Add or subtrack1. -16 +212.-13--123.-23-(-8)

Mathematics

2 answers:

Paul [167]3 years ago

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konstantin123 [22]3 years ago

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

marysya [2.9K]

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.