There is a question in Munkres' Topology which has me a little confused:
Let X have topologies $\mathfrak{T}$, $\mathfrak{T'}$, and Y have a topologies $\mathfrak{U}$,$\mathfrak{U'}$. Show that if $\mathfrak{T} \subset \mathfrak{T'}$ and $\mathfrak{U} \subset \mathfrak{U'}$, then the product topology on X$\times$Y under $\mathfrak{T'},\mathfrak{U'}$ (I'll denote by $\mathfrak{T'}\ast\mathfrak{U'}$) is finer than the product topology under $\mathfrak{T},\mathfrak{U}$ (I'll denote by $\mathfrak{T}\ast\mathfrak{U}$). Is the converse true?
So the first part was fine. The second I thought might be true and proved it, but I immediately found a counterexample. My question is where is the wrong step in the "proof" I got down.
Let $\mathfrak{B},\mathfrak{B'}$ be bases for $\mathfrak{T},\mathfrak{T'}$, respectively, and $\mathfrak{C},\mathfrak{C'}$ for $\mathfrak{U},\mathfrak{U'}$, respectively. For $x\in B\in \mathfrak{B}$ and $y\in C\in \mathfrak{C}$, then $(x,y)\in B\times C$ which is a basis element for $\mathfrak{T}\ast\mathfrak{U}$. Since it's finer, there is a basis element $U\times V$ of $\mathfrak{T'}\ast\mathfrak{U'}$ such that $(x,y)\in U\times V\subset B\times C$ ($U$ open in X, $V$ open in Y under $\mathfrak{T'},\mathfrak{U'}$, resp). So $x\in U, y\in V$, and since they are open, there exist $B'\in \mathfrak{B'}, C'\in \mathfrak{C'}$ with $x\in B'\subset U$ and $y\in C'\subset V$. Since $U\subset B$ and $V\subset C$ then $x\in B'\subset B$, $y\in C'\subset C$. Then $\mathfrak{T'}$ is finer than $\mathfrak{T}$, and $\mathfrak{U'}$ is finer than $\mathfrak{U}$.
As I said, I know something must be wrong. I know it's stupid, but I actually cannot find the wrong step here.
Thanks in advance.
EDIT: the counter-example I found was: X= {a,b,c}, Y={1,2,3}, $\mathfrak{T}$={X,{a},{a,c}}, $\mathfrak{T'}$={X,{a},{b},{a,b}}, $\mathfrak{U}$={Y,{1}}, $\mathfrak{U'}$={Y,{1},{2},{1,2}}. Of course, now I see this just confirms the contrapositive of the converse, so I was wrong about being wrong…
Best Answer
The converse is actually true. Your proof appears OK--you must be using the criterion given in Lemma 13.3? In this case I think there is a nicer proof: Take $U\in \mathfrak{T}$ and $V \in \mathfrak{U}$. Then $U \times V \in \mathfrak{T} \ast \mathfrak{U}\subseteq \mathfrak{T}' \ast \mathfrak{U}'$. But the projections $X \times Y \to X$ and $X \times Y \to Y$ are open maps (with respect to $\mathfrak{T}' \ast\mathfrak{U}'$, $\mathfrak{T}'$, and $\mathfrak{U}'$), so $U \in \mathfrak{T}'$ and $V \in \mathfrak{U}'$.