Closed AND open subspaces of a normed vector space Let $E$ be a finite dimension normed vector space.
How can I show that $E$'s only both closed and open (norm-wise) subsets are $\emptyset$ and $E$ ?
 A: Here's a sketch of both the metric spacey and topologyey proof (that Daniel Fischer sketched a sketch of in the comments).
Suppose once and for all that $U$ is neither the whole plane nor empty. First, we need a preliminary:
Let $x \in U, y \in U^c$. Then there is a path between $x$ and $y$; that is, a continuous map $f: [0,1] \to E$ such that $f(0) = x$, $f(1) = y$. We're in a vector space, so this is particularly easy: let $f(t) = (1-t)x+ty$. (I'm not going to show that this is continuous, but you should check that it is.) 
Metric spacey proof
Now consider $f^{-1}(U)$. By the least upper bound property, there is a least upper bound $x$ of $f^{-1}(U)$. This might be in $f^{-1}(U)$ or not; it doesn't matter. Consider $f(u)$. Now check that $B(f(u),\varepsilon)$ intersects both $U$ and $U^c$ for all $\varepsilon$. This contradicts the fact that $U$ is open, and thus either $U$ was empty or $U^c$ was empty all along. 
Topologyey proof
Because $U$ is open and closed, and for an open set $U$ and continuous map $f$, $f^{-1}(U)$ is open, we see that $f^{-1}(U)$ and $f^{-1}(U^c)$ are both open. Then $[0,1] = f^{-1}(U) \cup f^{-1}(U^c)$; this contradicts the fact that $[0,1]$ is connected unless $f^{-1}(U)$ and $f^{-1}(U^c)$ are both empty, which contradicts that $f(0) \in U$ and $f(1) \in U^c$; so this must not have been true. When you learn about topological spaces in general, you'll see that this argument generalizes to the fact that path-connected spaces are connected.
Why is $[0,1]$ connected? Well, you have to start somewhere, and the proof looks essentially like the metric space proof above. You can find the full proof here; you have to be a little bit careful in general.
A: Let $U \subset E$ be both open and closed.  Take any $x\in E\setminus U$ and note that there exists a $z_0 \in U$ for which $|x-z_0| = \inf_{z\in U} |x-z|$ since $U$ is closed. O.t.o.h. since $U$ is open, it contains a Ball of radius $\epsilon>0$ centered at $z_0$. This contradicts the minimizing property.
