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Interior of a Subspace

How do we show that any proper subspace of a normed linear space is not open. I know that for nay finite dimensional normed linear space $(X,||.||)$ any proper subspace is closed but I am not sure how to show this?

Thanks for any help

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hmmmm
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2 Answers2

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It is enough to show that the subspace does not contain any open ball centered on $0$. Assume we have such a ball $B_\delta(0)$.

Since the subspace is proper we can choose some vector $v$ outside the subspace. It is nonzero because $0$ is in every subspace, so its norm is nonzero too. Therefore by appropriate scaling there is a scalar $\lambda$ such that $\|\lambda v\|=\delta/2$. Can $\lambda v$ be in the subspace?

hmakholm left over Monica
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  • No, as then the space is not closed under scalar multiplication, right? – hmmmm Oct 09 '12 at 21:30
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    Exactly! ${}{}$ – hmakholm left over Monica Oct 09 '12 at 21:30
  • @HenningMakholm That's a nice proof, plus 1. I've been thinking about the following: complete implies closed and a finite dimensional linear space (over R or C) is complete. Then I was planning to assume that both the space and the subspace are infinite dimensional. Do you see any way to finish from there? My plan was to construct a sequence in the subspace whose limit is not in the subspace. But now I'm stuck. – Rudy the Reindeer Oct 10 '12 at 16:19
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    @Matt: Not clear to me what your goal is here -- to find a subspace that is not closed? That's a different question (and I don't thin infinite algebraic dimension is enough for that. But infinite dimension plus completeness of the larger space may be enough). – hmakholm left over Monica Oct 10 '12 at 17:40
  • Dear @HenningMakholm, no actually, I was working on an alternative proof (before I read yours). By observing that finite-dimensional spaces are closed we rule out the finite-dimensional case: if the space or the subspace are finite-dimensional, we're done since then the subspace is closed. So if we can show that if both the space and the non-empty, proper subspace have infinite dimension that then the subspace must be open, we have also proved the claim. But I'm stuck with finishing. I'm not sure there is a way to finish at all, so I was wondering if you see any way. – Rudy the Reindeer Oct 18 '12 at 07:48
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    @Matt: It sounds vaguely like you're conflating "open" with "not closed". Or do you have a connectedness argument in mind? – hmakholm left over Monica Oct 18 '12 at 13:27
  • @HenningMakholm Did you mean "conflating closed with not open"? Because that's what I'm going. But I don't see how that's wrong. Once we have that the proper subspace is closed, it cannot be open (since it's non-empty) and the only sets in a space that are both closed and open are the space and the empty-set. No? I'm not sure how to use connectedness to my ends in this context. – Rudy the Reindeer Oct 18 '12 at 13:30
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    @MattN: "The only sets that are both closed and open are the space itself and the empty set" is the definition of "connected" for a topological space. So since you're equating "closed" with "not open" you're implicitly depending on the fact that a normed vector space over $\mathbb R$ is connected. – hmakholm left over Monica Oct 18 '12 at 13:45
  • Dear @HenningMakholm Thank you! Sometimes I wonder whether G*d would owe me a brain if he existed. – Rudy the Reindeer Oct 18 '12 at 13:47
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Hint: If $Y$ is an open subspace of $X$, then $0$ is its inner point, so $Y$ contains one of its neighborhoods, say $B_\varepsilon(0)=\{x\mid ||x||<\varepsilon\}$.

Berci
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