In course of self-studying uniform space I have been stuck in some fundamental question:

Let $(X,\mathcal U)$ be a uniform space and $Y\subset X.$ Then $\mathcal U_Y=\{U\cap(Y\times Y):U\in\mathcal U\}$ forms an uniformity on $Y$ to be called the relative uniformity on $Y.$

Claim: Suppose $\mathcal B$ be a base for $\mathcal U.$ Then is $\mathcal B_Y=\{B\cap(Y\times Y):B\in\mathcal B\}$ forms a base for $\mathcal U_Y?$

I think this is true, but to prove my claim it suffices to show two things:

  • [1] $\mathcal B_Y$ forms a base for some uniformity on $Y$,

  • [2] $V\in\mathcal U_Y\implies B\subset V$ for some $B\in\mathcal B_Y.$

[2] is immediate since $V\in\mathcal U_Y\implies V=U\cap(Y\times Y)$ for some $U\in\mathcal U\implies$ $V\supset A\cap(Y\times Y)$ for some $A\in\mathcal B.$

But I could not prove [1].That is why I am a little skeptical about my claim.

Is my claim correct?

In case it is, how to prove [1]?


You don't need to prove (1). You already know $\mathcal{U}_Y$ is a uniformity. To see $\mathcal{B}_Y$ is a base for it, you just note that indeed $\mathcal{B}_Y \subseteq \mathcal{U}_Y$ (true by definition) and if $U \cap (Y \times Y)$ is a member of $\mathcal{U}_Y$ then for some $B \in \mathcal{B}$ we have that $B \subseteq U$ as $\mathcal{B}$ is a base for $\mathcal{U}$. But then clearly $\mathcal{B}_Y \ni B \cap (Y \times Y) \subseteq U \cap (Y \times Y)$ as required.

This is all one needs.


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