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I started reading about uniform spaces in Bourbaki, and the closure operation has been defined as such:

$$\bar{A}=\bigcap_{V\in\mathcal{U}}V(A)$$

Where $\mathcal{U}$ is the uniform structure of the space, and $V(A)$ is the set of all left-relatives of $A$. One thing I don't understand is why $\bar{A}=\bar{\bar{A}}$, as is necessary for a closure operation. In particular, I don't see why $\bar{\bar{A}}\subseteq\bar{A}$.

After playing around with it for a while, I have this. Take some $s\in\bar{\bar{A}}$. So $s\in V(\bar{A})$ for all $V\in\mathcal{U}$. So taking any $V$, there is some $r\in\bar{A}$ such that $(s,r)\in V$. But $r\in V(A)$, so $(r,t)\in V$ for some $t\in A$. Is there some why to show that $(s,t)\in V$, to conclude that $s\in V(A)$? At best I can see that $(s,t)\in V\circ V$.

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up vote 4 down vote accepted

For any $V\in {\cal U}$ choose $U\in {\cal U}$ such that $U\circ U\subset V$. Then you proceed with your proof above with $U$ (instead of $V$).

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Ah, I should have seen that! Thanks TCL. –  yunone Nov 30 '10 at 4:44
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