let $X$ be an arbitrary Hausdroff space, then is $X\setminus\{x_0\}$ always dense?

Question

I know $\mathbb{R}\setminus\{x_0\}$ is dense subset of $\mathbb{R}$. Is this result true for any arbitrary Hausdroff Topological space.

I mean, let $X$ be an arbitrary Hausdroff space, then is $X\setminus\{x_0\}$ always dense?

Answer 1

No. The set $X\setminus\{x_0\}$ is dense if and only if $x_0$ is not an isolated point. So, take $\mathbb Z$ with the usual topology, for instance, and any $x_0\in\mathbb Z$.

Answer 2

No.

Take $X=\{x_1, x_2\}$ with the discrete metric.

Answer 3

It is not true. For example, taking $X=[0,1]\cup\{2\}$, we can see that $X\setminus\{2\}$ is not dense.

$X\setminus \{x_0\}$ is dense if and only if its closure is $X$. This means that it is dense if and only if it is not closed.

Remember, in a Hausdorff space, every singleton is closed, which means $X\setminus\{x_0\}$ must be open. This, along with the statement above, means that $X\setminus\{x_0\}$ is dense if and only if $x_0$ is not an isolated point.