A question related to product topology 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.
 A: 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.
