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Let X be a topological space such that for every $x \in X$, the set $\{x\} \subset X$ is closed.

I have to prove that every neighborhood of $x_{0} \in X$ contains infinitely many different points from A $\subset$ X iff $x_{0} \in X$ is a limit point of A.

$(\Rightarrow)$ Assume that every neighborhood of $x_{0} \in X$ contains infinitely many different points from A $\subset$ X. I think I have to pick a row in X. I think I can then say that $x_{0}$ is an accumulation point of X. But does this imply that $x_{0}$ is a limit point of A?

$(\Leftarrow)$ Assume that $x_{0} \in X$ is a limit point of A. Then every neighborhood U of $x_{0}$ contains at least one point $x \in A$ with $x \neq x_{0}$. But this is just one point and not infinitely many points. I don't know how to continue then.

And since the set $\{x\} \subset X$ is closed, the set $X \setminus \{x\}$ is open. I get the feeling that I also need this?

Laura van Leuven
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  • for $(\Leftarrow)$, you could have a look at https://math.stackexchange.com/a/2383462/904944 – dfeg Apr 06 '21 at 14:50

2 Answers2

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$X$ has the property that $\{x\}$ is closed for all $x$. This is key.

So for any finite set of $\{x_1,....., x_k\}$ is a finite union of close sets. so $ \{x_1,......,x_k\}^c= \{x\in X: x\not \in \{x_1,....., x_k\}\}$ is open.

Suppose $x_0$ is a limit point of $A$ and $N$ is a neighborhood of $x_0$ that contains only finitely many $x1,x_2,....,x_k$ (all not equal to $x_0$) points in $A$.

Then $N$ is open and $\{ x_1, x_2,..... x_k\}^c$ is open and so $N\cap \{x_1,...., x_k\}^c$ is open. And $x_0\in N\cap \{ x_1, x_2,..... x_k\}^c$ so $N\cap \{ x_1, x_2,..... x_k\}^c$ is a neighborhood of $x_0$. But any point of $A$ not equal to $x_0$ that could be in $N\cap \{ x_1, x_2,..... x_k\}^c$ would have to be one of the finite $x_1, x_2,... x_k$. But none of these are in $\{ x_1, x_2,..... x_k\}^c$.

So that contradicts that $x_0$ is a limit point.

So any limit point (if there are any-- there might not be) can't have any neighborhood with finite number of points (and obviously can't have a neighborhood with no points) of $A$ not equal to $x_0$.

Suppose there is a neighborhood of $x_0$ that contains only finitely many points of $A$ not equal to $x_0$.

$\{x_0\}$ is closed so $\{x_0\}^c = \{x\in X| x\ne x_0\}$ is open.

fleablood
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If every neighbourhood of $x_0$ cointains infinitely many points of $A$, then at least one of those is different from $x_0$.

Conversely, if you take a neighbourhood $U$ of $x_0$ and finitely many points $y_1,\ldots,y_k$ of $A\cap U$, what can you say about $U\setminus\left\{y_i:i=1,\ldots,k\right\}$?

Luiz Cordeiro
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