# If $X$ is infinite dimentional Banach space, $S_{X}$ is dense $G_{\delta}$ set in $(B_X,\omega)$

I'm trying to prove that if $X$ is infinite dimentional Banach space, $S_{X}$ (the unit sphere) is a dense $G_{\delta}$ set in $(B_X,\omega)$ (where $\omega$ is the weak topology). From here I'm trying to conclude that in an infinite dimensional space the norm is never $\omega$-continuous.

I showed that $S_X$ is $\omega$-dense in $B_X$ by showing that for every point $x_0\in B_X$ and every $\omega$-neighborhood $U$, $U\cap S_X\neq \varnothing$, using that the intersection $\cap_{i=1}^nf^{-1}(0)$ of linear functionals in infinite dimention space contains a nonzero element.

After that, to show that the norm is not $\omega$-continuous, I used that $\lVert \cdot\rVert^{-1}[(-1,1)]$ is bounded and therefore not $\omega$-open.

I'm having trouble proving that $S_X$ is a $G_{\delta}$ set. The obvious candidates would be $G_n=\{x\in B_X\big| \lVert x\rVert>1-\frac{1}{n}\}$ and $S_X=\cap_{n=1}^\infty G_n$, but I'm not sure that $G_n$ are $\omega$-open.

• What is $S_X{}$? Jun 12, 2018 at 13:10
• And $\omega$, for that matter? Jun 12, 2018 at 13:12
• The unit sphere in X, $S_X=\{x\in X \big| ||x||=1\}$, and $\omega$ is the weak topology, generated by all the continuous linear functionals on X Jun 12, 2018 at 13:12
• I made some edits to the last two lines to fix some things that seemed like obvious typos. You might want to check my edits reflect what you meant. Jun 12, 2018 at 14:22
• It is often a good strategy to show that a set is open (with respect to some topology) by showing that its complement is closed. (That works quite well here.) Jun 12, 2018 at 14:24

Let $f$ be a norm $1$ functional. Then $G_{f,n}:=f^{-1}{\big(}(1-1/n,\infty){\big)}$ is weak open in $X$ and a subset of $\tilde G_n=\{x\in X\mid \|x\|>1-1/n\}$. For any $x\in \tilde G_n$ you can find via Hahn-Banach a norm $1$ functional $f_x$ with $f_x(x)=\|x\|$, so $x\in G_{f_x,n}$. For that reason $\tilde G_n=\bigcup_{f\in X^*,\|f\|=1} G_{f,n}$ and $\tilde G_n$ is open.
Notice that $G_n = H_n \cap B_X$ where $$H_n = \{x \in X : \|x\| > 1 - \frac1n\}.$$ So it will suffice to show that $H_n$ is a weak-open subset of $X$. This follows from weak lower-semicontinuity of the norm. A function $f:X \to \mathbb{R}$ is lower-semicontinuous at every point in the space if and only if $\{x : f(x) > \alpha \}$ is open for every $\alpha \in \mathbb{R}$.
• Still I'll need to use the Hahn-Banach theorem to show that there is a linear functional $f\in S_{X^*}$ such that $f(x)=\lVert x \rVert$ for every $x\in B_X$ Jun 12, 2018 at 18:07
• @user3701033 Why? The function that you want to be weakly lower-semicontinuous is $x \mapsto \|x\|$. Jun 12, 2018 at 18:09
• @user3701033 In that case, I wouldn't worry too much about them. My point was that any proof that the $G_n$ (or rather the $H_n$) are weakly open will boil down to proving that the norm is weakly lower-semicontinuous, by the definition pretty much. I just wanted to make you aware that you would be proving/using this more general result along the way. Jun 12, 2018 at 18:23