Weak topology is not metrizable I'm now reading some properties of weak topology, I have some problems which may related to the topology property in non-metrizable space ($E $  is a Banach Space):

*

*I know that $E^*$ with weak* topology is not a metrizable space, but the classical results shows that $B_{E^*} = \{f:\|f\|_{E^*} \leq 1 \}$ is metrizable in weak* topology provided $E$ is separable. How can we explain this contradiction?


*Let $S = \{x: \|x\| = 1, x \in E\}$, then the closure of $S$ in weak topology $\sigma(E,E^*)$ is $B_E = \{x:\|x\| \leq 1, x \in E\}$. However a result shows that If we want to find a sequence $\{x_n\}$ s.t. $x_n \rightharpoonup 0$,$\|x_n\| = 1$, we often need the extra condition that $E^*$ be separable or reflexive.  This two results seem a contradiction to me.


*Prove the following result: if $M \subset E$ is a linear subspace, $f_0 \in E^*$, then there exists some $g_0 \in M^\perp$ such that:
\begin{equation*}
\inf_{g \in M^\perp} \|f_0 - g\| = \|f_0 - g_0\|
\end{equation*}
 A: *

*That the unit ball, $ B_{E^*} $  is metrizable, does not mean that $E^*$ is. Actually, all balls in $E^*$ are metrizable, but you cannot define a global metric, that generates the weak* topology. So there is really no contradiction here.


*Remember that $E$ endowed with the weak topology is not first countable, and thus it is not sufficient to consider sequences when finding closures. In general, there only needs to be a net converging weakly to $0$.


*Let $(g_n)_{n\geq 1}\subset M^\perp$ be a sequence such that $$
\lim_{n\to\infty}\lVert f_0-g_n \rVert=\inf_{g\in M^\perp}\lVert f_0-g \rVert.
$$
Clearly $g_n$ is norm bounded, and we may assume, by possibly passing to a subnet, $(g_\alpha)_{\alpha\in A}$, that $(g_\alpha)_{\alpha\in A}$ converges to some $ g_0\in M^\perp$ in the weak* topology. Now using that the norm is weak* lower semicontinuous we have $$
\inf_{g\in M^\perp}\lVert f_0-g \rVert=\lim_{\alpha\in A}\lVert f_0-g_\alpha \rVert=\liminf_{\alpha\in A}\lVert f_0-g_\alpha \rVert\geq \lVert f_0-g_0 \rVert.
$$
On the other hand, it is obvious that $\inf_{g\in M^\perp}\lVert f_0-g \rVert\leq
\lVert f_0-g_0 \rVert$, since $g_0\in M^\perp$. So $ \inf_{g\in M^\perp}\lVert f_0-g \rVert= \lVert f_0-g_0 \rVert. $
