13.4 Separable Normed Vector Spaces
Proposition 13.4.1.label Let $K \in \RC$ and $\dpn{E, F}{\lambda}$ be a duality of normed vector spaces over $K$ with $E$ being separable, and $S = \bracsn{y \in F|\ \norm{y}_F \le 1}$ be the closed unit ball of $F$, then
- (1)
$S$ is separable with respect to the $\sigma(F, E)$-topology.
- (2)
$S$ is metrisable with respect to the $\sigma(F, E)$-topology.
- (3)
For any $A \subset F$, $A$ is separable with respect to the $\sigma(F, E)$-topology.
- (4)
If the duality is norming, then there exists $\seq{y_n}\subset F$ such that for each $x \in E$, $\norm{x}_{E} = \sup_{n \in \natp}|\dpn{x, y_n}{\lambda}|$.
Proof. (1): Let $\seq{x_n}\subset E$ be a dense subset. For each $N \in \natp$, let
Since $\real^{N}$ is separable, $T_{N}(S)$ is separable by Proposition 8.1.2. Thus there exists $\bracs{y_{N, k}}_{k = 1}^{\infty} \subset S$ such that $\bracs{T_Ny_{N, k}}_{k = 1}^{\infty}$ is dense in $T_{N}(S)$.
Let $y \in S$, then for each $N \in \natp$, there exists $k_{N} \in \natp$ such that for each $1 \le n \le N$,
Thus for each $N \in \natp$, $\dpn{x_n, y_{N, k_N}}{\lambda}\to \dpn{x_n, y}{\lambda}$ as $N \to \infty$. Since $y_{N, k_N}\to y$ pointwise on a dense subset of $E$ and $\bracsn{y_{N, k_N}|N \in \natp}\subset S$ is uniformly equicontinuous, $y_{N, k_N}\to y$ in the $\sigma(F, E)$-topology by Proposition 11.13.5.
(2): Let $\seq{x_n}\subset E$ be a dense subset, then by Proposition 11.13.5, the $\sigma(F, E)$-topology on $S$ is induced by $\seq{x_n}$, and hence metrisable by Theorem 6.3.10.
(3): For any $A \subset E$, $A = \bigcup_{n \in \natp}A \cap nS$. By Proposition 8.1.2, $A \cap nS$ is separable for each $n \in \natp$. Therefore $A$ is also separable.$\square$
Proposition 13.4.2.label Let $E$ be a separable normed vector space, then the Borel $\sigma$-algebra on $E$ is generated by the following families of sets:
- (1)
Open sets in $E$ with respect to the strong topology.
- (2)
$\bracs{B(x, r)|x \in E, r > 0}$.
- (3)
$\bracsn{\ol{B(x, r)}|x \in E, r > 0}$.
- (4)
Open sets in $E$ with respect to the weak topology.
Proof. (1) $\Leftrightarrow$ (2) $\Leftrightarrow$ (3): By Proposition 21.2.5.
(4) $\subset$ (1): Every weakly open set is strongly open.
(2) $\subset$ (4): By Proposition 12.9.7, $\norm{\cdot}_{E}: E \to [0, \infty)$ is Borel measurable with respect to the weak topology. For any $x \in E$, let
then $\phi_{x}$ is Borel measurable with respect to the weak topology, so $B(x, r) = \bracs{\phi_x < r}$ is a Borel set with respect to the weak topology.$\square$
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