Equivalence of definitions for closed set. I was given definition number $1$ and it'd extremely useful to me if I could prove the equivalence of the following two:


*

*A set $K$ is closed if and only if its complement is open.


*A set $K$ is closed if every convergent sequence $(x_n)\subseteq K$ converges to $x\in K$.

A set is $A$ is open if every $p\in A$ has a neighborhood $N_r(p)\subseteq A$.
I don't know for sure if these two are equivalent, but I do think so.
Could someone help me out?
 A: Suppose $K$ is closed according to definition 1. So we know $X \setminus K$ is open (I call the whole space $X$, for definiteness). So for every $x \notin K$ we have an $r > 0$ such that $N_r(x) \subseteq X \setminus K$, or equivalently, $N_r(x) \cap K = \emptyset$.
Now suppose we have a sequence $(x_n)$ where all $x_n \in K$, and this sequence converges to $x$. If $x \notin K$, use the $r > 0$ from above to see that a tail of the sequence lies in $N_r(x)$, and so does not lie in $K$. Contradiction, so $x \in K$.
Now suppose $K$ is closed according to definition 2. So we have to show, for $x \notin K$, an $r >0 $ as above (to see $X \setminus K$ is open etc.). Suppose no such $r$ exists, then for every $n$, $r = \frac{1}{n} > 0$ cannot work, so $N_r(x)$ intersects $K$; pick $x_n \in N_{\frac{1}{n}}(x) \cap K$. But then we can show (do this!) that $x_n \rightarrow x$, and $x \notin K$, while all $x_n \in K$, so this contradicts definition 2. So such an $r>0$ does exist.
A: Hint: Prove that if a set contains all of limit points (closed) then the complement is open (for each point in complement, there exists a Nbhd that is contained entirely in the complement). 
