if $X$ is perfect and Hausdorff, then each open neighborhood of $x$ contains infinitely many points other than $x$ itself.

Question

The question is:

Show that if $X$ is perfect and Hausdorff, then each open neighborhood of $x$ contains infinitely many points other than $x$ itself. Give an example of a space with finitely many points that is perfect.

Definition.

A topological space $X$ is perfect if every point of $X$ is an accumulation point of $X.$

Definition.

A point $x$ in a topological space $X$ is an accumulation point of $X$ if every open neighborhood $U$ of $x$ contains at least one point $y$ of $X$ besides $x$ itself.

Still, I do not know how to formulate a proof for the above question. Could anyone help me in doing so, please?

Answer 1

This already holds for a $T_1$ space $X$: $X$ is $T_1$ iff each singleton $\{x\}$ is closed in $X$ iff each finite subset is closed in $X$. Hausdorff implies $T_1$.

If $U$ is a neighbourhood that contains finitely many points of $X$ besides $x$, then $U$ is just a finite set containing $x$. So $U \setminus \{x\}$ is closed, being finite. Then $\{x\}= U\cap (X\setminus (U \setminus \{x\}))$ is open and a neighbourhood of $x$ that contradicts perfectness. So no such finite $U$ can exist when $X$ is perfect.

$X=\{0,1\}$ with the indiscrete topology $\{\emptyset, X\}$ is perfect and not $T_1$, so a finite perfect space can exist.

Answer 2

Corrected

HINT: Let $x$ be any point of $X$, and let $U$ be an open nbhd of $x$. Suppose that $U$ is finite; since $X$ is perfect, $U\setminus\{x\}\ne\varnothing$, and we can let $U\setminus\{x\}=\{x_1,\ldots,x_n\}$. $X$ is Hausdorff, so for $k=1,\ldots,n$ there are disjoint open sets $U_k$ and $V_k$ such that $x\in U_k$ and $x_k\in V_k$. Now consider the set $U\cap\bigcap_{k=1}^nU_k$ to get a contradiction, thereby showing that $U$ cannot be finite.

The second part of the question can be answered with a space that has only two points.