Theorem 15.1 of Munkres Topology

If $\mathcal{B}$ is a basis for the topology on $X$ and $\mathcal{C}$ is a basis for a topology on $Y$, then the collection $\mathcal{D} = \{B \times C ~|~ B, \in \mathcal{B}, C \in \mathcal{C}\}$ is a basis for the topology on $X \times Y$.

The set $X$ & $Y$ are equipped with what topology? What topology on $X\times Y$ Munkres talking about? If set $X\times Y$ equipped with arbitrary topology($\mathfrak{I}_{X\times Y}$) on $X\times Y$, then how can one show $\mathcal{D}\subseteq \mathfrak{I}_{X\times Y}$. “$\mathcal{B}$ is a basis for the topology on $X$“ exact same wording used to define set basis. While defining basis we didn’t know anything about the topology, but in this case we know about the topology. Here is a link: Section 13: Basis for a Topology. Definition and lemma.

After reading proof of this theorem. I conclude, let set $X$ & $Y$ equipped with $\mathfrak{I}_{X}$ and $\mathfrak{I}_{Y}$ topology, respectively. So now talking about product topology on $X\times Y$, I will denote with $\mathscr{I}_{p}$, makes sense(because the basis, I will denote with $\mathscr{B}_{p}$, which generates product topology on $X\times Y$ is well defined). Thus set $X\times Y$ equipped with $\mathscr{I}_{p}$ topology.

Question: (1) Since we have to show set $\mathcal{D}$ is the basis of product topology on $X\times Y$ (or $\mathscr{I}_{p}$) and basis of product topology is $\mathscr{B}_{p}=\{U\times V | U\in \mathfrak{I}_{X}, V \in \mathfrak{I}_{Y}\}$, can we show $\mathcal{D}=\{ B\times C | B\in \mathcal{B}, C\in \mathcal{C}\}= \mathscr{B}_{p}$. If not, then it shows that, there can be more than one basis for a given topology. Which I think is a nice result to keep in mind.

(2) Munkres didn’t show $\mathcal{D} \subseteq \mathscr{I}_{p}$ condition for some reason. Anyway here’s my proof. Proof: let $B\times C \in \mathcal{D}$. So $B\in \mathcal{B}\subseteq \mathfrak{I}_{X}$ and $C\in \mathcal{C}\subseteq \mathfrak{I}_{Y}$. Thus $B\times C \in \mathscr{B}_{p} \subseteq \mathfrak{I}_{p}$. Hence $\mathcal{D} \subseteq \mathscr{I}_{p}$.

Solution 1:

The proof is straightforward. Of course $X$ and $Y$ have topologies (we're even given bases $\mathcal{B}$ and $\mathcal{C}$ for them!).

Just check the two Munkres conditions for $X \times Y$ and $\mathcal{D}$. This shows that $\mathcal{D}$ generates some topology on $X \times Y$.

First let $x,y) \in X \times Y$. As $\mathcal{B}$ is a base for $X$, for some $U_x \in \mathcal{B}$ we have $x \in U_x$. Ditto for $V_y \in \mathcal{C}$ and $y \in V_y$. Note that $U_x \times V_y \in \mathcal{D}$ and contains $(x,y)$ by definition.

Second condition: suppose $B_1 \times C_1, B_2 \times C_2 \in \mathcal{D}$ and suppose $(z_1,z_2) \in (B_1 \times C_1) \cap (B_2 \times C_2)$. So in particular $z_1 \in B_1 \cap C_1$ and $z_2 \in B_2 \cap C_2$ and $B_i \in \mathcal{B}, i=1,2$ and $C_i \in \mathcal{C}, i=1,2$.

We can apply the second property of bases for $\mathcal{B}$ and $\mathcal{C}$ and get $B_3 \in \mathcal{B}, C_3 \in \mathcal{C}$ so that $z_1 \in B_3 \subseteq B_1 \cap B_2$ and $z_3 \in C_3 \subseteq C_1 \cap C_2$. It follows that

$$(z_1,z_2) \in B_3 \times C_3 (\in \mathcal{D}) \text{ and } B_3 \times C_3 \subseteq (B_1 \times C_1) \cap (B_2 \times C_2)$$

as required.

It's a simple "follow the nose" proof. No real thought or originality is required. Just know the conditions and the relevant facts from the text.

Trivially, the topology generated by $\mathcal{D}$ is a subset of the product topology on $X \times Y$ by definition. That they are equal is also clear after some thought and applying the definitions: let $O$ be open in $\mathcal{T}_p$. Let $(x,y) \in O$. Then there are open sets $O_x$ of $X$ and $O_y$ of $Y$ so that $(x,y) \in O_x \times O_y \subseteq O$. Again there are $B_x \in \mathcal B$ and $C_y \in \mathcal C$ so that $x \in B_x \subseteq O_x$ and $y \in C_y \subseteq O_y$. Then $B_x \times C_y \in \mathcal D$ and

$$(x,y) \in B_x \times C_y \subseteq O_x \times O_y \subseteq O$$

and we see that $\mathcal D$ is a basis for $\mathcal T_p$. This last direct proof might be the fastest way.