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Prove that a set is closed iff it contains all its accumulation points.

I have no clue on how to approach the above problem. At first I would appreciate hints on how to get started in either direction $\Leftarrow$ or $\Rightarrow$.

This is how we have def. an accumulation point:

A point $x \in X$ is called an accumulation point of $A \subset X$ if for every $U \in U_x$ there is $y$ s.t $y \not = x$ with $y \in U \cap A$. Where $U_x =\{ U \in X: U \text{ neighbour of } x \in X\}$

We have def. a closed set to be the complement of an open set.

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  • $\begingroup$ there is no y? or there is at least one? $\endgroup$
    – Doug M
    Commented Oct 26, 2016 at 22:56
  • $\begingroup$ There is no $y$ such that $y \not =x$, i.e for every $y$ in the intersection between $U$ and $A$, $y$ must be equal to $x$, if $x$ is an accumulation point @DougM $\endgroup$
    – Olba12
    Commented Oct 26, 2016 at 23:09
  • $\begingroup$ Check your definition. You have it nearly backwards x is an accumulation point in A, if every neighborhood of x contains a point (other than x) in A. $\endgroup$
    – Doug M
    Commented Oct 26, 2016 at 23:23
  • $\begingroup$ I dont understand, do you mean that my comment or def. In the post is wrong? @DougM $\endgroup$
    – Olba12
    Commented Oct 26, 2016 at 23:51
  • $\begingroup$ your definition is wrong. It should say "there is" and not "there is no" $\endgroup$
    – Doug M
    Commented Oct 26, 2016 at 23:54

2 Answers 2

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$A\subset X$ is closed $\implies A$ contains all its accumulation points.

Proof by contradiction:

$A\subset X$ is closed $\implies X\setminus A$ is open.

$X\setminus A$ is open $\implies \forall x \in X\setminus A, \exists U_x$ such that $\forall y\in U_x\implies y\in X\setminus A$

Suppose $x$ is an accumulation of $A$ that is not in $A$

$\forall U\in U_x, \exists y\ne x$ with $y \in U\cap A$

$y \in U\cap A \implies y\notin X\setminus A$ -- Contradiction.

$A\subset X$ contains all of its accumulation points $\implies A$ is closed

The other way -- also by contradiction:

Suppose $A\subset X$ contains all of its accumulation points.

Suppose $X\setminus A$ is not open.

There exists an $x \in X\setminus A$ such that $\forall U\in U_x$ there exists $y \in U$ that is also in $A.$

$x$ is an accumulation point, which contradicts the premise.

${\rm QED}$

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  • $\begingroup$ How do you define $X-A$? $\endgroup$
    – Olba12
    Commented Oct 26, 2016 at 23:23
  • $\begingroup$ The compliment of A. I don't know how to make a "\" in latex. $\endgroup$
    – Doug M
    Commented Oct 26, 2016 at 23:25
  • $\begingroup$ @DougM It should be "\backslash". (If you still don't know after three years.) $\endgroup$
    – justadzr
    Commented Nov 22, 2019 at 12:29
  • $\begingroup$ @DougM On the second to last line, didn't you mean "$\forall U_x$ there exists $y$..." (and not "$\forall U \in U_x$...")? Seems to me that $U_x$ is your notation for neighbourhood of x. $\endgroup$ Commented Apr 28, 2021 at 2:44
  • $\begingroup$ Also, it's not very hard to adapt the proofs by contradiction to proofs by conttrapositive, cf. here math.stackexchange.com/questions/162018/… $\endgroup$ Commented Apr 28, 2021 at 2:47
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I will provide a more general proof for a set $S \subset \mathbb{R}^n $. That is, we prove that

$\mathbf{Statement:}$ if $S \subset \mathbb{R}^n, S$ is closed iff it contains all its limit (a.k.a. accumulation) points.

$\mathbf{Proof:}$

($\Rightarrow $) Consider that $S$ is closed, that is $S \cup \delta S = S $; consider a limit point $x$ in $S$. THere exists a sequence $\{x_n\} \in S $ such that its limit point is $x$. Suppose by contradiction that $x \notin S$; therefore for any open ball $B_r(x)$ and every $n>N$ with $N$ sufficiently large, $ x_n \in B_r(x) \cap S \neq \emptyset$ and $B_r(x) \cap S^c \neq \emptyset$; so that $x \in \delta S$ but $x \notin S$ which is a contradiction.

($\Leftarrow $) Assume now that the set of all limit points in contained in $S$ and prove that $S=\delta S \cup S$. Take $x \in \delta S$; therefore $\forall\ n \in \mathbb{N}, \exists\ x_n $ such that $x_n \in B_\frac{1}{n} \cap S$. Consequently, either $x$ is a limit point for $S$ or $x \in S$.

This completes the proof.
$Q.E.D.$

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    $\begingroup$ This is actually less general because OP was asking about general topological spaces. Your proof only works in metric spaces :( $\endgroup$
    – Levi
    Commented Nov 4, 2019 at 11:09

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