I presume you must show that $\emptyset$ and the vector space itself are the only open and closed subsets. Where do you begin?
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Under what topology? – Paul Nov 22 '17 at 02:09
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@Paul The weakest topology. This metric is defined in the natural way: the distance between two vectors u and v is given by ‖u−v‖. – Darkdub Nov 22 '17 at 02:11
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@Darkdub;Did my answer serve you.In case so you can accept it or ask for clarifications – Learnmore Nov 22 '17 at 12:29
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If $V$ is a normed space then for any $x,y\in V$ ,$tx+(1-t)y\in V$
hence considering the path $f:[0,1]\to V$ by $f(t)=tx+(1-t)y$, $V$ is path connected.
Alternatively:To show that $V$ is connected it is enough to show that $V$ has no non-empty open set.
Let $S(\neq \{0\},\neq V)$ be an open subset of $V$. Let $a\in S\implies \exists r>0$ such that $B(a,r)\subset S\implies \{y:||a-y||<r\}\subset S$.
Now if $z\in V$ consider the point $y=a+\frac{r}{2||z||}z\implies y\in S\implies z\in S\implies S=V$
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How would you prove this without showing V is path connected first? i.e. To directly prove that V is connected. – Darkdub Nov 22 '17 at 02:05
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problem with your proof is. You just assume that $S$ is an open set. unless it is a subspace. $y\notin S$ – Dec 26 '18 at 05:49
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The alternative approach is wrong. In fact what you have stated is false. For instance consider the real line $\Bbb R$ with standard Euclidean norm. According to your claim it will have no non-empty proper open subset. Right? But what about $(0,1)\ $? The correct statement should be A normed space cannot have any non-empty proper open subspace. – Anacardium Oct 16 '20 at 19:19