Proof on countably discrete subset of any Hausdorff space.

Question

I have this question for several days:

Let $X$ be a topological space and $X$ is Hausdorff. $C$ is an countable discrete subset of $X$. Then does there exist a disjoint family of open sets $\{U_x: x\in C\}$ of $X$ such that $x\in U_x$ for all $x\in C$?

My idea is this: Assume that $C= \{x_n:n\in N\}$. Induction on $N$. For $n=1$, for $C$ is discrete, we choose an open set $U_1$ such that only $x_1 \in U_1$; now we assume for n=k, we have got $\{U_i: i=1,2,...,k\}$ such that they are disjoint. When $n=k+1$, for $x_{n+1}$, because $X$ is Hausdorff, so I can choose an open set $U_{n+1}$ which disjoint with any $U_i$ for any $i=1,2,...k$. Therefore I can get such disjoint family.

Am I right? Thanks for any help.

Answer 1

Your argument does not work. Suppose that $X=\Bbb R$ and $C=\Bbb N$. You might take $U_0=(-1,1)$: this is an open nbhd of $0$ that does not contain any other member of $\Bbb N$, but no open nbhd of $1$ is disjoint from $U_0$.

In fact the result is false. Let $X=\Bbb Q\times\Bbb Q$, and let $C=\Bbb Q\times\{0\}$. Points of $X\setminus C$ are isolated. For $r>0$ and $q\in\Bbb Q$ let

$$B(q,r)=\big\{\langle q,0\rangle\big\}\cup\left\{\langle x,y\rangle\in X\setminus C:(x-q)^2+y^2<r^2\right\}\;;$$

$B(q,r)$ is what’s left of the usual open ball of radius $r$ centred at $\langle q,0\rangle$ after all of the $x$-axis except $\langle q,0\rangle$ has been removed. The sets $B(q,r)$ for $r>0$ are a local base at $\langle q,0\rangle$ in $X$.

It’s easy to check that $X$ with this topology is Hausdorff, and $C$ is certainly a countable discrete subset of $X$. ($C$ is even a closed discrete subset of $X$.) However, there is clearly no function $\varphi:\Bbb Q\to\Bbb R^+$ such that $\left\{B\big(q,\varphi(q)\big):q\in\Bbb Q\right\}$ is a pairwise disjoint family of open nbhds of the points of $C$.