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 February 13th, 2018, 01:56 PM #1 Newbie   Joined: Dec 2017 From: vienna Posts: 9 Thanks: 1 Functional analysis in a normed space V can any proper W subspace be an open in V?
 February 13th, 2018, 08:33 PM #2 Senior Member   Joined: Oct 2009 Posts: 555 Thanks: 179 What is the span of a ball centered around 0?
 February 14th, 2018, 02:43 AM #3 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,445 Thanks: 2499 Math Focus: Mainly analysis and algebra $\mathbb R^2 \subset \mathbb R^3$ Thanks from topsquark
February 14th, 2018, 04:34 AM   #4
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Quote:
 Originally Posted by v8archie $\mathbb R^2 \subset \mathbb R^3$
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.

February 14th, 2018, 05:43 AM   #5
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Quote:
 Originally Posted by v8archie $\mathbb R^2 \subset \mathbb R^3$
$\mathbb{R}^2$ 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.

February 14th, 2018, 09:08 AM   #6
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Quote:
 Originally Posted by SDK $\mathbb{R}^2$ 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 09:14 AM.

 February 14th, 2018, 06:14 PM #7 Member   Joined: Jan 2016 From: Athens, OH Posts: 89 Thanks: 47 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. Thanks from Joppy
February 18th, 2018, 03:51 AM   #8
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Quote:
 Originally Posted by johng40 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"?

 February 18th, 2018, 08:50 AM #9 Member   Joined: Jan 2016 From: Athens, OH Posts: 89 Thanks: 47 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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