Every element of the uniformity include a cube of a symmetric element

Question

The following questions is from the book "General Topology" written by John L.Kelly and located in Page 179 in the proof of Thm 6.6.

Given ($X, \tau$), for definition of uniform space built up from $X$ and the uniform topology, please refer to the wiki link.

Given ($X, \tau$) and its uniformity $U_X$, define a product "$\circ$" between two elements in $U_X$ by

$U \circ V =$ {($x, z$)|$\exists y \in X$ such that ($x, y) \in U$ and ($y, z) \in V$} and by saying "$U$ cube", I mean $U \circ U \circ U$.

For any $U \in U_X$, define $U^{-1}$ = {($y, z$)|($x, y$) $\in U$}. An element $U \in U_X$ is symmetric iff $U$ = $U^{-1}$.

Once we have ($X, \tau$), $U_X$ and the uniform topology (say $\tau_{U}$), we can equip $X \times X$ with the product uniform topology.

Used facts:

1. Thm 6.4 on Page 178: If we equip $X$ with the uniform topology, then the interior of $A \subset X$ is {$x \in X$|$\exists U \in U_X$ such that $U[x] \subset A$}

2. Lemma 6.1 on Page 176: If $V$ is symmetric, then $V \circ U \circ V$ = $\cup${$V[x] \times V[y]$|($x, y$) $\in U$}

Below I will present the complete proof of Thm 6.

Thm 6: Let ($X, \tau$) be a topological space and define $U_X$ to be $X$'s uniformity. Let $U$ be an element of $U_X$ and then show that the interior of $U$ (relative to the uniform topology) is also a member of $U_X$. Consequently the family of all open symmetric members of $U_X$ is a base for $U_X$.

Proof: Equip $X \times X$ with the product uniform topology and the interior of a subset $M$ of $X \times X$ is the set of all ($x, y$) such that, for some $U$ and some $V$ in $U_X$, $U[x] \times V[x] \subset M$. Since $U \cap V \in U_X$, the interior of $M$ is {($x, y$)|$V[x] \times V[y] \subset M$}. If $U \in U_X$ there is a symmetric member $V \in U_X$ such that "$V$ cube" $\subset M$ and then by lemma 6.1, $V$ cube = $\cup${$V[x] \times V[y]$|($x, y$) $\in V$}. Hence, every point of $V$ is an interior point of $U$ and, since the interior of $U$ contains $V$, it is a member of $U_X$.

Here the bold sentence in the proof is where I get confused. Please believe that before Thm 6.6, there are no proofs for this statement and how can I prove that for any $U \in U_X$ there is a symmetric $V \in U_X$ such that "$V$ cube" $\subset U$.

Answer 1

In the list of axioms that you link to, say $U$ is in the uniformity, then by axiom 4 there is $V$ (all sets listed will be in the uniformity) such that $V^2\subseteq U$. By axioms 5 and 3 we have that $V\cap V^{-1}$ is in the uniformity, so we may replace $V$ with $V\cap V^{-1}$, the latter set is symmetric, and so we may assume there is a symmetric $V$ with $V^2\subseteq U$. Could you push this a bit further, to get that there is a symmetric set in the uniformity say $W$, such that $W^3\subseteq U$?

Hint. Try $W^2\subseteq V$.