Question

A space is totally disconnected if its connected spaces are one-point-sets.Show that a finite Hausdorff space is totally disconnected.

Answer 1

Step-by-step explanation:

If X is a finite Hausdorff space then every two points of X can be separated by open neighborhoods. Say the points of X are $x_1, x_2, ..., x_n$. So there are disjoint open neighborhoods $U_{12}$ and $U_2$, of $x_1$ and $x_2$ respectively (that's the definition of Hausdorff space). There are also open disjoint neighborhoods $U_{13}$ and $U_3$ of $x_1$ and $x_3$ respectively, and disjoint open neighborhoods $U_{14}$ and $U_4$ of $x_1$ and $x_4$, and so on, all the way to disjoint open neighborhoods $U_{1n}$, and $U_n$ of $x_1$ and $x_n$ respectively. So $U=U_2 \cup U_3 \cup ... \cup U_n$ has every element of $X$ in it, except for $x_1$. Since $U$ is union of open sets, it is open, and so $U^c$, which is the singleton $\{ x_1\}$, is closed. Therefore every singleton is closed.

Now, remember finite union of closed sets is closed, so $\{ x_2\} \cup \{ x_3\} \cup ... \cup \{ x_n\}$ is closed, and so its complemented, which is $\{ x_1\}$ is open. Therefore every singleton is also open.

That means any two points of $X$ belong to different connected components (since we can express X as the union of the open sets $\{ x_1\} \cup \{ x_2,...,x_n\}$, so that $x_1$ is in a different connected component than $x_2,...,x_n$, and same could be done with any $x_i$), and so each point is in its own connected component. And so the space is totally disconnected.