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I have a question... If given a set $A$ the closure $\bar{A}$ is $\bigcap_{i\in I} A_i$, where $A_i$ is closed and for all $i$, $A \subset A_i$, how to prove the equivalence with the definition that $\bar{A} = A \cup \left\{ \text{limit points for $A$}\right\}$?

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  • $\begingroup$ From wiki: ("Left subset")Suppose x is in the closure of S. If x is in S, we are done. If x is not in S, then every neighbourhood of x contains a point of S, and this point cannot be x. In other words, x is a limit point of S and x is in L(S). ("Right subset") If x is in S, then every neighbourhood of x clearly meets S, so x is in the closure of S. If x is in L(S), then every neighbourhood of x contains a point of S (other than x), so x is again in the closure of S. $\endgroup$ – GAVD Aug 13 '15 at 9:55
  • $\begingroup$ Just for "logic", assuming x is in the closure of S and x is not in S why "every neighbourhood of x contains a point of S"? $\endgroup$ – user8469759 Aug 13 '15 at 9:59
  • $\begingroup$ $x$ is a point of closure of $S$ if every neighbourhood of $x$ contains a point of $S$. $\endgroup$ – GAVD Aug 13 '15 at 10:05
  • $\begingroup$ And... how come the union of all closure points for S is equivalent to the intersection of all closed sets containing S? $\endgroup$ – user8469759 Aug 13 '15 at 10:14
  • $\begingroup$ Uhm, you have the equivalent definitions, see this (Wolfram)[mathworld.wolfram.com/SetClosure.html]. Perharps it helps. $\endgroup$ – GAVD Aug 13 '15 at 10:20
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For a set $B$ let $B'$ denote the set of limitpoints of $B$.

Set $B$ is closed if and only if $B'\subseteq B$.

Let $F_{0}$ be a closed set with $A\subseteq F_{0}$. Then $A'\subseteq F_{0}'\subseteq F_{0}$ so that $A\cup A'\subseteq F_{0}$.

Proved is now that: $$A\cup A'\subseteq\cap\left\{ F\mid F\text{ is closed and }A\subseteq F\right\} $$


Conversely $b\notin\cap\left\{ F\mid F\text{ is closed and }A\subseteq F_0\right\} $ implies the existence of some closed $F_{0}$ with $A\subseteq F_0$ and $b\notin F_{0}$.

Then $F_{0}^{c}$ is an open set with $b\in F_{0}^{c}$ and $F_{0}^{c}\cap A=\varnothing$. This implies $b\notin A\cup A'$.

Proved is now that: $$\cap\left\{ F\mid F\text{ is closed and }A\subseteq F\right\} \subseteq A\cup A'$$

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