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Thread: unit sphere in an infinite dimensional Banach space

  1. #1
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    unit sphere[solved by myself]

    Let X be an infinite dimensional Banach space and show that the unit sphere $\displaystyle S_X=\{||x||=1\}$ is a $\displaystyle G_\delta$ set (i.e. a countable intersection of open sets) in the unit ball $\displaystyle B_X=\{||x||\leq1\}$, when the latter is endowed with the relative topology inherited from the weak topology of X.

    I know that the unit sphere is dense in the unit ball in the above topology, but how to show it is $\displaystyle G_\delta$? And I know that the open set in X under the weak topology of X are all unbounded, and the complement of unit ball is open since unit ball is closed under the weak topology. Thanks in advance!
    Last edited by Jameson; May 3rd 2009 at 01:16 PM. Reason: restored original question
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  2. #2
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    Quote Originally Posted by frankmelody View Post
    Let X be an infinite dimensional Banach space and show that the unit sphere $\displaystyle S_X=\{||x||=1\}$ is a $\displaystyle G_\delta$ set (i.e. a countable intersection of open sets) in the unit ball $\displaystyle B_X=\{||x||\leq1\}$, when the latter is endowed with the relative topology inherited from the weak topology of X.

    I know that the unit sphere is dense in the unit ball in the above topology, but how to show it is $\displaystyle G_\delta$? And I know that the open set in X under the weak topology of X are all unbounded, and the complement of unit ball is open since unit ball is closed under the weak topology. Thanks in advance!
    Do you know the theorem that a closed convex subset of a Banach space is weakly closed? That implies that for $\displaystyle n\in\mathbb{N}$ the set $\displaystyle \{x\in X:\|x\|\leqslant1-\tfrac1n\}$ is weakly closed, and so the set $\displaystyle G_n = \{x\in B_X:\|x\|>1-\tfrac1n\}$ is weakly open. Since $\displaystyle \textstyle S_X = \bigcap_{n\in\mathbb{N}}G_n$, it follows that $\displaystyle S_X$ is a $\displaystyle G_\delta$ set.

    If you don't know the "closed and convex implies weakly closed" theorem then you'll need to show from the definition that the set $\displaystyle G_n$ is weakly open. This follows from the Hahn–Banach theorem, which tells you that if $\displaystyle \|x\|>1-\tfrac1n$ then there is a functional f in the unit sphere of the dual space such that $\displaystyle |f(x)|>1-\tfrac1n$.
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