How to prove that $ \left \{ \prod_{i=1}^{n} U_i: U_i \text{ are open in } X_i \right \} $ is a base for the product topology?

Asked Nov 12 '23 at 18:15

Active Nov 12 '23 at 18:42

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I'm trying to prove that

$$ B = \left\{ \prod_{i=1}^{n} U_i : U_i \text{ are open in } X_i \right\} $$

is a basis of the product topology.

I was trying to use the subbase of the product topology, that is sets of the form:

$ p_i ^ {-1} (G_i) $ where $G_i$ is an arbitrary open set in $X_i$. Where $p_i((x_1, x_2, \dots, x_n)) = x_i$.

edited Nov 12 '23 at 18:42

Didier

asked Nov 12 '23 at 18:15

John

3

Hint: Write each set $\prod_{i=1}^{n} U_i$ as a finite intersection of sets $p_i^{-1}(G_i)$ [and vice versa]. – GEdgar Nov 12 '23 at 18:30
1

@GEdgar you mean to show that $\prod_{i=1}^{n} G_i = \bigcap_{i=1}^{n} p_i^{-1} (G_i)$? – John Nov 12 '23 at 19:19
Have you tried writing the basis coming from the subbase you mentioned? – Keen-ameteur Nov 14 '23 at 14:14

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