Conditions for Comparability of Product Topologies (2.16.5)

James Munkres Topology, chapter 2.16, exercise 5:

Let $X$ and $X'$ denote the same set under the topologies $J$ and $J'$ respectively; let $Y$ and $Y'$ denote the same set under the topologies $T$ and $T'$ respectively.

(a) Show that if $J⊆J'$ and $T⊆T'$ then the topology of $X'×Y'$ is finer than the topology of $X×Y$.
(b) Does the converse of (a) hold?
(c) What can you say if $J⊆J'$ and $T \subset T'$?

Proof: (a) The general basis element for $X×Y$ is $U×V$ when $U⊆X$ and $V⊆Y$ are open. Since the topologies of $X'$ and $Y'$ are finer than $X$ and $Y$ respectively, $U⊆X'$ and $V⊆Y$ are open and $U×V⊆X'×Y'$ is open. Hence the generated topology of $X×Y$ is contained in that of $X'×Y'$.

(b) Suppose $U⊆X$ and $V⊆Y$ are open, with $x∈U$ and $v∈V$. Then $(u,v)∈U×V⊆X×Y$ is contained in an open set, so since the topology of $X'×Y'$ is finer than that of $X×Y$ we have $(u,v)$ is contained in a basis element of $X'×Y'$, so write $(u,v)∈U'×V'⊆U×V$. Hence $u∈U'⊆U$ and $v∈V'⊆V$ showing $U$ and $V$ are also open in $X$ and $Y$ respectively.

(c) Suppose $X$ is nonempty. Then if the topology of $Y'$ is strictly finer than that of $Y$, let $V$ be open in $Y'$ and not in $Y$. As such, choose $v∈V$ such that there is no open $W⊆Y$ such that $v∈W⊆V$. Choose some $x∈X$. Then $(x,v)$ is contained in the open set $X×V⊆X'×Y'$. Now there cannot be a basis element $U×W⊆X×V⊆X×Y$ with open $U⊆X$ and $W⊆V⊆Y$ containing $(x,v)$ by specific choice of $v$. Hence the topology of $X'×Y'$ is strictly finer than that of $X×Y$.$~\square$