Understanding the proof that if a set is closed its complement has to be open. The statement proven here is that: $A \text{ is closed} \implies A^{c} \text{ is open}$.
The proof given in class is this:
Suppose $A^{c}$ is not open, then it must be the case that for some $a \in A^c \implies \nexists N_{\delta}(a) \in A^c $. That is, for such $a \in A^c$ we cannot find a neighborhood of $a$, for all $\delta > 0$, where the neighborhood belongs to the set $A^c$.
What I don't understand from this proof is that why is it the case we can conclude, that $N_{\frac{1}{n}}(a)$ contains points $x_n \in A$, and $x_n$ approaches the point $a$ because $|x_n - a| < \frac{1}{n} \rightarrow 0$, thus concluding that $a$ is a limit point of $A$.
I know that once you prove $a$ is a limit point of $A$, because $A$ is closed, it forces $a \in A$ and hence it creates a contradiction. I guess I'm still confused on this neighborhood concept, what is a good way to explain this proof?
 A: This proof relies on choosing an element from each $N_{\frac{1}{n+1}}(a)$, this sequence must converge to $a$; but as you assumed that each element of the sequence is on $\mathbb{R}\setminus A$, you get an absurd as the set $A$ is closed and a convergent sequence in $\mathbb{R}$ must have an unique limit. Proceed as follows.
If $\mathbb{R}\setminus A$ is not open, then one can find an element $a\in \mathbb{R}\setminus A$ such that for every $\varepsilon > 0$, $N_\varepsilon (a) \not\subset A$. As $\varepsilon$ is arbitrary, for every $n\in\omega$ the set $N_{\frac{1}{n+1}}(a)$ has some element $x_n$ such that $x_n\not\in \mathbb{R}\setminus A$, and, thus is an element of $A$; as $A$ is a closed set, the sequence $\langle x_n \rangle _{n\in\omega}$ converges to some point $x\in A$ which is an absurd, as $x_n\to a.$
A: By definition of a closed set A, the set A contains all its limit points. 
Let t be a limit point and an element of A', i.e. the complement set of A. (note: it is necessary that we make this limit point t an element of A' as the definition of an open set (the set S is open if for every limit point t that is an element of the set S, there exists a delta neighborhood such that the set S contains this specific delta neighborhood), and when we negate this definition to provide a proof by contradiction the assumption that the limit point is the element of the set is still there)
Assume A' is not an open set then for all delta A' does not contain any delta neighborhoods of t. This means that the intersection of the delta neighborhoods of t and the set A is not an empty set. By the definition of a limit point (t is a limit point of S for every delta neighborhood of t iff the intersection of the various delta neighborhoods of t and set A is not an empty set) t is a limit point of set A. Since A is a closed set, t is an element of A which means t is not an element of A' therefore a contradiction. 
