# Separation axioms in uniform spaces

I have some problems understanding the proof of the following lemma:

Lemma: Let $x \in X, \ \ \ U, W \in \mathcal{U}, \ \ \ \mathcal{T(U)}$ is the topology on $X$. If there exists $V \in \mathcal{U}$ such that $U \circ V \subset W$, then $U[x] \subset intW[x]$ and $\overline{U[x]} \subset W[x]$.

$\mathcal{U}$ is a uniform structure on $X$

$\mathcal{U} \subset 2^{X \times X}$ is a uniform structure on $X$. It is a filter on $X \times X$ satisfying:

$(1) \ \forall U \in \mathcal{U}: \Delta(X) \subset U, \\ (2) \ \forall U \in \mathcal{U}: \ U^{-1} \in U, \\(3) \ U \in \mathcal{U} \Rightarrow \exists V \in \mathcal{U} : \ V \circ V \subset U$.

$U[x] = \{ y \in X \ | \ (x,y) \in U\}$

If $\mathcal{B}$ is a basis of $\mathcal{U} \ \ \ \$ ($\ \forall U \in \mathcal{U} \ \exists B \in \mathcal{B} : \ B \subset U \$),

then $\mathcal{B_x}: = \{U[x] \ | \ U \in \mathcal{B}\}$ satisfies all conditions of basis for the neighbourhood system of $x \in X$ and $\mathcal{T(U)}$ mentioned above is a topology for which $\mathcal\{{B_x}\}_{x \in X}$ is a basis for the neighbourhood system in this topology.

I hope my question is clear now.

Could you help me understand the proof of this lemma?

My problem is that I don't have a clear picture of what int$W[x]$ or $\overline{U[x]}$ should be.

Here is the proof:

Let $y \in U[x], \ z \in V[y]$. Then $(x,z) \in U \circ V$, so $(x,z) \in W$. Thus $U[x] \subset intW[x]$.

Let $y \in \overline{U[x]}$. Then $\forall Z \in \mathcal{U} \ \ \exists x_Z \in U[x] \cap Z[y]$, which means that $(x, x_Z) \in U, \ (y, x_Z) \in Z$.

We choose $Z$ such that $Z = Z^{-1}$ and $Z \subset V$. Then $(x,y) \in U \circ Z \subset W$.

• I'm sorry it took so long. I haven't noticed your comment. I've already edited my question. I hope it's comprehensible now. – Bilbo Sep 8 '13 at 16:11
• Yes, I guess it's a very nice and simple way to put it. I just wanted to be as clear as possible :) – Bilbo Sep 8 '13 at 17:03

## 1 Answer

The proof given is some basic information about neighborhoods in a uniform space. Probably it's a hint.

well I prefer the normal notation $V\circ U$ for your $U\circ V$. The first part seems trivial: $$V\circ U\subseteq W \to$$ $$V[U[x]]=(V\circ U)[x]\subseteq W[x] \to$$ $$\bigcup_{y\in U[x]}V[y]\subseteq W[x] \to$$ $$U[x]\subseteq W[x]^o$$

Last $\to$ is because $V[y]$ is a neighborhood of $y$ as said in the hint.

For the second part, I prefer to use nets.

• I'm sorry, I forgot to explain what I mean by $\circ$. We use different order for composition of functions and different for relations. Thank you for your answer. How would you prove the second part using nets? My knowledge about uniform spaces is really basic. – Bilbo Sep 8 '13 at 16:52
• Maybe it's a stupid question, but I'll ask it anyway. When it comes to the second part of the original proof, here is how I understand it. $y \in \overline{U[x]}$, so $U[x]$ intersects every neighbourhood of $y$ which is $Z[y]$. My question is - why can we choose $Z$ satisfying the conditions mentioned in my post? – Bilbo Sep 8 '13 at 17:01
• Oh, right, indeed. Everything is ok now, I think. Thank you a lot. – Bilbo Sep 8 '13 at 17:06
• you're welcome. – user79193 Sep 8 '13 at 17:07