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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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marked as duplicate by Nate Eldredge, userNaN, tomasz, rschwieb, sdcvvc Oct 12 '12 at 21:27

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up vote 7 down vote accepted

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?

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No, as then the space is not closed under scalar multiplication, right? –  hmmmm Oct 9 '12 at 21:30
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Exactly! ${}{}$ –  Henning Makholm Oct 9 '12 at 21:30
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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). –  Henning Makholm Oct 10 '12 at 17:40
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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? –  Henning Makholm Oct 18 '12 at 13:27
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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. –  Henning Makholm Oct 18 '12 at 13:45

Hint: If $Y$ is an open subspace of $X$, then $0$ is its inner point, so $Y$ contains one of its neighborhoods, say $B_\varepsilon(0)=\{x\mid ||x||<\varepsilon\}$.

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