unit sphere in an infinite dimensional Banach space

• May 2nd 2009, 03:30 AM
frankmelody
unit sphere[solved by myself]
Let X be an infinite dimensional Banach space and show that the unit sphere $S_X=\{||x||=1\}$ is a $G_\delta$ set (i.e. a countable intersection of open sets) in the unit ball $B_X=\{||x||\leq1\}$, 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 $G_\delta$? 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!
• May 3rd 2009, 12:15 AM
Opalg
Quote:

Originally Posted by frankmelody
Let X be an infinite dimensional Banach space and show that the unit sphere $S_X=\{||x||=1\}$ is a $G_\delta$ set (i.e. a countable intersection of open sets) in the unit ball $B_X=\{||x||\leq1\}$, 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 $G_\delta$? 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 $n\in\mathbb{N}$ the set $\{x\in X:\|x\|\leqslant1-\tfrac1n\}$ is weakly closed, and so the set $G_n = \{x\in B_X:\|x\|>1-\tfrac1n\}$ is weakly open. Since $\textstyle S_X = \bigcap_{n\in\mathbb{N}}G_n$, it follows that $S_X$ is a $G_\delta$ 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 $G_n$ is weakly open. This follows from the Hahn–Banach theorem, which tells you that if $\|x\|>1-\tfrac1n$ then there is a functional f in the unit sphere of the dual space such that $|f(x)|>1-\tfrac1n$.