Prove that complement of the interior equals closure of the complement

I tried as follows:

• If x is not in the interior of set E, then no neighborhood of x is contained in E. So for every neighborhood of x there is a point y which is not in E. That is, every neighborhood of x contains a point y which lies in complement of E. So x is a limit point of complement of E.

• -

On the other hand, if y is in closure of complement of E, then y is in E complement or set of all limit points of it. In the first case we are done since Int(E) is contained in E.

• In the second case, y is the limit point of complement of E. So every neighborhood of y contains a point which is not in E. Hence no neighborhood of y completely lies in E. Thus y is not an interior point of E.

Is this correct?? Thanks in advance.

• Your proof is kind of hard to read. It seems like it could use some paragraph breaks or something. – Chris Custer Jan 12 at 4:32
• I think it's correct. – Chris Custer Jan 12 at 4:36

Let X be discrete, E = X - {a}. Assume x in interior E.
In your 1st proof, you conclude x is a limit point of the
complement of E, namely {a} which has no limit points.

That example shows your proof is wrong.
You do however, have the right idea.
Use the right definition of closure:
the set of all adherence points.
a is an adherance point of A when for all
open U nhood a, U $$\cap$$ A is not empty.

• Oh yes. You are right. Whether this proof works?: Let x be not in Int(E). On the contrary assume that x is not in closure of E complement. Then there is a nbhd U of x such that U and complement of E have no points in common. Then U is contained in E so that x will be in Int(E) which is a contradiction – USK Jun 22 at 6:16
• @USK. Now you know the folly of limit points which topologically are a minor topic. – William Elliot Jun 22 at 6:26