# The subbase theorem for total boundedness

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) :

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 for completely regular spaces from following facts:

1. 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.
2. A uniform space is compact if and only if it is totally bounded and complete.
3. The product of complete spaces is complete.
4. The product of uniform spaces is totally bounded if and only if each coordinate space is totally bounded. (Only this fact needs to use above theorem)

Any and all help is appreciated.

## 1 Answer

It seems the following.

The answer is positive. Let $\mathcal S$ be a subbase satisfying the condition and $U\in\mathcal U$ be an arbitrary entourage. Then there exists a finite subfamily $\mathcal V=\{V_1,V_2\dots, V_n\}$ of the family $\mathcal S$ such that $\bigcap\mathcal V\subset U$. For each member $V_i$ of the family $\mathcal V$ there exists a finite cover $\mathscr A_i$ of the set $X$ such that $A\times A\subset V_i$ for each $A\in\mathscr A_i$. Put $\mathscr A=\{A_1\cap A_2\cap\cdots\cap A_n: A_i\in \mathscr A_i$ for each $i\}$. Then $\mathscr A$ is a finite cover of the set $X$ such that $A\times A\subset V_i$ for each $A\in\mathscr A$ and each $i$. So $A\times A\subset \bigcap\mathcal V\subset U$ for each $A\in\mathscr A$.

• Nice. Note that this proof doesn't require the axiom of choice while the full version of Alexander's theorem does. – Ramiro de la Vega May 2 '15 at 21:57