If $E$ is a normed space and $F$ is a subspace of $E$, how to prove that if $F\neq\{0\}$ then $F$ is not compact?
I begin by this let $x\in F$ then $F=\bigcup_{x\in F} B(x,\varepsilon)$
how to say that we can't deduce a finite cover ?
Thank you
2026-04-24 12:11:29.1777032689
Compactness in a vector space
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In any metric space a compact sub space is closed and bounded. (The reciprocal is not true in general). Take $0\neq x\in F$ and $n\in\Bbb{N}$. One has
$$\|n\cdot x\|=n\cdot\|x\|$$
And the sequence $y_n=n\cdot x\in F$ is not bounded ; therefore $F$ is not bounded henceforth not compact.