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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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The closure of $X=\{(x,y) : xy<1\}$ is not $\{(x,y) : xy< 2\}$. It is $\{(x,y) : xy\le 1\}$, which is indeed closed. Also note that you cannot prove that a set is not closed just by showing that it is open; it might be both closed and open. –  MJD Oct 24 '12 at 3:38
    
@MJD, it is $\{(x,y) : xy\le 1\}$, but isn't $\{(x,y) : xy\le 1\}$ also contained in $\{(x,y) : xy < 2\}$ anyways? –  sidht Oct 24 '12 at 3:42
    
$\{(x,y):xy\le 1\}$ is contained in $\{(x,y):xy\lt 2\}$, but it is not equal to $\{(x,y):xy\lt 2\}$, and $\{(x,y):xy\lt 2\}$ is not the closure of $\{(x,y):xy\lt 1\}$ . –  MJD Oct 24 '12 at 3:50
    
Why isn't $\{(x,y) : xy < 2\}$ the closure? It contains $\{(x,y) : xy\le 1\}$ doesn't it? –  sidht Oct 24 '12 at 4:11
    
Yes, but it also contains a lot of other stuff. For example, it contains the point $(1, 1.5)$, which is not a limit point of $X$, since a small neighborhood of $(1, 1.5)$ is disjoint from $X$. –  MJD Oct 24 '12 at 4:14

2 Answers 2

up vote 2 down vote accepted

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.

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