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I've recently encountered the concept of uniform space in topology. I'm not familiar with it at all and I am trying to understand things a bit better.

Let $U$ be a uniform structure on $X$. For $x\in X$ and $W\in U$, let $W(x) = \{y:(x,y)\in W\}$. Let $\Sigma$ be the family of sets $S$ such that, for every $x\in S$, there is $W_x\in U$ satisfying $W_x(x)\subseteq S$. Then, it is not hard to check that $\Sigma$ is closed under finite intersection and arbitrary union. So, it defines a topology on $X$.

Now, fix $x\in X$. Let $N_x = \{W(x)\,:\,W\in U\}$. Then, from my understanding, $N_x$ should be a neighbourhood basis for $x$. It is clear to me that by definition of $\Sigma$, every neighbourhood of $x$ must contain an element of $N_x$. However, what I am struggling to show is that each $W(x)$ is actually a neighbourhood of $x$, which is necessary for $N_x$ to be a neighbourhood basis.

In practice, I need to show that for each $W(x)$, we can find a set $S\subseteq W(x)$ containing $x$ and such that, for any $x'\in S$, there is $W'\in U$ that satisfies $W'(x')\subseteq S$. I suspect that one can pick $S= W(x)$ and that this will be in $\Sigma$, but I am not sure and I don't see how to find such $W'$ for a generic $x'\neq x$ in $W(x)$.

ECL
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    In your definition of "uniform structure", there must be some axiom of the form "for every $U$ there exists $V$ with $V\circ V\subset U$". That's what you have to use (maybe multiple times) to get $W'$ from $W$. – Stefan Jan 14 '25 at 19:57
  • Yep, the axioms for the uniformity here are that for any $W\in U$ the diagonal of $X\times X$ is in $W$, for any $W\in U$ there is $V$ such that $V\circ V\subseteq W$, and if $W$ is in $W$ then $W^{-1}$ (its symmetric) is in $U$. I guess these properties are needed (so far only the def of filter is used to define a topology...), but still I don't see it as straightforward how to use them. – ECL Jan 15 '25 at 12:06
  • Yes, less straightforward than I thought yesterday. Examine my answer critically. – Stefan Jan 15 '25 at 20:56

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Define $$S=\Big\{x'\in X:\exists V\in U\text{ with }V(x')\subset W(x)\Big\}.$$ Let $x'\in S$ and choose $V\in U$ with $V(x')\subset W(x)$. Then we can choose $W'\in U$ with $W'\circ W'\subset V$. If now $y\in W'(x')$, we have $$W'(y)\subset V(x')\subset W(x),$$ i.e. $y\in S$. So $$W'(x')\subset S.$$

Stefan
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  • Great! That works indeed. As also clearly $x\in S$ and $S\subseteq W(x)$ by construction. I guess there is also some way to construct $W'$ by iterated uses of the composition axiom (as you had suggested in your comment), as this is what wikipedia quickly seems to mention (without doing it explicitly). But I like the alternative approach you give here! – ECL Jan 16 '25 at 10:17