# Closure in topology proof?

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.

• 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? – user38268 Jan 21 '12 at 2:57

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.