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Let $\mathcal B_1, \mathcal B_2, \dots, \mathcal B_n$ be bases for topological spaces $(X_1, \mathcal T_1), (X_2, \mathcal t_2), \dots , (X_n, \mathcal T_n)$, respectively. Prove that the family $\mathcal B'=\{B_1\times B_2 \times \dots \times B_n:B_i\in \mathcal B_i\}$ is a basis for the product topology $(X, \mathcal T)= \prod_{i=1}^n(X_i, \mathcal T_i)$.

Here is my attempt:

We know, by definition of finite product topology $\mathcal T$ on $X=\prod_{i=1}^n(X_i, \mathcal T_i)$, that $$\mathcal B = \{O_1\times O_2\times \dots \times O_n: O_i \in \mathcal T_i\}$$ is a basis for $\mathcal T$.

Now let $U \in \mathcal T$ with $\bar{x}=(x_1,x_2, \dots,x_n)\in U$. Then, by definition of a basis, there exits a $B_\bar{x}=O_{1x}\times O_{2x}\times \dots \times O_{nx} \in \mathcal B$ such that $$\bar{x}\in B_\bar{x} \subseteq U.$$

Notice that, for each $i=1, 2, \dots , n$, $O_{ix} \in \mathcal T_i$ and so, since each $\mathcal B_i$ is a basis for $\mathcal T_i$, there exists a $B_i \in \mathcal B_i$ such that $$x_i \in B_i \subseteq O_{ix}.$$ Choose $B'_{\bar{x}} = B_1\times B_2\times \dots \times B_n$, then $$\bar{x}\in B'_{\bar{x}} \subseteq U.$$

That is, $\mathcal B'$ is a basis for $\mathcal T$.

Is this correct?

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    $\begingroup$ Looks good ${}{}$ $\endgroup$ – user384138 Nov 14 '16 at 21:48
  • $\begingroup$ @OpenBall. Thank you! My textbook left this proof in the notes as an exercise for the reader to prove, so I just wanted to double check if my attempt at proving it was correct :) $\endgroup$ – user860374 Nov 14 '16 at 21:51

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