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Suppose $(X,\mathcal{D})$ is a uniform space and $D\in\mathcal{D}$. Is it true that $$\overline{D}\subseteq D\circ D,$$?

here we use the product topology to define $\overline{D}$.

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If $D\subset X\times X$, then $\overline{D}=\bigcap\{U\circ D\circ U: U\in\mathcal{D}\}$. hence $\overline{D}\subset D\circ D\circ D$. I'm not sure but i guess $\overline{D}\subset D\circ D$ is not true. –  M.Sina Feb 11 '13 at 18:42
    
@M.Sina: Do you have a counterexample? –  user59671 Feb 11 '13 at 18:44
    
not yet. i think about it. –  M.Sina Feb 11 '13 at 18:48

1 Answer 1

I'm not sure but I think this is a counterexample:

Let $\mathcal D$ be the normal uniformity on $\Bbb R$ made by the Euclidean metric $d$. Let

$$D=U_d(1)\cup \Bbb Q^2$$ where $$U_d(1)=\{(x,y)\in \Bbb R^2 \mid d(x,y)\le 1 \}$$

Then $$(-\sqrt 2,\sqrt 2)\in \overline D$$ But $$(-\sqrt 2,\sqrt 2)\notin D\circ D$$

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Why are you not sure? –  Colin McQuillan Apr 3 '13 at 13:04

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