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This question is from topology of metric spaces by s.kumersean page-90 chapter-compactness .

Is a nonzero vector subspace of a nonzero NLS is compact?

Honestly I don't know how to show that because what i have read so far I don't find any link how to prove this. But if i consider (V,|| ||) is a NLS and because it is not provided wheather the vector space is finite-dimensional or not and if it is finite dimensional and if dimV=n then V is isomorphic to $R^n$. So the subspace have to be closed and bounded for this case. This is what i know so far. Intuitively i dont think so this is compact at all but don't find any way to show that.

Sinchan Bhattacharjee
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  • What is a NLS? Care to give a definition? – J. De Ro Jun 13 '20 at 10:55
  • Let V be a vector space over R or C. A norm on V is a function || ||: V —R satisfying the conditions (i)||x||$\geq$0 for all x in V (ii) ||ax||=|a| ||x|| where a is the element of the field and (iii) || || satisfies triangle inequality .The pair (V, II II) is called a normed linear space, or NLS in short – Sinchan Bhattacharjee Jun 13 '20 at 11:16
  • Ah normed linear space. – J. De Ro Jun 13 '20 at 11:18

1 Answers1

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No, this is never compact.

Fix a non-zero vector $x$ in your subspace. Then the sequence

$$x_n:= nx, \quad n \geq 1$$ has no convergent subsequence. One sees this for example because the sequence has no bounded subsequence.

Alternatively, let $W$ be the given non-zero subspace. Put $$B_W(r) := \{x \in W: \Vert x \Vert < r\}, \quad r > 0$$ Then $$\{B_W(n): n \geq 1\}$$ is an open cover of the subspace $W$ without finite subcover.

J. De Ro
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