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?

  • $\begingroup$ 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

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 $n$.

  • $\begingroup$ 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

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