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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 $E$ ?

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    $\begingroup$ Show that $E$ is (path-)connected. $\endgroup$ – Daniel Fischer Dec 10 '14 at 22:54
  • $\begingroup$ @DanielFischer I do not know much about norms. I know their defintion and some basic properties such as the equivalence of norms (in finite dimension) etc. I do not know what "path connected" refers to. $\endgroup$ – Hippalectryon Dec 10 '14 at 22:55
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    $\begingroup$ "Path connected" means you can join any two points with a path, i.e. for all $x,y\in E$ there is a continuous $\alpha\colon [0,1]\to E$ with $\alpha(0) = x$ and $\alpha(1) = y$. Path-connectedness implies connectedness (which is equivalent to $\varnothing$ and $E$ being the only clopen subsets). $\endgroup$ – Daniel Fischer Dec 10 '14 at 23:00
  • $\begingroup$ It seems that you have accepted Mike Miller's answer but not familiar with path connected or connectedness. Without any supplementary knowledge you can do something like this : Let $u\in A$ and $v\in E-A$. Denote $S:=\{\lambda v +(1-\lambda)u: \lambda \in [0,1]\}$ and $f:[0,1] \rightarrow S, \lambda \rightarrow \lambda v +(1-\lambda)u, B:=A\cap S, C:=(E-A)\cap S, I:=f^{-1}(B), J=f^{-1}(C)$. And we just need to prove that $I \cap J=\emptyset, I\cup J= [0,1], 0\in I, 1\in J$ and to conclude : $\sup(I)\in I$. $\endgroup$ – Krokop Dec 15 '14 at 20:57
  • $\begingroup$ @Krokop I was able to talk a bit with Mike Miller in the chat, he explained me the points I needed, but thanks for that alternative way. $\endgroup$ – Hippalectryon Dec 15 '14 at 20:59
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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.

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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.

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  • $\begingroup$ I intuitively understand where your first argument comes from, but how would you prove it ? ("there exists a $z_0\in U$ for which $|x−z_0|=\inf z\in U|x−z|$ since $U$ is closed". To me, a closed is just the complementary of an open.) $\endgroup$ – Hippalectryon Dec 10 '14 at 23:10

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