all $n\in \N$ there are infinitely many $x\in X$ such that $x+1,\cdots x+n\notin X$.
\end{definition}
It is then not difficult to prove that the exchange property implies that there is no definable expanding set:

\begin{prop}\label{EP implies no expanding set} Let $G\equiv \Z$. If $T=Th(G,\mathcal{L})$ satisfies EP then for all $\mathcal{M}\vDash T$, for all $X\subset M$ definable, $X$ is not expanding.
\end{prop}
\begin{proof}
Let $M$ be a model of $T$ and $X\subset M$ definable. Assume that $X$ is expanding. Then by compactness there is $M^*$ elementary expansion of $M$ and $x<y'\in X(M^*)$ such that $x\notin dcl(\varnothing)$, $x+n\notin X(M^*)$ for all $n\in \N$. Let $y$ be the successor of $x$ in $X(M^*)$. By compactness again there is an elementary extension $M'$ of $M^*$ and $z\in M'$ such that $x<z<y$ and $z\notin dcl(\{x\})$. Then, $x\in dcl(\{z\})$ as it is the only element in $M'$ that satisfies the formula $\Phi(t,z)\equiv t\in X\wedge t<z$.
\par So $x\in dcl(\{z\})\setminus dcl(\varnothing)$ while $z\notin dcl(\{x\})$. This contradicts the assumption that $T$ has the exchange property.
\end{proof}

In general it is not true that EP implies $\mathcal{L}_{Pres}$-minimality. For instance consider $G$ be a nonstandard model of Presburger and the expansion of $\mathcal{L}_{Pres}$ by a unary predicate interpreted in $G$ by a proper convex subgroup. In fact, the existence a proper definable subset closed under successor or predecessor is the only remaining obstruction to $\mathcal{L}_{Pres}$-minimality.

\begin{definition} $(G,\mathcal{L})$ satisfies DC (definable completeness) if any definable unary set has a supremum in $G\cup\{\infty\}$.
\end{definition}
\begin{Remark} DC is a first-order property. So it is also a property of the theory of $G$.
\end{Remark}

For the rest of this paper, we assume that $G\vDash DC$ and that $Th(G,\mathcal{L})$ satisfies EP (so by Proposition \ref{EP implies no expanding set}, there is no definable expanding set). We fix $X$ a $\mathcal{L}$-definable subset of $G$. Under these hypotheses, we shall prove that $X$ is $\mathcal{L}_{Pres}$-definable (Proposition \ref{X Lpres-def}). Then, the main theorem follows:
\begin{theorem} $Th(G,\mathcal{L})$ is $\mathcal{L}_{Pres}$-minimal iff $Th(G,\mathcal{L})$ satisfies EP and DC.
\end{theorem}
\begin{proof}
One direction is done in \cite{Cluckers}:  EP follows for instance from cell-decomposition and DC is an immediate consequence of $\mathcal{L}_{Pres}$-minimality.
The other direction will be proved in Proposition \ref{X Lpres-def}.
\end{proof}
The proof of Proposition \ref{X Lpres-def} has two main steps. First, we prove a non-standard version of Michaux-Villemaire \cite{M-V}. More precisely following their strategy we prove that if $X\subset G$ is definable then $X\cap\N$ is a finite union of points and of cosets of $dG$ (for some integer $d$). Then by DC, we can extend this property to an infinite interval $[0,g]$.
Finally by compactness and EP, we prove that $X$ is a finite union of $X_i$ were $X_i$ is an interval intersected with finitely many cosets of $d_iG$.
\begin{Remark} It is already known that the above theorem fails for generalisation of $\mathcal{L}_{Pres}$-minimality. For coset-minimal groups (in the sense of \cite{P-W}; note that $\mathcal{L}_{Pres}$-minimal groups are coset-minimal), there is an example in \cite{B-P-W} of coset-minimal group which does not have the exchange property.
\end{Remark}

\section{Proof of the main theorem.}

If $X$ is a subset of $M$ where $M\equiv G$ then for all $x\in X$, we denote the successor of $x$ in $X$ by $\sigma(x)$.
\begin{lemma}\label{non expanding set} Let $X$ be a non expanding set. Then there is $u\in \N$ such that for all $x\in X\cap\N$, if $\sigma(x)\in X\cap \N$. Then, $\sigma(x)-x\leq u$.
\end{lemma}
\begin{proof}
For we may assume that $X\cap\N$ is infinite (Otherwise the Lemma is trivial). By contradiction assume that for all $u\in \N$, there is $x\in X\cap\N$ such that $\sigma(x)-x>u$. Apply this assumption with $u=n$ and we get $x_0\in X\cap\N$ such that $\sigma(x_0)-x_0>n$. By induction we can construct $x_0<x_1<\cdots<x_{k}\in X\cap\N$ such that for all $i$, $\sigma(x_i)-x_i>n$. Indeed, we can apply the assumption with $u=\sigma(x_{k-1})+n$ (note that $u\in \N$ as $\sigma(x_{k-1})\in \N$ because $X\cap \N$ is infinite). Then we obtain $x_k$ with $\sigma(x_k)-x_k>u=\sigma(x_{k-1})+n$. In particular $\sigma(x_k)-x_k>n$ and $\sigma(x_k)>\sigma(x_{k-1})$ ie $x_k>x_{k-1}$. Let $U_n=\{x_k;\ k\in \N\}$. Then $U_n$ is an infinite subset of $X$. Therefore $X$ is expanding as for all $x\in U_n$, $x+1,\cdots, x+n\notin X$. Contradiction.
\end{proof}

 Set $G^+=\{x\in G:\ x\geq 0\}$ and $G^{>0}=G^+\setminus \{0\}$. For the rest of this paper, we will assume without loss of generality that $X\subset G^+$.
Let $a\leq b\in G^+$. Let $X[a,b]=\{x\in G^+:\ a+x\in X,\ a+x\leq b\}$.
\begin{lemma}\label{explicit X[a,b]} Let $a\leq b\in G^+$. Let $g\in G^+$. Then $X[a,b]=X[a+g,b+g]$ if and only if for all $a\leq x\leq b$, $x\in X$ iff $x+g\in X$.
\end{lemma}
\begin{proof}
Immediate from the definition.
\end{proof}

\begin{lemma}\label{increase}
For all $a\leq c\leq d\leq b\in G^+$ for all $g\in G^+$, if $X[a,b]=X[a+g,b+g]$ then $X[c,d]=X[c+g,d+g]$.
\end{lemma}
\begin{proof}
By Lemma \ref{explicit X[a,b]}, for all $a\leq x\leq b$, $x\in X$ iff $x+g\in X$. In particular, this is the case for all $c\leq x\leq d$. By Lemma \ref{explicit X[a,b]} again, $X[c,d]=X[c+g,d+g]$.
\end{proof}

\begin{definition}
$$
\begin{array}{rl}
\widetilde{d}:& G^+\rightarrow G^+\cup \{-1\}: \\
& n\longmapsto
\left\{\begin{array}{ll}
\min\{g>0:\ \exists a\in G^+ X[a,a+n]=X[a+g,a+g+n]\} & \mbox{if such $g$ exists;}\\
-1 & \mbox{ otherwise.}
\end{array}\right.
\end{array}
$$
\end{definition}

\begin{lemma} For all $n'<n\in G^+$, if $\widetilde{d}(n)\not=-1$ then $0< \widetilde{d}(n')\leq \widetilde{d}(n)$.
\end{lemma}
\begin{proof}
Let $n'\leq n\in G^+$ with $\widetilde{d}(n)>0$. Then by definition of $\widetilde{d}(n)$ there is  $a\in G^+$ such that $X[a,a+n]=X[a+\widetilde{d}(n),a+n+\widetilde{d}(n)]$.
So by Lemma \ref{increase} $X[a,a+n']=X[a+\widetilde{d}(n),a+n'+\widetilde{d}(n)]$. Therefore by definition of $\widetilde{d}$, $0<\widetilde{d}(n')\leq \widetilde{d}(n)$.
\end{proof}

\begin{lemma}\label{function d} $\widetilde{d}(\N)\subset \N$.
\end{lemma}
\begin{proof}
First we remark that for all $k,n\in \N$, $X[k,k+n]\subset \{0,\cdots n\}$. Therefore by the Pigeonhole principle for all $n\in \N$, there is $k<l\in \N$ such that $X[k,k+n]=X[l,l+n]$. Now by definition of $\widetilde{d}$, $0<\widetilde{d}(n)\leq l-k\in \N$.
\end{proof}

\begin{definition}
$$
\begin{array}{rl}
\widetilde{a}:&G^+\times G^{>0} \rightarrow G^+\cup \{-1\}:\\
& (n,d)\longmapsto
\left\{\begin{array}{ll}
\min\{a\in G^+:\  X[a,a+n]=X[a+d,a+n+d]\} & \mbox{if such $a$ exists;}\\
-1 & \mbox{ otherwise.}
\end{array}\right.
\end{array}
$$
\end{definition}

\begin{lemma}\label{function a} For all $n\in \N$, there is $d\in \N_0$ such that $\widetilde{a}(n,d)\in \N$.
\end{lemma}
\begin{proof}
By the Pigeonhole principle for all $n\in \N$, there is $k<l\in \N$ such that $X[k,k+n]=X[l,l+n]$. Then by definition of $\widetilde{a}$, $0\leq \widetilde{a}(n,l-k)\leq k\in \N$. Take $d=l-k$.
\end{proof}

The functions $\widetilde{a},\widetilde{d}$ come from Michaux-Villemaire \cite{M-V}. The authors prove that if $G=\Z$, $\alpha(n):=\widetilde{a}(n,\widetilde{d}(n))\in \N$. In fact, it is proved that for $n$ large enough, $X\cap [\alpha(n),\infty)$ is defined by congruences relations modulo $d(n)$. In our case this is not true anymore as $\alpha(n)$ may not be in $\N$. For instance, take $X= (3G\cap [0, g]) \cup (2G\cap [g+1,\infty)$ for some $g\in G\setminus \N$. Then for $n\in \N$, $\widetilde{d}(n)=2$ and $\alpha(n)=g+1$. What we would like to capture is $\widetilde{d}(n)=3$ and $\alpha(n)=0$. For we defined the below function $D$ which is a twisted version of $\widetilde{d}$. With this new function we will get that $D(n)=2$ and $A(n):=\widetilde{a}(n,D(k))=0$ for all $k,n\in \N$ large enough as required.

\begin{definition}
$$
\begin{array}{rl}
D:&G^+ \rightarrow G^+\cup \{-1\}:\\
& n \longmapsto
\left\{\begin{array}{ll}
\min\{d\in G^{>0}:\ \widetilde{a}(n,d)\not=-1\wedge\\
 \forall d'>d\ \widetilde{a}(n,d')\not=-1\rightarrow [\widetilde{a}(n,d)\leq \widetilde{a}(n,d') \vee \widetilde{a}(n,d')+d'>\widetilde{a}(n,d)+n]\}&\mbox{ if such $d$ exists}\\
-1 & \mbox{ otherwise.}
\end{array}\right.
\end{array}
$$
Let $A(n):=\widetilde{a}(n,D(n))$.
\end{definition}

\begin{lemma}\label{function D A} $D(\N)\subset \N$ and $A(\N)\subset \N$.
\end{lemma}
\begin{proof}
By Lemma \ref{function a}, for all $n\in \N$ there is $d\in \N_0$ such that $\widetilde{a}(n,d)\in \N$. Let $B_n:=\{d'\in \N_0:\ \widetilde{a}(n,d')\in \N\mbox{ and } \forall d''\in \N_0, \widetilde{a}(n,d'')\in \N\rightarrow \widetilde{a}(n,d')\leq \widetilde{a}(n,d'')\}$. This set is nonempty. For there is $d'$ such that $\widetilde{a}(n,d')$ is minimal in $\widetilde{a}(n,\N) \setminus\{-1\} $. Then as $\widetilde{a}(n,d)\in \N$,  $\widetilde{a}(n,d')\in \N$. By definition, $d'\in B_n$.
\par Let $d^*=\min\{d'\in B_n\}\in \N_0$. Set
$$E_n:=\{d\in G^{>0}: \widetilde{a}(n,d)\not=-1\wedge \forall d'>d\ \widetilde{a}(n,d')\not=-1\rightarrow [\widetilde{a}(n,d)\leq \widetilde{a}(n,d') \vee \widetilde{a}(n,d')+d'>\widetilde{a}(n,d)+n]\}.$$
Let $d'>d^*$ such that $\widetilde{a}(n,d')\not=-1$. First if $d'\in \N$ then $\widetilde{a}(n,d^*)\leq \widetilde{a}(n,d')$ (for either $\widetilde{a}(n,d')\notin \N$ and it is trivial as $\widetilde{a}(n,d^*)\in \N$ or $\widetilde{a}(n,d')\in \N$ and this follows from the definition of $B_n$). Second if  $d'\notin\N$ then $\widetilde{a}(n,d')+d'>k$ for all $k\in \N$. So $\widetilde{a}(n,d^*)+n<\widetilde{a}(n,d')+d'$. This proves that $d^*\in E_n$.
This implies that $D(n)\leq d^*$ by definition of $D$. So $D(\N)\in \N$.
\par Let $d\in \N$ such that $d=D(n)$. Assume that $A(n)=\widetilde{a}(n,d)\notin \N$. As $\widetilde{a}(n,d^*)\in \N$, $\widetilde{a}(n,d)>\widetilde{a}(n,d^*)$. As $D(n)\leq d^*$ we get that $d<d^*$. On the other hand, as $\widetilde{a}(n,d^*)+d^*\in \N$ and $\widetilde{a}(n,d)\notin \N$, we have that $\widetilde{a}(n,d)+n>\widetilde{a}(n,d^*)+d^*$. Now by definition of $D(n)$, $d\not=D(n)$. We get a contradiction. Therefore $A(n)\in \N$.
\end{proof}

\begin{definition} Let $U\subset V\subset G$. We say that $U$ is cofinal in $V$ if for all $v\in V$ there is $u\in U$ such that $v\leq u$.
\end{definition}

\begin{lemma}\label{f nondecreasing}
Let $f:U\rightarrow G^+$ definable such that $U_0:=U\cap \N$ is cofinal in $\N$ and $f(U_0)\subset \N$ then there is $U'\subset U$ definable such that $U'\cap \N$ is cofinal in $\N$ and $f$ is nondecreasing on $U'$.
\end{lemma}
\begin{proof}

\par First, if $f(U_0)$ is finite: Then by the Pigeonhole principle there is $s\in Im\ f\cap U_0$ such that $f^{-1}(s)\cap U_0\subset G^+$ is infinite. Set $U'= f^{-1}(s)\cap U$. By definition of $U'$, $f$ is constant on it. Also $U'\cap \N$ is cofinal in $\N$ as it is infinite. Clearly $U'$ is definable.

\par Second, if $f(U_0)$ is infinite. In that case, set $U':=\{x\in U: \forall y\in U y<x\rightarrow f(y)< f(x)\}$. By definition $f$ is nondecreasing of $U'$ and $U'$ is definable. It remains to prove that $U_0':= U'\cap \N$ is an infinite set. We remark that $U_0'$ is non empty as $\min\{n:\ n\in U_0\}\in U_0'$. Let $y\in U_0'$. Assume that for all $t\in U_0$ with $t>y$, $f(t)\leq f(y)$. So, $f(U_0)= [0,f(y)]\subset\N$ is a finite set. This contradicts the assumption that $f(U_0)$ is an infinite subset of $\N$. So, there is $t>y$, $t\in U_0$ such that $f(y)<f(t)$. Let $t^*=\min\{t>y:\ t\in U,\ f(y)<f(t)\}$. As $t^*\leq t$, $t^*\in \N$. Let $z<t^*$. If $y\leq z<t^*$ then $f(z)\leq f(y)$ and $f(y)<f(t^*)$ by definition of $t^*$. If $z<y$ then $f(z)< f(y)$ (by definition of $U'$) and $f(y)<f(t)$ (by definition of $t^*$). So, by definition of $U'$, $t^*\in U'$. Therefore $t^*\in U_0'$. This proves that $U_0'$ is infinite and concludes the proof of the lemma.
\end{proof}

\begin{lemma}\label{D A nondecreasing}
There is $\widetilde{U}\subset G^+$ definable such that $\widetilde{U}\cap \N$ is cofinal in $\N$ and $D,A$ are nondecreasing on $\widetilde{U}$.
\end{lemma}
\begin{proof}
By Lemma \ref{function D A} and Lemma \ref{f nondecreasing}.
\end{proof}

\begin{lemma}\label{DA definable} $D,A: \widetilde{U}\rightarrow G^+$ are definable maps.
\end{lemma}
\begin{proof}
By definition, $D,A: G^+\rightarrow G^+$ are definable maps. By Lemma \ref{D A nondecreasing}, $\widetilde{U}$ is definable.
\end{proof}

\begin{lemma}\label{d constant}
Assume that $\widetilde{U}$, $Im\ D$ and $Im\ A$ are nonexpanding. Then, there is $d\in \N_0$, such that for all $n\in \N$, there is $a\in \N$ such that $X[a,a+n]=X[a+d,a+d+n]$.
\end{lemma}

\begin{proof}
Let us remark that it  is sufficient to prove that there is $d\in \N_0$, such that for all $n\in \N$, there is $a,m\in \N$ such that $m\geq n$ and $X[a,a+m]=X[a+d,a+d+m]$.	For in that case, by Lemma \ref{increase}, $X[a,a+n]=X[a+d,a+d+n]$.
\par By Lemma \ref{non expanding set} and $D$ is nondecreasing, there is $u\in \N$ such that for all $n\in \widetilde{U}_0:=\widetilde{U}\cap \N$, $0\leq D(\sigma(n))-D(n)\leq u$. Also there is $v\in \N$ such that for all $n\in \widetilde{U}_0$, $0\leq A(\sigma(n))-A(n)\leq v$.
\begin{claim}
There are $V,N\in \N$ such that  for all  $n\in \N$,  $n\geq N$, there are $d(n)\in \{0,\cdots, V\}$ and  $a\in \N$ with  $X[a,a+n]=X[a+d(n),a+d(n)+n]$.
\end{claim}
\begin{proof}
	First if $(D(n), n\in\widetilde{U}_0)$ is eventually constant, take $V'=\lim D(n)$, $N=\min\{n: \forall m\geq n D(n)=D(m)\}$, $d(n)=V'$
		and $a=A(n,D(n))$.	By definitions of $A,D,d$, we are done.
	\par Otherwise, $\forall n'\in \widetilde{U}_0\ \exists k'\in\widetilde{U}_0,\ n'<k'$ and $D(n')<D(k')$ (as $D$ is nondecreasing).
	Apply this with $n'=n+v$. Then for all $n$ there is $k'$ such that $D(k')>D(n+v)$. Take $k=\max\{k^*\in\widetilde{U}_0: D(k^*)=D(n+v)\}$. Then $D(\sigma(k))>D(n+v)=D(k)$. We remark that by definition of $A,D$
	$$ (1) X[A(k), A(k)+k]=X[A(k)+D(k),A(k)+D(k)+k] \mbox{ and, }$$
	$$(2) X[A(\sigma(k)), A(\sigma(k))+\sigma(k)]=X[A(\sigma(k))+D(\sigma(k)),A(\sigma(k))+D(\sigma(k))+\sigma(k)].$$

\par We have that $A(k)+k-A(\sigma(k))=k-(A(\sigma)(k)-A(k))\geq n+v-v=n$. So,
 $$A(k)\leq A(\sigma(k))\leq A(k)+k\qquad \mbox{ and }\qquad A(\sigma(k))\leq A(k)+k\leq A(\sigma(k))+\sigma(k).$$
 Therefore by Lemma \ref{increase}, (1) and (2)
	$$ (1') X[A(\sigma(k)), A(k)+k]=X[A(\sigma(k))+D(k),A(k)+D(k)+k] \mbox{, and }$$
	$$(2') X[A(\sigma(k)), A(k)+k]=X[A(\sigma(k))+D(\sigma(k)),A(k)+D(\sigma(k))+k].$$
We combine $(1')$ and $(2')$ to get
$$X[A(\sigma(k))+D(k),A(k)+D(k)+k]=X[A(\sigma(k))+D(\sigma(k)),A(k)+D(\sigma(k))+k].$$
Take $a=A(\sigma(k))+D(k)$ and $d=d(k)= D(\sigma(k))-D(k)$. We remark that $a\in \N$, that $d\leq u$ (by definition of $u$) and that $X[a,a+n]=X[a+d,a+d+n]$ (for remark that $a\leq a+n\leq A(k)+D(k)+k$ and apply Lemma \ref{increase} and the above equality). Set $V=u$ and $N=0$.
\end{proof}
By the claim and the Pigeonhole principle there is $d\in \{1,\cdots, V\}$ such that $E=\{n:\ d(n)=d\}$ is cofinal in $\N$. Then, for all $n\in \N$ there is $m\in E$ such that $m\geq n$. So, by the above claim, there is $a$ such that $X[a,a+m]=X[a,a+d+m]$. This shows that there is $d\in \{1,\cdots, V\}$ such that for all $n$ there is $a,m\in \N$ such that $m\geq n$ and $X[a,a+m]=X[a+d,a+d+m]$. By the remark at the beginning of the proof we are done.
\end{proof}

\begin{definition}\label{alpha}Let $d$ given by Lemma \ref{d constant}. We define
$$
\begin{array}{rl}
\alpha:& \widetilde{U}\rightarrow G^+\cup \{-1\}: \\
& n\longmapsto
\left\{\begin{array}{ll}
\min\{a\geq 0:\ X[a,a+n]=X[a+d,a+n+d]\} & \mbox{if there is at least one such $a$;}\\
-1 & \mbox{ otherwise.}
\end{array}\right.
\end{array}
$$
\end{definition}

By Lemma \ref{d constant}, for all $n\in \widetilde{U}\cap \N$, $\alpha(n)\in \N$. Note that $\alpha$ is definable.

\begin{lemma}\label{alpha nondecreasing} There is $U\subset \widetilde{U}$ definable such that $U_0:= U\cap\N$ is cofinal in $\N$ and $\alpha$ is non decreasing on $U$.
\end{lemma}
\begin{proof}
By Lemma \ref{f nondecreasing}.
\end{proof}

From now, we will assume that $\alpha$ is restricted to $U$. We set $U_0:=U\cap \N$.

\begin{lemma}\label{loc definable} If $Im\ \alpha$, $\widetilde{U}$, $Im\ D$ and $Im\ A$ are nonexpanding, then there is $g\in G^+\cup\{\infty\}\setminus \N$ and $N\in \N$ such that $X\cap [N,g]$  is a finite union of coset of $dG$ intersected with $[N,g]$ (so is $\mathcal{L}_{Pres}$-definable).
\end{lemma}
\begin{proof}
By Lemma \ref{non expanding set}, there is $v$ such that for all $n$, $0\leq \alpha(\sigma(n))-\alpha(n)\leq v$. So for all $n\geq l:=\max\{d,v\}$, $\alpha(\sigma(n))\leq \alpha(n)+v\leq \alpha(n)+n$.
Set $N= \alpha(l)$.
\par We have that for all $k\in \N$, $k\geq N$, there is $n\in U_0$, $n\geq l$ such that $k\in [\alpha(n),\alpha(n)+n]$. For there are two cases: first if there is $n\in U_0$ such that $\alpha(n)\leq k\leq\alpha(\sigma(n))$. In that case $k\in [\alpha(n), \alpha(\sigma(n))]\subseteq [\alpha(n),\alpha(n)+n]$ and we are done. Second if for all $n\in U_0$ $\alpha(n)<k$. In that case by Lemma \ref{alpha nondecreasing}, $\alpha$ is non decreasing. Furthermore, $\alpha(U_0)\subset \N$ (by Lemma \ref{d constant}). So as it is bounded by $k$, $\alpha$ is eventually constant in $U_0$ i.e., there is $M\in \N$ such that for all $n\in U_0$, $n>M$, $\alpha(n)=\alpha(\sigma(n))$. Let $n\in U_0$ such that $n>\max\{M, k\}$ (such $n$ exists as $U_0$ is cofinal in $\N$ see Lemma \ref{alpha nondecreasing}). Then, $\alpha(n)=\alpha(M)<k<\alpha(M)+n=\alpha(n)+n$.
\par Let $x\in [N,\infty)\cap \N$. By the above argument we know that there is $n$ such that $x\in [\alpha(n),\alpha(n)+n]$. So by definition of $\alpha$, $X[\alpha(n),\alpha(n)+n]=X[\alpha(n)+d,\alpha(n)+n+d]$. By Lemma \ref{explicit X[a,b]}, this implies that $x\in X$ iff $x+d\in X$. Therefore $X\cap [N,\infty)\cap \N= (X\cap [\alpha(l),\alpha(l)+d-1])+d\N$. Take $\{a_1,\cdots, a_k\}$ be the set $X\cap [\alpha(l),\alpha(l)+d-1]$. Then $X\cap [N,\infty)\cap \N=\bigcup_i (a_i+dG)\cap [N,\infty)\cap \N$.
\par Let $Y=\{x\in G^+: x\geq N \wedge \forall y\leq x,\ N\leq y\rightarrow  (y\in X\rightleftarrow \vee_i y\equiv_d a_i)\}$. This is a $\mathcal{L}$-definable set. Therefore by DC there $g\in G^+\cup\{\infty\}$ such that $g\in Y$ and either $g=\infty$ or $g+1\notin Y$. So we get that $X\cap [N,g)=\bigcup_i (a_i+dG)\cap [N,g)$ and $g$ is maximal for this property. As $X\cap [N,\infty)\cap \N=\bigcup_i (a_i+dG)\cap [N,\infty)\cap \N$, $g\notin \N$.

\end{proof}

We can now prove the generalisation of the result of Michaux-Villemaire.
\begin{theorem} Let $G\equiv \Z$ and $\mathcal{L}\supseteq \mathcal{L}_{Pres}$. Assume that $Th(G,\mathcal{L})$ admits EP and DC. Then for all $X\subset G$ definable. $X\cap \N$ is a finite union of points and of cosets of $d\N$ for some $d\in \N$.
\end{theorem}
\begin{proof}
	By Proposition \ref{EP implies no expanding set}, $Im\ \alpha$, $\widetilde{U}$, $Im\ D$ and $Im\ A$ are nonexpanding. So the result is an immediate consequence of Lemma \ref{loc definable}.
\end{proof}

This property is also true for $X\cap x+\N$ for all $x\in G^+$.

\begin{prop}\label{relative local definability} For all $x\in X$ there is $N(x)\in \N$ minimal, $d(x)\in \N_0$ and $g(x)\in G^+\cup\{\infty\}\setminus \N$ maximal such that $[x+N(x), x+g(x)]\cap X$ is a finite union of classes of $d(x)G$ intersected with $[x+N(x), x+g(x)]$.
\end{prop}
\begin{proof}By Lemma \ref{loc definable} applied with the set $X_x=\{y-x: y\in X\}$. For by Proposition \ref{EP implies no expanding set}, $Im\ \alpha$, $\widetilde{U}$, $Im\ D$ and $Im\ A$ are non expanding.
\end{proof}
\begin{Remark} We do not claim nor need that $N(x),d(x)$ or $g(x)$ are definable functions.
\end{Remark}

\begin{definition}
$$
\begin{array}{rl}
M:&G^+\times G^+ \rightarrow G^+\cup \{\infty,-1\}:\\
& (x,n)\longmapsto
\left\{\begin{array}{ll}
\max\{h\geq x+n:\ X[x,h-n]=X[x+n,h]\} & \mbox{if such $h$ exists in $G^+\cup \{\infty\}$;}\\
-1 & \mbox{ otherwise.}
\end{array}\right.
\end{array}
$$
\end{definition}

\begin{lemma}\label{bound on classes} Let $x\in X$ and $N=N(x),d=d(x),g=g(x)$ as given in Proposition \ref{relative local definability}. Then for all $x+N\leq y\leq x+N+\N$, if $n$ is finite and $d$ divides $n$ then $M(y,n)=x+g$. If $x\leq t<x+N(x)$ then $M(t,n)<x+N+n$.  Furthermore, for all $n$ finite, $M(x,n)\leq x+g$.
\end{lemma}
\begin{proof}
\par 1) First we prove that $M(y,n)=x+g$: By Proposition \ref{relative local definability} and definition of $X[x+N,x+g]$, $X[x+N,x+g]=\bigcup_{i<s}b_i+dG\cap [0,g-x]$ where $b_i$ is a representative for the classe of $a_i-x$. As $y\in x+N+\N$, there is $r\in \N$ maximal such that $x+N+rd\leq y$. So,
$$\begin{array}{rll}
		[y,y+(d-1)]\cap X	&= & [x+N+rd+j,x+N+rd+(d-1)]\\
											&  & \bigcup[x+N+(r+1)d, x+N+(r+1)d+(j-1)]\cap X \mbox{ for some $j<d$}\\
											&= & \{a'_i+dr_i:\ i\leq s\}\\
											& & \mbox{ where  $r_i$ is either $r$ or $r+1$ and $a'_i\equiv_d b_i\equiv_d a_i$}
	\end{array}
$$
Therefore by the above description of $X[x,g]$, $[y,x+g]\cap X=(\{a'_i+dr_i:\ i<s\}+dG^+)\cap [y,x+g]$. So we remark that by definition of $X[y,x+g-n]$ (resp. $X[y+n, x+g]$)
$$X[y,x+g-n]=(X[y,y+(d-1)]+dG^+)\cap [0,x+g-n-y]\mbox{ and }$$
$$X[y+n,x+g]=(X[y+n,y+n+(d-1)]+dG^+)\cap [0,x+g-n-y].$$
Also by definition of $X[y,y+(d-1)]$, $t\in X[y,y+(d-1)]$ iff $y+t=a_i+dr_i$ for some $i$. This means that $X[y,y+(d-1)]=\{a'_i+dr_i-y,\ i<s\}$. Similarly $X[y+n,y+n+(d-1)]=\{a'_i+dr_i-y,\ i<s\}$.
This implies that $X[y,x+g-n]=X[y+n,x+g]$ i.e., $M(y,n)\geq x+g$. If $g=\infty$ we are done by definition of $M$. Otherwise assume that $M(y,n)>x+g$ i.e., assume that there is $g'>g$ $X[y,x+g'-n]=X[y+n,x+g']$.  By Lemma \ref{explicit X[a,b]}, $x+g+1-n\in X$ iff $x+g+1\in X$. So (as $d$ divides $n$) $[N+x,x+g+1]\cap X=\bigcup_{i<s}a_i+ dG\cap [N+x,x+g+1]$. This contradicts the maximality of $g$ in Proposition \ref{relative local definability}. So $M(y,n)=x+g$.
\par 2) $M(t,n)<x+N+n$: Assume that  $h:=M(t,n)\geq x+N+n$. By definition of $M(t,n)$, $X[t,h-n]=X[t+n,h]$. By Lemma \ref{increase} $X[x+N-1,h-n]=X[x+N(x)-1+n,h]$. So by Lemma \ref{explicit X[a,b]}, $x+N-1\in X$ iff $x+N-1+n\in X$. Then by Proposition \ref{relative local definability} $[x+N-1,x+g(x)]\cap X=\bigcup_i (a_i+dG)\cap[x+N-1,x+g(x)]$. This equality contradicts the minimality of $N$ in Proposition \ref{relative local definability}.
\par 3) $M(x,n)\leq x+g$: Assume $h=M(x,n)>x+g$. By definition of $M(x,n)$, $X[x,h-n]=X[x+n,h]$. So by Lemma \ref{increase} $X[x+N,h-n]=X[x+N+n,h]$. By definition of $M(x+N,n)$ we get that $M(x+N,n)\geq h$. On the other hand by Lemma \ref{increase} again $X[x+N,x+g-n]=X[x+N+n,x+g]$. Then by Lemma \ref{explicit X[a,b]}, for all $x+N\leq y\leq x+g-n$, $y\in X$ iff $y+n\in X$. By Proposition \ref{relative local definability} this proves that $d$ divides $n$. Now as $M(x+N,n)\geq h>x+g$, we get a contradiction with step 1). So, $h=M(x,n)\leq x+g$.
\end{proof}

\begin{prop}\label{X Lpres-def}
$X$ is $\mathcal{L}_{Pres}$-definable.
\end{prop}
\begin{proof}
By Proposition \ref{relative local definability}, for all $x\in X$, there is $N(x), d(x)\in \N$ and $g(x)\in G^+\setminus \N$ maximal such that $X\cap [x+N(x),x+g(x)]$ is a union of classes of $d(x)G$ restricted to $[x+N(x),x+g(x)]$. So if there is $x_1,\cdots, x_k\in G^+$ such that $X\subseteq \bigcup_i [x_i,x+g(x_i)]$ we are done. Therefore for a contradiction we assume that for all $k\in \N$, for all $x_1,\cdots, x_k$ $X\not= \bigcup_i [x_i, x_i+g(x_i)]\cap X$.
\par For all $n\in\N$, there is $x_n$ such that for all $k\geq d(x_n)$, $M(x_n+N(x_n),k!)=x_n+g(x_n)$ and furthermore, if $n\not= n'$ there is $N$ such that for all $k\geq N$, $M(x_n+N(x_n),k!)\not=M(x_{n'}+N(x_{n'}),k!).$ Indeed, we can construct the $x_n$'s by induction: Take $x_0=0$. Then, by Lemma \ref{bound on classes} for all $k\geq d(0)$, $M(0+N(0),k!)=g(0)$. For all $n$, set $x_{n}:=x_{n-1}+g(x_{n-1})+1$. First remark that we may assume $x_n<\infty$. Otherwise, $x_{n-1}+g(x_{n-1})=\infty$. So, $G^+=\bigcup_{i<n}[x_n,x_n+g(x_n)]$. This contradicts our hypothesis. Now by Lemma \ref{bound on classes}, for all $k\geq d(x_n)$, $M(x_n+N(x_n),k!)=x_n+g(x_n)$. Also for all $k\geq \max\{d(x_i),\ i\leq n\}$,  $M(x_n+N(x_n),k!)=x_n+g(x_n)>x_n=x_{n-1}+g(x_{n-1})+1>x_i+g(x_i)=M(x_i+N(x_i),k!)$.
\begin{claim}For all $k\geq d(x_i)$, $M(x_i+N(x_i),k!)$ is not at finite distance from its predecessor in $Im\ M(\cdot, k!)$ (if any).
\end{claim}
Assume that the claim is true. Then we will build an elementary expansion of $G$ such that $Im\ M(\cdot, T)$ is an expanding set in this model (for some $T$). Then we get a contradiction with the exchange property (by Proposition \ref{EP implies no expanding set}) and the proposition is proved.
\par Let $G^*$ be an ultrapower of $G$ on a nonprincipal ultrafilter over $\N$. Let $T$ be the class of $(n!)_{(n\in \N)}$ and $y_i$ be the class of $(x_i+N(x_i))_{(n\in \N)}$ (the constant sequence). By construction of $x_i$ and \L os Theorem, if $i\not=j$, $M(y_i,T)\not= M(y_j,T)$. By the claim and \L os Theorem, for all $i,n\in \N$, $M(y_i,T)-n\notin Im\ M(\cdot, T)$ i.e., for all $i$, $M(y_i,T)$ is not at finite distance form its predecessor in $Im\ M(\cdot, T)$. This proves that $\{M(y_i,T),\ i\in \N\}$ is an infinite subset of $Im\ M(\cdot, T)$ and that each point in this set is not at finite distance from its predecessor in $Im\ M(\cdot, T)$. So, $Im\ M(\cdot, T)$ is an expanding set.

\par We give now a proof of the claim: By Lemma \ref{bound on classes} as $d(x_i)$ divides $k!$, $M(x_i+N(x_i),k!)=x_i+g(x_i)\not=-1$.  Let $t$ such that $M(t,k!)< M(x_i+N(x_i),k!)$. We have to prove that the distance between these two elements is not finite. There are three possible cases:
\par 1) if $x_i+N(x_i)<t$: in that case we may assume that $t< M(x_i+N(x_i), k!)-k!$. For if $t\geq M(x_i+N(x_i), k!)-k!$ then $M(t,k!)\geq t+k!\geq M(x_i+N(x_i), k!)+k!-k!$: contradiction with the choice of $t$. By definition of $M$, $X[x_i+N(x_i),M(x_i+N(x_i),k!)-k!]=X[x_i+N(x_i)+k!,M(x_i+N(x_i),k!)]$. So by Lemma \ref{increase}, $X[t,M(x_i+N(x_i),k!)-k!]=X[t+k!,M(x_i+N(x_i),k!)]$. This proves that $M(t,k!)\geq M(x_i+N(x_i), k!)$: contradiction. So case 1) never occurs.
\par 2) $x_i\leq t<x_i+N(x_i)$: Then by Lemma \ref{bound on classes}, $M(t,k!)<x_i+N(x_i)+k!$. As $M(x_i+N(x_i),k!)=x_i+g(x_i)$, $N(x_i),k!\in \N$ and $g(x_i)\notin \N$, we are done.
\par 3) $t<x_i$: Assume that $M(t,k!)$ is a finite distance from $M(x_i,k!)$. By Lemma \ref{bound on classes}, $M(t+N(t),k!)\leq t+g(t)$ where $N(t), g(t)$ are given by Proposition \ref{relative local definability}. Then $[t+N(t), t+g(t)]\cap [x_i+N(x_i), x+g(x_i)]$ is an infinite interval. Let $r=\max\{x_i+N(x_i), t+N(t)\}$, $s=\min\{x_i+g(x_i), t+g(t)\}$. By Proposition \ref{relative local definability},  $X\cap [r,s]=\bigcup_i a_i+d(x_i)G\cap [r,s]$ for some $0\leq a_i<d(x_i)$ and $X\cap [r,s]=\bigcup_i b_i+d(t)G\cap [r,s]$ for some $0\leq b_i<d(t)$. So we may replace $d(t)$ by $e=gcd(d(x_i),d(t))$. Also, $t+g(t)=x+g(x)$ (by maximality of $g(x),g(t)$). Now $d(t)$ divides $k!$. Then by Lemma \ref{bound on classes} $M(t+N(t),k!)=t+g(t)=x+g(x)=M(x_i+N(x_i),k!)$: contradiction.

\end{proof}

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\noindent Nathana\"el Mariaule\\
Universit\'e de Mons, Belgium\\
\emph{E-mail address: Nathanael.MARIAULE@umons.ac.be}
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