Let E be a set of real nunbers. If E is not closed, show that E is an infinite set. This question was from the previous midterm and I have no idea where to start. I was thinking of using the theorem that a closed set contains all of its limit points or if a set is closed then its compliment is open. 
 A: If $E$ is not closed, there is a limit point of $E$ which is not contained in $E$. Most importantly, there is a limit point of $E$. Finite sets have no limit points. 
A: Probably the simplest argument is that a singleton point set in $\mathbb{R}$ is closed, and finite unions of closed sets are closed.  Therefore every finite set is closed, and the contrapositive of this says:

If set $E$ (in the real numbers with usual topology) is not closed, then $E$ is not a finite set.

It is fairly easy to show a singleton point set $\{x\} \subset \mathbb{R}$ is closed.  One argument would be to note that the complement $\mathbb{R} \setminus \{x\}$ is open, e.g. the union of two open intervals $(-\infty,x)\cup (x,+\infty)$.
Alternatively, if the finitely many points of a set $E$ were ordered:
$$ x_1 \lt x_2 \lt \ldots \lt x_n $$
then we could argue $\mathbb{R}\setminus E = (-\infty,x_1)\cup (x_1,x_2) \cup \ldots \cup (x_n,+\infty)$ is open (as the union of open intervals).  Then $E$ is again closed.
A: Welp, the easy way is that every singleton is closed. (Although it wouldn't hurt to argue why that is.) So ever finite union of singletons is closed.  Every finite set is a finite union of singletons and thus closed. So if a set isn't closed it can't be finite.
But if you didnt think of that, as I didn't.  The direct way is:
If $E$ is not closed then there is limit point $p$ of $E$ with $p \not \in E$. Let $\epsilon > 0$.  We will construct a sequence of $\epsilon_i$ and $q_i \in E$ so that $\epsilon_0 = \epsilon$.  As $p$ is a limit point there is a $q_0 \in N(\epsilon_0, p)\cap E$.  We will define $\epsilon_1 = |p-q_0|$ so $\epsilon_0 > \epsilon_1 > 0$.  Inductively we will select $q_i \in N(\epsilon_i, p)\cap E$ and inductively we will set $\epsilon_{i+1} = |p-q_i|$.  Thus by induction $\epsilon_0 > \epsilon_2 > .... > 0$ and there are infinite $q_i$.  So $\{q_i\}\subset E$.  So $E$ is infinite.
