Confusion regarding the proof steps of the topology generated by a subbasis
Munkres, page 88, Theorem 15.2
The author outlines the proof (given above) that the subbasis $S$ generates the topology $\tau'$ which is same as the product topology $\tau$. The proof can be broken down into two steps. First: every element of $S$ belongs to $\tau$. So $\tau' \subset \tau$. Second step: every elements of the basis (let's call it $B$) of $\tau$ is a finite intersection of the elements of $S$. So, $\tau \subset \tau'$.
My question is why do we need two steps? Isn't it sufficient to prove that the basis $B'$ corresponding to the subbasis $S$ is the same as the basis $B$ that generates $\tau$? In that case, the two basis sets ($B$ and $B'$) are equal and hence the topology generated by the two basis sets should also be the same. Is it a good line of proof?
$\mathcal{S}$ is indeed a subbase for some topology $\mathcal{T}'$ on $X \times Y$. In general we can write the members of this $\mathcal{T}'$ as "arbitrary unions of (finite intersections from $\mathcal{S}$)", so as the result of a two step process:
- Take all finite intersections from $\mathcal{S}$, so form $$\mathcal{B}(\mathcal{S}) = \{\bigcap_{i=1}^n S_i\mid n=1,2,3,4,\ldots, S_1, \ldots S_n \in \mathcal{S}\}$$
note that in particular $\mathcal{S}\subseteq \mathcal{B}$.
- $\mathcal{T'}$ is the unique topology that has $\mathcal{B}(\mathcal{S})$ as a basis, so
$$\mathcal{T'}= \{\bigcup \mathcal{B}'\mid \mathcal{B}' \subseteq \mathcal{B}(\mathcal{S})\}$$ and this is well-defined because $\mathcal{B}(\mathcal{S})$ is a basis for some topology (it obeys the two conditions given by Munkres when he first discusses a basis for a topology).
In this case, $\mathcal{B}(\mathcal{S})$ is just the set $\{U \times V\mid U\subseteq X \text{ open }, V \subseteq Y \text{ open }\}$, i.e. the standard basis for the product topology that Munkres gives as a definition.
So by unicity of the generated topology from a basis, $\mathcal{T}'$, generated from $\mathcal{S}$ in these steps is exactly the product topology.
So what you propose is enough, don't worry. What Munkres does is to avoid the proof that "$\mathcal{B}(\mathcal{S})$ is just the set $\{U \times V\mid U\subseteq X \text{ open }, V \subseteq Y \text{ open }\}$", which is true but a bit tedious notationwise, and replace it by a slightly more abstract argument involving both $\mathcal{T}'$ and the product topology $\mathcal{T}$.
I hope this helps.