Section 16: Problem 5 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 $X$ and $X'$ denote a single set in the topologies $\mathcal{T}$ and $\mathcal{T}'$ , respectively; let $Y$ and $Y'$ denote a single set in the topologies $\mathcal{U}$ and $\mathcal{U}'$ , respectively. Assume these sets are nonempty.

(a) Show that if $\mathcal{T}'\supset\mathcal{T}$ and $\mathcal{U}'\supset\mathcal{U}$ , then the product topology on $X'\times Y'$ is finer than the product topology on $X\times Y$ .

(b) Does the converse of (a) hold? Justify your answer.

(a) Every basis set in $X\times Y$ is a basis set in $X'\times Y'$ (if one of the sets is empty, then, both topologies are trivial).

(b) Yes (assuming the sets are nonempty). If $U$ is open in $X$ , $x\in X$ , $V$ is open in $Y$ , $y\in Y$ , then $U\times V$ is open in $X\times Y$ and, therefore, open in $X'\times Y'$ . Therefore, there exists a basis set $A\times B$ in $X'\times Y'$ such that it is a subset of $U\times V$ and it contains $x\times y$ . Therefore, there are open sets $A\in\mathcal{T}'$ and $B\in\mathcal{U}'$ such that $x\in A\subset U$ and $y\in B\subset V$ . So, $U$ is open in $X'$ and $V$ is open in $Y'$ .

If $\mathcal{T}'\supsetneq\mathcal{T}$ and $\mathcal{U}'\supset\mathcal{U}$ , then, assuming $Y\neq\emptyset$ , the product topology on $X'\times Y'$ is strictly finer than the product topology on $X\times Y$ . Indeed, suppose $U$ is open in $X'$ but not in $X$ . Then, there is some $x\in U$ such that for every set $W$ open in $X$ , $x\in W$ implies $W\not\subset U$ . Now, $U\times Y$ is open in $X'\times Y'$ . If it were open in $X\times Y$ , there would be a basis element $A\times B\subset X\times Y$ such that $x\times y\in A\times B\subset U\times Y$ for some $y\in Y$ , and then we would have $x\in A\subset U$ where $A$ is open in $X$ , contradicting the assumption on $x$ .

In fact, all of these could have been easily proved using the following lemma.

If $A$ and $B$ are nonempty, then $A\times B\subset X\times Y$ is open iff $A$ and $B$ are open in $X$ and $Y$ , respectively.

This lemma should have been given in the text or exercises, unless I missed it.

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Section 16: Problem 5 Solution

Section 16: Problem 5 Solution