1
$\begingroup$

I am writing down a sketch of the proof to the following problem in words and would appreciate your 2cent!

Given an arbitrary metric space $(X,δ)$, show that if a set $Y \subset X$ is open, then its complement is closed.

Sketch of the proof (in words):

(1) Given the arbitrary metric space, suppose $Y\subset X$ is open.

(2) Suppose $Y^c$, the complement of $Y$, is not closed.

(3) By (2), there exists a limit point of $Y^c$ that is not contained in $Y^c$. Denote this point $y^*$.

(4) $y^*$ is in $Y$, and there exists a sequence in $Y^c$ that converges to $y^*$.

(5) By (4), $\forall\epsilon>0$, $\exists y_c\in Y^c$, $y_c\in B_\epsilon(y^*)$.

(6) But we assumed Y is open.

This is a contradiction. It must be that $Y^c$ is closed. QED.

$\endgroup$
  • 1
    $\begingroup$ A closed set is by definition the complement of an open set. There's nothing to prove. $\endgroup$ – Evan Aad Oct 27 '16 at 21:05
  • $\begingroup$ It is a lemma not a definition. $\endgroup$ – Frank Swanton Oct 27 '16 at 21:15
  • $\begingroup$ what is your definition of closed? containing every limit points? $\endgroup$ – user251257 Oct 27 '16 at 21:31
  • $\begingroup$ I guess, $A$ is open if $\forall a\in A\,\exists\epsilon>0: B_\epsilon(a)\subseteq A$, and $B$ is closed if all its limit points belong to $B$. $\endgroup$ – Berci Oct 27 '16 at 21:34
  • $\begingroup$ In my very first analysis class, we defined an open set as one in which every point is an interior point and a closed set in which every boundary point is in the set. $\endgroup$ – Nitin Oct 27 '16 at 21:38
2
$\begingroup$

So far so good.
Maybe you should elaborate why (5) contradicts $Y$ being open.

Otherwise, if done, I also suggest to translate the same to a direct proof:
We want to prove that $Y^\complement$ is closed, so take an arbitrary limit point $y^*$ of $Y^\complement$, and try to deduce that $y^*\notin Y$.

$\endgroup$
  • $\begingroup$ Is it possible to prove the "A set is closed if and only if its complement is open" without using the notion of metricspace ? $\endgroup$ – NewBornMATH Apr 11 '19 at 7:36
  • 1
    $\begingroup$ Yes, moreover in general topological space, it is already defined so. $\endgroup$ – Berci Apr 11 '19 at 12:30

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.