TITLE: Question from Munkres Topology Regarding the Product Topology (Section 16, Exercise 5)
QUESTION [5 upvotes]: Let $X$ and $X'$ denote a single set in the topologies $\mathscr T$ and $\mathscr T'$ respectively; let $Y$ and $Y'$ denote a single set in the topologies $\mathscr U$ and $\mathscr U'$ respectively.  Assume these sets are non-empty.
a) Show that if $\mathscr T'\supseteq\mathscr T$ and $\mathscr U'\supseteq\mathscr U$, then the product toplogy on $X'\times Y'$ is finer than the topology on $X\times Y$.
Having trouble making sense of this problem.  How can I compare the topologies if they are on different sets.?

REPLY [6 votes]: The notation is a bit confusing. As a topological space is just a pair $(X, \mathcal{T})$ it would have been better stated, I think, as follows:
Suppose we have non-empty topological spaces $(X,\mathcal{T})$ and $(X,\mathcal{T}')$ and another pair of non-empty topological spaces $(Y,\mathcal{U})$ and $(Y,\mathcal{U}')$. 
As an auxiliary notation, denote the product topology of two topologies $\mathcal{T}$ and $\mathcal{U}$ by $\mathcal{T} \otimes \mathcal{U}$. As you probably know, this is the smallest topology on $X \times Y$ that contains the set $\{U \times V: U \in \mathcal{T}, V \in \mathcal{U} \}$.
If $\mathcal{T} \subseteq \mathcal{T}'$ and $\mathcal{U} \subseteq \mathcal{U}'$, then the product topological space $(X \times Y, \mathcal{T}' \otimes \mathcal{U}')$ has a finer topology than $(X \times  Y, \mathcal{T} \otimes \mathcal{U})$, or more succinctly: $\mathcal{T} \otimes \mathcal{U} \subseteq \mathcal{T}' \otimes \mathcal{U}'$.