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I was asked to prove the next claim:

let $(V,\|\cdot\|)$ be a normed vector space, and let $U \subseteq V$ be an open and bounded set. Prove: $\partial U \neq \varnothing$. (where $\partial U$ denotes the boundary of $U$).

I've been trying to prove this for some time but with no success.

I also know this claim is false in a general metric space, but cant seem to understand why taking $V$ to be a normed space makes this true.

Hints and suggestions will be highly appreciated!

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Hint: if $\partial U = \emptyset$, then this would imply that the space is disconnected.

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  • $\begingroup$ We haven't learned about connectivity yet... so I assume I should not use it in my proof. But I appreciate the comment. $\endgroup$ – Mike. R. Nov 8 '17 at 10:28
  • $\begingroup$ I was hoping that it might give you a related problem that could lead to a proof. One way to show that a normed linear space is connected is to show it's path-connected. You can form a path between $x \in U$ and $y \notin U$ by the line segment $tx + (1 - t)y$, for $t \in [0, 1]$. If $U$ and $V \setminus U$ are both open, then what happens at the changeover points? What about $\inf \lbrace t \in [0, 1] : tx + (1 - t)y \in U \rbrace$? $\endgroup$ – Theo Bendit Nov 8 '17 at 11:13
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Claim fails if $U = \varnothing$. Otherwise take $u \in U \setminus \{ 0 \}$ and observe that the set $\{ \alpha \in \mathbb{R} : \alpha \cdot u \in U \}$ is a bounded subset of $\mathbb{R}$, so it has a least upper bound $s$. It remains to prove that $s \cdot u \in \partial{U}$.

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  • $\begingroup$ But I can't understand why the boundnes of the above subset of $\mathbb{R}$ follows from the boundnes of U? $\endgroup$ – Mike. R. Nov 8 '17 at 8:50
  • $\begingroup$ the deffinition I have for boundnes is : $U$ is bounded (in $V$) if $\text{diam}(U)$ = $\sup\{\|x-y\|:\forall\ x,y\in U \}<\infty$ or: if there exsiset $v$ $\in$ $V$ and $r$ $\gt$ $0$ such that $U$ $\subseteq $ $B_r(v)$. $\endgroup$ – Mike. R. Nov 8 '17 at 9:28
  • $\begingroup$ @Mike.R. Yeah, take the second definition for instance. If we have $v \in V$ and $r > 0$ such that $U \subseteq B_r(v)$, then since $\| \alpha \cdot u - v \| \geqslant |\alpha| \| u \| - \| v \|$, we see that $\alpha \cdot u$ can only be in $u$ if $|\alpha| \| u \| - \| v \| \leqslant r$, that is: $|\alpha| \leqslant \frac{r+\|v\|}{\|u\|}$. $\endgroup$ – Adayah Nov 8 '17 at 19:07

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