# In metric space, show if a set is open, then its complement is closed.

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.

• A closed set is by definition the complement of an open set. There's nothing to prove. – Evan Aad Oct 27 '16 at 21:05
• It is a lemma not a definition. – Frank Swanton Oct 27 '16 at 21:15
• what is your definition of closed? containing every limit points? – user251257 Oct 27 '16 at 21:31
• 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 $B$. – Berci Oct 27 '16 at 21:34
• 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. – Nitin Oct 27 '16 at 21:38

Maybe you should elaborate why (5) contradicts $Y$ being open.
We want to prove that $Y^\complement$ is closed, so take an arbitrary limit point $y^*$ of $Y^\complement$, and try to deduce that $y^*\notin Y$.