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Is it true that the closure of any set is closed? I am just assuming this fact from the word closure. My whole proof based on this fact

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Proof

Let $A_1 =(a_n)_{n \in\mathbb{N}} = \{ a_n :n \geq 1\}$ and $A_2 = \{ a\}$ where $A_2$ contains all (and the only one) the limit points of $A_1$

Hence the closure of $A_1$ is $\bar{A_1} = A_1 \cup A_2$ and $\bar{A_1}$ is closed

I am thinking that I should even omit that silly last conclusion that "closure is closed"

EDIT: I just came up with a counterexample to my own argument. What if

$X = \{ (x,y) : xy < 1\}$. Technically $\bar{X} = \{ (x,y) : xy < 2\}$ is open, but it is also the closure

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    No, there's only one closure of $X$, and it's the *smallest* set that contains $X$ and all the limit points of $X$. If it could contain extra stuff, it wouldn't be "the" closure, it would be "a" closure.2012-10-24

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Well, since (the or one of the equivalent) definition of closure of a set $\,A\subset X\,$ , in a topological space $\,X\,$ is

$\overline A:=\bigcap_{A\subset B\,,\,B\,\text{closed}}B$

and any intersection of closed sets is closed, then $\,\overline A\,$ is closed, too.

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Depends on how you define closure vs. how you define closed. If you define $\bar{A}$ as the smallest set $B \supset A$ which is closed, then it's obvious that the closure is closed.

But there are conceivable definitions of closure for which that isn't immediatly obvious. Like maybe defining the closure of $A$ as $A$ plus all limit points of $A$, or something like that, while at the same time defining closed as "the complement is open". Still, showing that the closure is indeed closed would then probably be one of the very first collorary of the definition, if only to prove that the name of the operation is actually sensible.