# Is $\mathcal B_Y=\{B\cap(Y\times Y):B\in\mathcal B\}$ forms a base for $\mathcal U_Y?$

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.