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

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.

• there is no y? or there is at least one? Commented Oct 26, 2016 at 22:56
• 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 Commented Oct 26, 2016 at 23:09
• 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. Commented Oct 26, 2016 at 23:23
• I dont understand, do you mean that my comment or def. In the post is wrong? @DougM Commented Oct 26, 2016 at 23:51
• your definition is wrong. It should say "there is" and not "there is no" Commented Oct 26, 2016 at 23:54

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

$$A\subset X$$ 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.

$${\rm QED}$$

• How do you define $X-A$? Commented Oct 26, 2016 at 23:23
• The compliment of A. I don't know how to make a "\" in latex. Commented Oct 26, 2016 at 23:25
• @DougM It should be "\backslash". (If you still don't know after three years.) Commented Nov 22, 2019 at 12:29
• @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. Commented Apr 28, 2021 at 2:44
• Also, it's not very hard to adapt the proofs by contradiction to proofs by conttrapositive, cf. here math.stackexchange.com/questions/162018/… Commented Apr 28, 2021 at 2:47

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.
$$Q.E.D.$$

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