Closure of $A$ with respect to uniform topology

Question

Theorem: Let $(X,D)$ be a uniform space and $A\subset X$. The closure of A with respect to the uniform topology is equal to the intersection of the uniform neigborhoods of A. That is,

$\overline A=\bigcap_{U\in D} U(A)$.

Proof: We know that $x\in U(y) \iff y\in U^{-1}(x)$ and for all $U\in D$, $U^{-1}\in D$.

$x\in\overline A$ $\iff$ $\forall U\in D$, $U(x)\cap A\neq\emptyset$

$\iff$ $\forall U\in D$, $\exists y\in A : y\in U(x)$

$\iff$ $\forall U\in D$, $\exists y\in A : x\in U^{-1}(y)$

$\iff$ $\forall U\in D$, $x\in U^{-1}(A)$

(*)$\iff$ $\forall U\in D$, $x\in U(A)$

This proof is in my textbook. But I do not understand the (*) part. Why does it look like the sets $U(A)$ and $U^{-1}(A)$ are equal? Does the set $U$ have to be symmetric?

Thank you..

Answer 1

Suppose that

$$\forall U \in D: x \in U^{-1}(A) (\ast)$$

We want to show that $\forall U \in D: x \in U(A)$. So pick any $U \in D$. Then $U^{-1} \in D$, so we apply $\ast$ to $U^{-1}$ and we get that $x \in (U^{-1})^{-1}(A)$. But he latter means exactly that $x \in U(A)$ and as $U$ was arbitrary we are done. The reverse is entirely similar.

Answer 2

Each individual set $U(A)$ may well be different from $U^{-1}(A)$. But we are talking about all elements of $D$ here. And, for each $U\in D$, $U^{-1}\in D$.