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I'm taking an introductory course on topology and we have just defined the product topology as follows:

Let $(X,T_{X})(Y,T_{Y})$ be topological spaces, then the product topology of $X$ and $Y$ is the one generated by the base $B_{X \times Y}$ over the cartesian product $X \times Y$, where $B_{X \times Y}= \lbrace A \times B | A \in T_X \land B \in T_Y \rbrace$.

Now, I have this proposition, which I'm not sure if it is right and in case it is if the proof I am giving is ok:

$\textbf{Proposition:}$ Let $(X,T_{X})(Y,T_{Y})$ be topological spaces with bases $B_X$ and $B_Y$ respectively, then the product topology of those spaces equals the topology over $X \times Y$ generated by $B_X \times B_Y$

$\textbf{Proof:}$ Let $C$ be open in $X \times Y$ with the product topology, then $C=\bigcup_{i \in I} (A_i \times B_i)$, where $A_i$ and $B_i$ are open sets in $X$ and $Y$ respectively. It is also tbe case that $\forall i \in I (A_i=\bigcup_{j \in I_i} A_j^i \land B_i=\bigcup_{l \in J_i} B_l^i)$ where $\forall i \in I \ \forall j \in I_i \ \forall l\in J_i (A_j^i \in B_X \land B_j^i \in B_Y)$. So, we have $$C=\bigcup_{i \in I}[(\bigcup_{j \in I_j} A_j^i) \times (\bigcup_{l \in J_i} B_j^i)]=\bigcup_{i \in I} [\bigcup_{(j,l) \in I_i \times J_i}(A_j^i \times B_j^i)]$$ which is a union of sets of $B_X \times B_Y$, hence $C$ is in the topology generated by $B_X \times B_Y$. The rest of the proof follows directly from the fact that $B_X \times B_Y \subset B_{X \times Y}$

Is this right?

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The proof as it stands is fine. This method is the "index-juggling" method, so to say.

You could also use a more "point-centric" approach, which comes down to the same: suppose $O$ is open in the product topology on $X \times Y$. Suppose $(p,q) \in O$. Then there are open sets $O_p \subseteq X$ and $O_q \subseteq Y$ such that $(p,q) \in O_p \times O_q \subseteq O$.

As $p \in O_p$ we have a set $B_p \in \mathcal{B}_X$ with $p \in B_p \subseteq O_p$. Similarly we find $B_q \in \mathcal{B}_Y$ with $q \in B_q \subseteq O_q$.

Then $B_p \times B_q \in \mathcal{B}_X \times \mathcal{B}_q$ and $(p,q) \in B_p \times B_q \subseteq O_p \times O_q \subseteq O$.

So $\mathcal{B}_X \times \mathcal{B}_q$ is a base for the product topology (because the members are all open by the definition of the product topology).

The above uses the equivalent definition for a base: If $(X,\mathcal{T})$ is a space then $\mathcal{B} \subseteq \mathcal{T}$ is a base for the topology iff for every open set $O \subseteq X$ and every $x \in O$, there exists some $B_x \in \mathcal{B}$ such that $x \in B_x \subseteq O$.

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