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February 13th, 2018, 02:56 PM   #1
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Functional analysis

in a normed space V can any proper W subspace be an open in V?
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February 13th, 2018, 09:33 PM   #2
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What is the span of a ball centered around 0?
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February 14th, 2018, 03:43 AM   #3
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$\mathbb R^2 \subset \mathbb R^3$
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February 14th, 2018, 05:34 AM   #4
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Quote:
Originally Posted by v8archie View Post
$\mathbb R^2 \subset \mathbb R^3$
What is that supposed to show? The OP didn't ask whether every subspace is open, the OP asked whether it is possible that there is SOME space which has SOME subspace that is open.
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February 14th, 2018, 06:43 AM   #5
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Originally Posted by v8archie View Post
$\mathbb R^2 \subset \mathbb R^3$
$\mathbb{R}^2$ is not open in $\mathbb{R}^3$. The answer to the question is no. As micromass pointed out, every subspace contains 0 so an open subspace must contain the span of an open ball centered at 0 but this is necessarily all of $V$ so it can't be proper.
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February 14th, 2018, 10:08 AM   #6
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Quote:
Originally Posted by SDK View Post
$\mathbb{R}^2$ is not open in $\mathbb{R}^3$. The answer to the question is no. As micromass pointed out, every subspace contains 0 so an open subspace must contain the span of an open ball centered at 0 but this is necessarily all of $V$ so it can't be proper.
In the indiscrete topology, the only open set containing 0 is the entire space. This can't happen in a normed space, which is a detail needing a bit of explanation.

Last edited by Maschke; February 14th, 2018 at 10:14 AM.
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February 14th, 2018, 07:14 PM   #7
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Isn't this almost obvious? Let W be an open subspace of V and $0\neq x\in V$. Since there is an open ball around 0 contained in W, some scalar multiple of x is in W. Hence x itself is in V.
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February 18th, 2018, 04:51 AM   #8
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Isn't this almost obvious? Let W be an open subspace of V and $0\neq x\in V$. Since there is an open ball around 0 contained in W, some scalar multiple of x is in W. Hence x itself is in V.
What? You start with the hypothesis that $x\in V$ and conclude "Hence x itself is in V"? Did you mean "Hence x itself is in W"?
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February 18th, 2018, 09:50 AM   #9
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Yes, of course. When I saw the typo, it was too late to edit, but I thought it would be clear that I meant $x\in W$.
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