If $X$ is a Hausdorff space, and $x_1, x_2,\ldots, x_n$ are distinct points, do they have pairwise disjoint open neighborhoods?

Question

A quick question about Hausdorff spaces:

If $X$ is a Hausdorff space and $x_1, x_2,\ldots, x_n$ are distinct points, is it possible to find pairwise disjoint open sets $U_{x_1},\ldots, U_{x_n}$ with $x_i\in U_{x_i}$?

Answer 1

Hint: The intersection of finitely many open sets is an open set.

Answer 2

Yes, you can prove this using induction. The base case holds by definition. Now assume for some $n > 2$, you can find pairwise disjoint open neighborhoods for $n$ distinct points. Given points $x_1, \dots, x_n, x_{n+1} \in X$, by the inductive hypothesis, there exists pairwise disjoint neighborhoods $U_i$ of $x_i$, and there exists disjoint neighborhoods $V_i$ of $x_i$ and $N_i$ of $x_{n+1}$ for $1 \le i \le n$. Now let

$$U_{x_{n+1}} = \bigcap_{1 \le i \le n} N_i$$

and let $U_{x_{i}} = V_i \cap U_i$ for $1 \le i \le n$. This completes the inductive step.

Answer 3

There always exist pairwise disjoint open sets for the finite points in Hausdorff space. But it is not true that there always exist pairwise disjoint open sets for the countable points in Hausdorff space. But we have following claim:

Claim 1: There always exist pairwise disjoint open sets for the countble points in regular space.