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

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.