Let $\{ (X_\alpha, Y_\alpha): \alpha \in A\}$ be a family of topological spaces for some index set $A$. Prove that the product topology $\mathcal T$ on $X= \displaystyle \prod_{\alpha \in A} X_\alpha$ is the coarsest topology on which the projection mapping $p_\alpha:\displaystyle \prod_{\alpha \in A} X_\alpha \to X_\alpha$ is continuous for each $\alpha \in A$.

Firstly, in my notes, the product topology $\mathcal T$ is defined as the topology on $X$ which has as a basis the collection $$\mathcal B = \{O_\alpha: O_\alpha \in \mathcal T_\alpha \text{ for all but a finite number of } \alpha \in A \}.$$

My attempt:

Suppose that, for each $\alpha \in A$, $p_\alpha$ is continuous on some topology $\mathcal T'$ of $X$. We must show that $\mathcal T'\supseteq \mathcal T$.

Let $W=W_{\alpha_1}\times W_{\alpha_2}\times W_{\alpha_n}\times X_{\alpha_{n+1}}\times \cdots$ be a basis element of the product topology $\mathcal T$ on $X$, then $W \in \mathcal T$.

Now (this is the part I'm not sure is correct) $$W=\bigcap_{k=1}^n\left(W_{\alpha_k}\times\prod_{\alpha_k \in A, \alpha_k \neq k}X_{\alpha_k}\right) = \bigcap_{k=1}^n \left(p_{\alpha_k}^{-1}(W_{\alpha_k})\right)$$ Then, since $p_{\alpha_k}$ is continuous on $X$ w.r.t. $\mathcal T'$ (by hypothesis), we must have that $p_{\alpha_k}^{-1}(W_{\alpha_k}) \in \mathcal T'$ and so we also have that $W \in \mathcal T'$, since $\mathcal T'$ is closed under finite intersection.

That is, $\mathcal T' \supseteq \mathcal T$.

Is the proof given above correct? If not, can anyone please point out to me where I went wrong?


1 Answer 1


Your ideas are good, but some notations are clumsy, for instance $$\prod_{\alpha_k \in A, \alpha_k \neq k}X_{\alpha_k}$$ has some typo. It should be something like $$\prod_{\alpha \in A, \alpha \not \in \{\alpha_1,\dots,\alpha_n\}}X_{\alpha}$$

Otherwise, it seems to be fine: any basis element of the product topogy $\mathcal T$ can be written as $$\bigcap_{k=1}^n \left( p_{\alpha_k}^{-1}(W_{\alpha_k})\right),$$ which is open w.r.t to $\mathcal T'$.

You should not write $$\bigcap_{k=1}^n \left( \bigcup_{\alpha_k \in A}p_{\alpha_k}^{-1}(W_{\alpha_k})\right)$$ because $\alpha_k$ is fixed, so it is unclear what you mean by $\bigcup_{\alpha_k \in A}$.

Remember that $p_{\alpha_k}^{-1}(W_{\alpha_k})$ is the product $$\prod_{a \in A} U_a$$ where $U_a = X_a$ if $a \neq \alpha_k$, and $U_{\alpha_k}:=W{\alpha_k}$.

  • $\begingroup$ Thank you so much for pointing that out! :). I noticed after putting in the $\bigcup_{\alpha_k \in A}$ that it did not make sense. So I removed it :) $\endgroup$
    – user860374
    Nov 16, 2016 at 13:32
  • $\begingroup$ I most certainly will accept! :). $\endgroup$
    – user860374
    Nov 16, 2016 at 13:56
  • $\begingroup$ No problem! Have fun with topology, then ;-) $\endgroup$
    – Watson
    Nov 16, 2016 at 13:59

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