0
$\begingroup$

We have been asked to show "Let $X_1, \ldots, X_n$ be topological spaces. Show that the product topology is the unique topology on $X_1 \times \cdots \times X_n$ with the property that, for any topological space $Y$ and map $f : Y \to X_1 \times \cdots \times X_n$, $f$ is continuous if and only if each component $\pi_i \circ f : Y \to X_i$ is continuous.

I have managed to show this. But I was wondering what would happen if we take $n \to \infty$. ie. we want to show there is a topology $X_1 \times \cdots \times X_n \times \cdots$ with the property that for any topological space $Y$ and map $f : Y \to X_1 \times \cdots \times X_n \times \cdots $, $f$ is continuous if and only if each component $\pi_i \circ f : Y \to X_i$ is continuous.

I am inclined to believe the product topology should still have this property except the proof of the backward implication uses the infinite intersection of open sets which I know can be closed. Can anyone think of another topology with this product over an infinite amount of topological spaces?

$\endgroup$
  • $\begingroup$ Maybe this link can help. http://en.wikipedia.org/wiki/Product_topology $\endgroup$ – Crostul Oct 25 '14 at 21:28
  • $\begingroup$ I think the link here might be helpful too; I'm not quite sure whether this question defines the box topology or the product topology on infinite sets. $\endgroup$ – Milo Brandt Oct 25 '14 at 21:29
1
$\begingroup$

The product topology has this property. Your difficulty about "infinite intersection of open sets" might have resulted from forgetting that a basis for the product topology is given by products $U_1\times U_2\times\cdots$ where all the $U_n$ are open in the corresponding $X_n$ and $U_n=X_n$ for all but finitely many $n$.

$\endgroup$
  • $\begingroup$ Also, your mention that an infinite intersection of open sets "can be closed" suggests to me that you're thinking of "closed" as synonymous with "not open"; it isn't synonymous. $\endgroup$ – Andreas Blass Oct 25 '14 at 21:33
  • $\begingroup$ Great thank you this confirms the conclusion I just came to (thanks to @Crostul above). Fundamental factor missing that only finitely many $U_i$ can not be equal to entire component space. Also 'not open' is just carelessness on my part, apologies $\endgroup$ – yhu Oct 25 '14 at 21:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.