Section 32: Problem 6 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

A space $X$ is said to be completely normal if every subspace of $X$ is normal. Show that $X$ is completely normal if and only if for every pair $A$ , $B$ of separated sets in $X$ (that is, sets such that $\bar{A}\cap B=\emptyset$ and $A\cap\bar{B}=\emptyset$ ), there exist disjoint open sets containing them. [Hint: If $X$ is completely normal, consider $X-(\bar{A}\cap\bar{B})$ .]

$\Rightarrow$ Suppose $A,B$ is a pair of separated subsets of $X$ . Then $Y=X-(\overline{A}\cap\overline{B})$ is an open subset of $X$ that contains both $A$ and $B$ . $\overline{A}_{Y}\cap\overline{B}_{Y}=Y\cap\overline{A}\cap\overline{B}=\emptyset$ . Thus, $\overline{A}_{Y}$ and $\overline{B}_{Y}$ can be separated by open neighborhoods in $Y$ . Since $Y$ is open, these neighborhoods are also open in $X$ .

$\Leftarrow$ Given two disjoint sets closed in a subspace $Y$ , neither contains a limit point of the other in $X$ , hence, they are separated in $X$ , and can be separated by open neighborhoods in $X$ , and, hence, in $Y$ .

Please note that, as discussed with Indrayudh Roy below in comments, Munkres gives a definition of a normal space that also requires it to be a $T_{1}-$ space, so he should have also added this requirement to the $\Leftarrow$ direction. The fact stated in the problem is correct both ways if a normal space is not required to be a $T_{1}-$ space.

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Section 32: Problem 6 Solution

Section 32: Problem 6 Solution