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The following is a theorem from Dugundji's "Topology".

In the space $\Pi_{\lambda\in\Lambda}Y_\lambda$, $\forall\lambda\in\Lambda,$ let $\Sigma_\lambda$ be a subbasis for the topology $\tau_\lambda$ of $Y_\lambda$.Then the family $\{\pi_{\beta}^{-1}(V_\beta)|\ all\ V_\beta\in\Sigma_\beta;\ all\ \beta\in\Lambda\}$ is also a subbasis for the cartesian product topology in $\Pi_{\lambda\in\Lambda}Y_\lambda$.

I think it suffices to prove that the aforementioned product topology is the smallest topology on that product such that it contains the set $\{\pi_{\beta}^{-1}(V_\beta)|\ all\ V_\beta\in\Sigma_\beta;\ all\ \beta\in\Lambda\}$. This result seems to be an "easy-to-prove" result. But I cannot find a way to write one such rigorous proof. Could someone please help? Thank you.

Janitha357
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1 Answers1

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Suppose $\mathcal{T}$ contains $\Sigma:= \{\pi_{\beta}^{-1}(V_\beta)| V_\beta\in\Sigma_\beta;\, \beta\in\Lambda\} \subset \mathcal{T}_{\text{prod}}$, and is minimally so. As $\mathcal{T}_{\text{prod}}$ contains $\Sigma$, we immediately get $\mathcal{T} \subseteq \mathcal{T}_{\text{prod}}$ for free.

Then if $O$ is open in $Y_\beta$, we can write

$$O = \cup \{\cap_{j \in F(i)} V_{j,i} \mid \text{ all } V_{j,i} \in \Sigma_\beta, F(i) \text{ finite } ,i \in I\}\text{,}$$

for some index set $I$. ($O$ is a union of basic elements formed from the subbase)

But then, as inverse images preserve unions and intersections:

$$\pi_\beta^{-1}[O] = \cup \{\cap_{j \in F(i)} \pi_\beta^{-1}[V_{j,i}] \mid \text{ all } V_{j,i} \in \Sigma_\beta, F(i) \text{ finite } ,i \in I\}\text{,}$$

and we know all $\pi_\beta^{-1}[V_{j,i}] \in \Sigma \subset \mathcal{T}$.

It follows (as topologies are closed under unions of finite intersections) that $\pi_\beta^{-1}[O] \in \mathcal{T}$ for all open $O \subset Y_\beta$ and all $\beta \in \Lambda$. So as $\mathcal{T}_{\text{prod}}$ is minimal with this property, (beause now $\mathcal{T}$) by definition makes all projections continuous e.g.) $\mathcal{T}_{\text{prod}} \subseteq \mathcal{T}$, and we have equality of topologies.

Henno Brandsma
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  • @HennoBrandsma how do you know that ${\pi_{\beta}^{-1}(V_{\beta})|V_{\beta}\in \Sigma_{\beta}; , \beta \in \Lambda } \subset \mathcal{T}{\text{prod}}$? And what does it mean "and is minimally so"? Do you mean $\mathcal{T}$ is the minimal topology that contains this set? Do you mean $\mathcal{T}{\text{prod}}$ is the minimal topology that contains this set? The language is not clear to me. –  May 15 '17 at 00:03
  • @ALannister because the product topology is the minimal topology that contains all sets of the form $\pi_\alpha^{-1}[O_\alpha]$ where $O_\alpha \subseteq Y_\alpha$ is open, and all members of this family are of this form so are certainly in the product topology. – Henno Brandsma May 15 '17 at 19:33
  • @ALannister By minimally so, I mean that $\mathcal{T}$ is the smallest topology that contains the family under discussion (so the topology generated by that subbase). The goal is to show $\mathcal{T} =\mathcal{T}{\text{prod}}$. That $\mathcal{T}$ is the minimal topology that contains the family and $\mathcal{T}{\text{prod}}$ is a topology containing it, gives the desired inclusion from left to right for this equality. – Henno Brandsma May 15 '17 at 19:33