February 13th, 2018, 02:56 PM  #1 
Newbie Joined: Dec 2017 From: vienna Posts: 12 Thanks: 1  Functional analysis
in a normed space V can any proper W subspace be an open in V?

February 13th, 2018, 09:33 PM  #2 
Senior Member Joined: Oct 2009 Posts: 696 Thanks: 235 
What is the span of a ball centered around 0?

February 14th, 2018, 03:43 AM  #3 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,561 Thanks: 2562 Math Focus: Mainly analysis and algebra 
$\mathbb R^2 \subset \mathbb R^3$

February 14th, 2018, 05:34 AM  #4 
Senior Member Joined: Oct 2009 Posts: 696 Thanks: 235  
February 14th, 2018, 06:43 AM  #5 
Senior Member Joined: Sep 2016 From: USA Posts: 535 Thanks: 306 Math Focus: Dynamical systems, analytic function theory, numerics  $\mathbb{R}^2$ 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, 10:08 AM  #6 
Senior Member Joined: Aug 2012 Posts: 2,135 Thanks: 621  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. 
February 14th, 2018, 07:14 PM  #7 
Member Joined: Jan 2016 From: Athens, OH Posts: 92 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.

February 18th, 2018, 04:51 AM  #8 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 895  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, 09:50 AM  #9 
Member Joined: Jan 2016 From: Athens, OH Posts: 92 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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