Exercise 16.5

Proof. Consider any $W\in T_p$ and any $(x,y)\in W$, noting that obviously $W\subset X\times Y$. Then there is a basis element $U\times V$ of $T_p$ such that $(x,y)\in U\times V$ and $U\times V\subset W$. By the definition of the product topology, we have that $U$ and $V$ are open sets in $T$ and $U$, respectively. Then we also have that $U\in T^{\prime }$ and $V\in U^{\prime }$ since $T\subset T^{\prime }$ and $U\subset U^{\prime }$. Hence $U\times V$ is also a basis element of $T_p^{\prime }$. Since we know that $(x,y)\in U\times V$, $U\times V\subset W$, and $(x,y)\in W$ was arbitrary, this suffices to show that $W$ is an open subset of $X^{\prime }\times Y^{\prime }$ and hence $W\in T_p^{\prime }$. This in turn shows that $T_p\subset T_p^{\prime }$ since $W$ was arbitrary. □

Proof. As a counterexample consider $A=\left\{a,b,c,d\right\}$ so that clearly

are topologies on $A$. Clearly also $T^{\prime }$ is not finer than $T$. Similarly let $B=\left\{1,2,3,4\right\}$ so that

are topologies on $B$, also noting that clearly $U^{\prime }$ is not finer that $U$. Now let $X=X^{\prime }=\left\{a,b\right\}$ and $Y=Y^{\prime }=\left\{1,2\right\}$ so that clearly $X$ and $X^{\prime }$ are in topologies $T$ and $T^{\prime }$, respectively, and $Y$ and $Y^{\prime }$ are in $U$ and $U^{\prime }$, respectively.

Then the bases for the product topologies $T_p$ on $X\times Y$ and $T_p^{\prime }$ on $X^{\prime }\times Y^{\prime }$ are then

respectively, since there are no subsets of $X$ in $T$ or $T^{\prime }$ other than $\text{∅}$ and $X$ itself, and similarly no subsets of $Y$ in $U$ or $U^{\prime }$ other than $\text{∅}$ and $Y$. Since their bases are the same, clearly $T_p=T_p^{\prime }$ so that it is true that $T_p^{\prime }$ is finer than $T_p$ (though not strictly so). □