Recall Tychonoff's Product Theorem:

Given a family of topological spaces $(X_{\alpha}, \mathcal{T}_{\alpha})_{\alpha \in \Lambda}$, $(\prod_{\alpha}X_{\alpha}, \mathcal{T}_{\text{prod}})$ (where $\mathcal{T}_{\text{prod}}$ is the Product topology) is compact if and only if each topological space $(X_{\alpha}, \mathcal{T}_{\alpha})_{\alpha \in \Lambda}$ is compact.

The proof of this theorem relies on Alexander's Theorem, which states that's a topological space on $X$ is compact if and only if every covering of $X$ by subbasic elements of $\mathcal{T}$ has a finite subcovering.

I was wondering what is the advantage to appealing only to subbasic sets in the proof of Tychonoff's Product Theorem? Does there also exist a proof that relies on the more traditional approach of every open covering implying a finite subcovering?

Thank you for indulging my curiosity!


Tacklings with covers often are based on structural properties of the covering sets. So very simple structure of cylinder sets forming a canonical subbase of the Tychonoff product allows to obtain an extremely short proof of Tychonoff theorem. For alternative proofs see this question.

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