Closure Operation on a Uniform Topology

I began reviewing consistent rooms in Bourbaki, and also the closure procedure has actually been specified thus :

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

Where $\mathcal{U}$ is the consistent framework of the room, and also $V(A)$ is the set of all left - loved ones of $A$. One point I do not recognize is why $\bar{A}=\bar{\bar{A}}$, as is essential for a closure procedure. Specifically, I do not see why $\bar{\bar{A}}\subseteq\bar{A}$.

After experimenting with it for some time, 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 kind of $V$, there is some $r\in\bar{A}$ such that $(s,r)\in V$. Yet $r\in V(A)$, so $(r,t)\in V$ for some $t\in A$. Exists some why to show that $(s,t)\in V$, in conclusion that $s\in V(A)$? At ideal I can see that $(s,t)\in V\circ V$.

0
2019-12-02 02:54:08
Source Share
Answers: 1

For any kind of $V\in {\cal U}$ pick $U\in {\cal U}$ such that $U\circ U\subset V$. After that you wage your evidence over with $U$ (as opposed to $V$).

0
2019-12-03 04:23:25
Source