Understanding a problem in Munkres This problem is from Chapter 2, Section 16, number 5 in Munkres' Topology. This is not a homework problem, but I'm trying to complete all problems from the sections covered in class. 
Let $X$ and $X'$ denote a single set in the topologies $\mathcal T$ and $\mathcal T'$, respectively; let $Y$ an $Y'$ 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 $X\times Y$.
The way that I would like to prove this is to show that a basis element of the product topology on $X \times Y$ is also a basis element of the product topology on $X' \times Y'$. A basis element for $X\times Y$ is of the form $(A\cap X) \times (B\cap Y) = (A\times B) \cap (X\times Y)$ for some $\mathcal T$-open set $A$ and some $\mathcal U$-open set $B$. I'm not sure how I can show that this is also a basis element of the product topology on $X' \times Y'$ if I don't know the relationship between $X$ and $X'$ or $Y$ and $Y'$. 
 A: When they say that $X$ and $X'$ are a single set in the topologies $\mathcal T$ and $\mathcal T',$ what they mean is that $X=X'$ as sets; when we denote it by $X$ as a topological space, we consider it with the topology $\mathcal T,$ and likewise with $X'$ and $\mathcal T'.$ The same idea goes for $Y,Y',\mathcal U,\mathcal U'.$ This makes your job almost trivial, since the subsets of $X$ and $X'$ are identical, as are the subsets of $Y$ and $Y'.$
A: I'm not completely sure, but I suggest the following resolution:
Let $U$ and $U^\prime$ be open sets of $\mathcal{T}$ and $\mathcal{T}^\prime$, respectively. And let $V$ and $V^\prime$ be open sets of $\mathcal{U}$ and $\mathcal{U}^\prime$, respectively. The collection of all the products $U \times V$ is a basis $\mathcal{B}$ for the product topology on $X \times Y$. And the collection of all the products $U^\prime \times V^\prime$ is a basis $\mathcal{B}^\prime$ for the product topology on $X^\prime \times Y^\prime$.
Since $\mathcal{T}^\prime \supset\mathcal{T}$, then, all the open sets $U \in \mathcal{T}$ are also in $\mathcal{T}^\prime$. And because $\mathcal{U}^\prime \supset\mathcal{U}$, then, all the open sets $V \in  \mathcal{U}$ belong to $\mathcal{U}^\prime$ as well. Thus, the elements of the basis $\mathcal{B}$ are also elements of the basis $\mathcal{B}^\prime$. Therefore, by Lemma 13.3, pag.81, from the Munkres's textbook, we can see that the product topology on $X^\prime \times Y^\prime$ is finer than the product topology on $X \times Y$.
