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I want to prove $A$ is closed iff $\overline{A}=A$;

I need to use the definition of neighbourhoods instead open sets and not use the complement to prove this.

So wondering how can you prove it just using closure I can't see how it is possible.

Edit it asks you to use the definition of closure directly. Suppose that X is a topological space and A is a subset of X. A point $x \in X$ is a closure points of A if for all neighbourhoods N of x, $N\cap A \not= \emptyset$. The set of closure points of A is the closure of A.

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And can you state the definition of a closed set? –  Srivatsan Jan 21 '12 at 1:13
    
"Compliment" is what you would tell someone in order to flatter them into solving the exercise for you. "Complement" is when you take "everything that is not in the set". –  Arturo Magidin Jan 21 '12 at 1:34
    
And what is the definition of closed that you want to use, since you don't want to use "complements"? –  Arturo Magidin Jan 21 '12 at 1:36
    
and what is the definition of $\overline{A}$? –  Damian Sobota Jan 21 '12 at 1:45
    
@simplicity Do you know that the closure of a subset A of a topological space X is the intersection of all closed sets that contain A and therefore is the smallest closed set containing A? –  fpqc Jan 21 '12 at 2:57
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up vote 3 down vote accepted

Suppose that $A$ is closed, so that it contains all of its accumulation points. Then $\bar{A}$, which is $A$ together with its accumulation points, is just $A$.

Conversely, if $A = \bar{A}$, then $A$ contains all of its accumulation points. By definition, $A$ is closed.

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