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The theorem which I want to show is the following:

For a Hausdorff space $X$ and a subset $A$ of $X$, $x$ is a limit point of $A$ if and only if every neighborhood of $x$ contains infinitely many points of $A$.

And this is my answer;

(⇒) Suppose not. Then there is a neighborhood of $x$, say $M$, which has finitely many points $a_1, \ldots, a_n$ of $A$. Since $M$ is an open set containing $x$, $M$ must contain a point of $A$ distinct from $x$. Then, there exist open sets $U_1 ,U_2, \ldots, U_n$ and $V$ in $X$ such that each $U_i$ contains in $a_i$ for $i=1,\ldots,n$ and $V$ contains $x$, and all of $U_1,\ldots,U_n$ and $V$ are disjoint, since $X$ is Hausdorff space. Then, since $V$ is an open set containing $x$, $V$ contains another point $a_{n+1}$ of $A$ distinct from $x$. Thus, $M$ has $n+1$ points of $A$; contradiction!

(⇐) Suppose not. Then $x$ is not a limit point of $A$; there is an open set $O$ containing $x$ and contains no point of $A$ but $x$. Then $O$ is a neighborhood of $x$. Thus $O$ has infinitely many points of $A$. This is contradiction.

Is it correct?

And if $A$ is a neighborhood of $x$ means interior of $A$ contains $x$; $A$ is an open set containing $x$?

Martin Sleziak
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Darae-Uri
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2 Answers2

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The right proof should use separation axiom of Hausdorff space to create an open set that contains no point of $A$ and get contradiction. It is as follow:

(=>) Suppose not. Then there is a neighborhood of $x$, says $M$, which has only finite many points $a_1,…,a_n$ of $A$.

By separation axiom of Hausdorff space, for each $a_k\in M$ and $x$ ($1\leqslant k\leqslant n$), there are open set $U_k$ and $V_k$ of $M$, $a_k\in U_k, x\in V_k$, such that $U_k\cap V_k=\varnothing$. Let $V=\bigcap \limits_{k=1}^{n}V_k$. Clearly $V$ is open set, and for each $k, 1\leqslant k\leqslant n$, $U_k\cap V\subset U_k\cap V_k$. And so $U_k\cap V=\varnothing$ or $a_k\notin V$.

Therefore $V$ is an open subset of $M$ that contains $x$ but contains no point of $A$, which means $x$ is not a limit point of $A$, contradiction.

(<=) same as yours.

Eugene Zhang
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Your proof is correct. Note that the second implication is true in general and follows directly from the definition. If a neighborhood of $x$ contains infinitely many points of $A$, then it contains a point of $A$ different than $x$, which is a definition of $x$ being in $A'$.

Also note that the first implication and thus the whole proposition holds even for $T_1$ spaces. If every neighborhood of $x$ contains a point of $A$ different than $x$, then $x ∈ \overline{A \setminus \{x\}}$. If there is a neighborhood $U$ of $x$ such that $U ∩ A$ is finite, then $x ∈ \overline{U ∩ A \setminus \{x\}} = U ∩ A \setminus \{x\}$, which is a contradiction.

For your last question, if $A$ is a neighborhood of $x$, it means, as you say, that $x$ is in the interior of $A$, but that doesn't imply that $A$ is open. If $U$ is an open set containing $x$, then any its superset is a neighborhood of $x$ as well.

user87690
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