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Let $(X,\mathcal U)$ be a uniform space and let $U\in \mathcal U$. Is this statement true? $$\forall V\in \mathcal U, \exists W\in \mathcal U, U\circ W\subseteq V\circ U$$

I think if the above statement is true then we can easily prove: $$\forall V\in \mathcal U, \exists W\in \mathcal U, W\circ U\subseteq U\circ V$$

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  • $\begingroup$ 1) I am a bit confused between the text and the title of your "question". $\endgroup$
    – András Bátkai
    Apr 2, 2013 at 14:21
  • $\begingroup$ What do you mean exactly by $VU$ and $UW$? The product of these entourages? Are you working in a topological group? $\endgroup$
    – Filippo Alberto Edoardo
    Apr 2, 2013 at 14:30
  • $\begingroup$ $UV$ is $UoV$. I'll edit my question. $\endgroup$
    – user31968
    Apr 2, 2013 at 14:50
  • $\begingroup$ Need some motivation. Why do you care about this result? Why is this "research-level?" $\endgroup$
    – David White
    Apr 2, 2013 at 14:59
  • $\begingroup$ I just tried many cases to find a counterexample for this. But I could not. I think finding a counterexample is not simple. $\endgroup$
    – user31968
    Apr 2, 2013 at 15:02

2 Answers 2

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The second statement follows from the first one by passing to inverses (i.e. reflecting along the diagonal).

Now consider the uniform structure on $\mathbb{R}$ consisting of all subsets of $\mathbb{R}^2$ that contain an open neighborhood of the diagonal.

Just to avoid confusion with left and right, I want to use the notation $U\circ V = \{(x,z)| \exists y: (x,y)\in U \wedge (y,z)\in V\}.$

Let $U:=\{(x,y)| |y-x|<1 \}\cup \mathbb{R}\times \{0\}$. $V:=\{(x,y)| |y-x|<1 \}$. Then $V\circ U=\{(x,y)| |y-x|<2 \}\cup \mathbb{R}\times \{0\}$.

Now let $W$ be any entourage and let $Z$ be some open neighborhood of $0$ such that $\{0\}\times Z\subset W$. Then $U\circ W$ will contain $\mathbb{R}\times Z$. Thus $U\circ W$ will not be contained in $V\circ U$. Since $W$ was arbitrary this should be a counterexample.

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  • $\begingroup$ My tries teached me: $U o W = (W^{-1}oU^{-1})^{-1}=(WoU)^{-1}$ and not $WoU$. too common mistake I made too! $\endgroup$
    – user31968
    Apr 3, 2013 at 14:02
  • $\begingroup$ And for second part $(100,0.5)\in VoU$ because $(100,0)\in U$ and $(0,0.5)\in V$ but $(100,0.5)\notin \{(x,y)| |y-x|<2 \}\cup (\mathbb{R}\times \{0\})$. So $$V\circ U \ne \{(x,y)| |y-x|<2 \}\cup (\mathbb{R}\times \{0\})$$ $\endgroup$
    – user31968
    Apr 3, 2013 at 14:14
  • $\begingroup$ @CC Thank you comments. I agree with your first comment. However your second comment suggests that your definition of $U\circ V$ is my definition of $V \circ U$. (To avoid this I included my definition of $U\circ V$). I still think this is a counterexample. $\endgroup$
    – HenrikRüping
    Apr 3, 2013 at 16:46
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I recently found an answer to a similar question. Suppose: $$(\forall V\in \mathcal U)(\exists W\in \mathcal U)(U\circ W\subseteq V\circ U)$$ By axiom of choice,for each $V \in \mathcal U$, there's some symmetric $D_V\in \mathcal U$ such that $$D_V\circ D_V\subseteq V$$

and there's some symmetric $W_V\in \mathcal U$ such that:

$$U\circ W_V \subseteq D_V\circ U$$ and $$W_V\subseteq D_V$$

so $$W_V\circ U\circ W_V \subseteq W_V\circ D_V\circ U\subseteq D_V\circ D_V\circ U\subseteq V\circ U$$

Therefore

$$\overline U=\bigcap_{W\in \mathcal U}W\circ U \circ W\subseteq \bigcap_{V\in \mathcal U}W_V\circ U \circ W_V\subseteq \bigcap_{V\in \mathcal U}V\circ U\subseteq U\circ U$$

Now see this thread.

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