Section 23: Problem 10 Solution

Working problems is a crucial part of learning mathematics. No one can learn topology merely by poring over the definitions, theorems, and examples that are worked out in the text. One must work part of it out for oneself. To provide that opportunity is the purpose of the exercises.

James R. Munkres

Let $\{X_{\alpha}\}_{\alpha\in J}$ be an indexed family of connected spaces; let $X$ be the product space $X=\prod_{\alpha\in J}X_{\alpha}.$ Let $\mathbf{a}=(a_{\alpha})$ be a fixed point of X.

(a) Given any finite subset $K$ of $J$ , let $Х_{K}$ denote the subspace of $X$ consisting of all points $\mathbf{x}=(x_{\alpha})$ such that $x_{\alpha}=a_{\alpha}$ for $\alpha\notin K$ . Show that $Х_{K}$ is connected.

(b) Show that the union $Y$ of the spaces $X_{K}$ is connected.

(a) It is homeomorphic to a finite product of connected spaces.

(b) They all have a common point ( $\mathbf{a}$ ).

(c) This is the main part of the proof. Take any point $\mathbf{x}\in X$ . Any its neighborhood in the product topology has only finite number of coordinates restricted by some proper open subsets of the corresponding coordinate spaces. A point that equals $\mathbf{x}$ at these finite number of coordinates, and equals $\mathbf{a}$ at all others coordinates, belongs to the neighborhood and to $Y$ . Therefore, the closure of $Y$ is the whole space $X$ , and by Theorem 23.4, $X=\overline{Y}$ is connected.

Subpages

Section 23: Problem 10 Solution

Section 23: Problem 10 Solution