I would appreciate if you could evaluate my sketch of the proof for the following problem:
For an arbitrary metric space $(X,d)$ and $Y\subset X$, show if $Y^c$ is closed, then $Y$ is open.
Sketch of the Proof:
(1) Given the metric space and $Y\subset X$, suppose $Y^c$ is closed.
(2) Suppose $Y$ is NOT open.
(3) By (2), $\exists y*\in Y$ s.t. $\forall\epsilon>0$ $B_\epsilon(y*)\not\subset Y$.
(4) By (3), $\forall\epsilon>0$ $\exists y_c\in Y^c$ s.t. $y_c\in B_\epsilon(y*)$.
(5) Since each open ball contains at least one member of $Y^c$, you can produce a sequence $(y_{c_n})\in Y^{c^\mathbb{N}}$ s.t. $y_{c_n}\rightarrow y*$.
(6) By (5), there exists a limit point $y^*$ that is not a member of $Y^c$.
(7) We assumed $Y^c$ to be closed so it contains all of its limit points.
(8) This is a contradiction.
(9) It follows that $Y$ must be open. QED.
Best Answer
It's correct, but you can avoid contradiction and simplify the proof.
Suppose $Y^c$ is closed and let $y\in Y$. Since $y\notin Y^c$ and $Y^c$ is closed, there is a ball $B_r(y)$ such that $B_r(y)\cap Y^c=\emptyset$.
Therefore $B_r(y)\subset Y$.
Hence $Y$ contains a ball centered at each of its points.