The uniform space analogue of Alexander's subbase lemma on compact subbase is (As we know, Alexander subbase lemma can be used to prove Tychonoff's theorem) :
Theorem: Let $(X,\mathcal{U})$ be a uniform space such that for each member $U$ of some subbase for $\mathcal{U}$ there is a finite cover $A_{1}$, $A_{2}$,...,$A_{n}$ of $X$ such that $A_{i}\times A_{i}\subset U $ for each $i$. Then the space $(X,\mathcal{U})$ is totally bounded.
I was wondering how to prove this theorem. Is the proof of this theorem similar to Alexander's theorem?
Moreover, if above theorem holds we can get a new proof of Tychonoff's theorem from following facts:
- the product of uniform spaces is totally bounded if and only if each coordinate space is totally bounded. (Only this conclusion uses above theorem)
- a uniform space is compact if and only if it is totally bounded and complete.
- A topology $\mathcal{T}$ for a set $X$ is the uniform topology for some uniformity for $X$ if and only if the topological space $(X,\mathcal{T})$ is completely regular.
Anyway, any and all help is appreciated. Thanks.