2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ And can you state the definition of a closed set? $\endgroup$ – Srivatsan Jan 21 '12 at 1:13
  • $\begingroup$ "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". $\endgroup$ – Arturo Magidin Jan 21 '12 at 1:34
  • $\begingroup$ And what is the definition of closed that you want to use, since you don't want to use "complements"? $\endgroup$ – Arturo Magidin Jan 21 '12 at 1:36
  • $\begingroup$ and what is the definition of $\overline{A}$? $\endgroup$ – Damian Sobota Jan 21 '12 at 1:45
  • $\begingroup$ @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? $\endgroup$ – user38268 Jan 21 '12 at 2:57
3
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.