14.4 Separable Normed Vector Spaces

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 12.13.4.

(2): Let $\seq{x_n}\subset E$ be a dense subset, then by Proposition 12.13.4, 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.$\square$

Proof. First note that if $0 \not\in A$ or $1 \not\in A$, $C(A; E)$ embeds isometrically into $C(A \cup \bracs{0, 1}; E)$ through extension by $0$. Thus assume without loss of generality that $A$ contains the endpoints $0$ and $1$.

Let $U = [0, 1] \setminus A$, then there exists $\seq{(a_n, b_n)}\subset [0, 1]^{2}$ such that $U = \bigsqcup_{n \in \natp}(a_{n}, b_{n})$. For each $f \in C(A; E)$, let

then the mapping $f \mapsto Tf$ is an isometric embedding into $E^{[0, 1]}$ with respect to the uniform norm.

Since $U$ is open and $Tf$ is affine on each component of $U$, $Tf$ is continuous on $U$. It remains to show that $Tf$ is continuous on $A$. Let $x \in A$ and $\eps > 0$, then there exists $\delta > 0$ such that $\norm{f(y) - f(x)}_{E} < \eps$ for all $y \in (x -\delta, x + \delta) \cap A$. Now, a case analysis:

1. (1)

   If there exists $y \in (x - \delta, x) \cap A$, then for any $z \in U \cap (y, x)$, there exists $n \in \natp$ such that $(a_{n}, b_{n}) \subset (y, x)$ and $z \in (a_{n}, b_{n})$. In which case, since $\norm{f(a_n) - f(x)}_{E} < \eps$ and $\norm{f(b_n) - f(x)}_{E} < \eps$, $\norm{Tf(z) - Tf(x)}_{E} < \eps$. Thus $\norm{Tf(z) - Tf(x)}_{E} < \eps$ for all $z \in (y, x)$.

2. (2)

   Otherwise, $x = 0$ or $Tf|_{(x - \delta, x)}$ is an affine function. Either way, there exists $y \in (x - \delta, x)$ such that $\norm{Tf(z) - Tf(x)}_{E} < \eps$ for all $z \in (y, x) \cap [0, 1]$.

Thus there exists $y \in (x - \delta, x)$ with $\norm{Tf(z) - Tf(x)}_{E} < \eps$ for all $z \in (y, x) \cap [0, 1]$. Similarly, there exists $y' \in (x, x + \delta)$ with $\norm{Tf(z) - Tf(x)}_{E} < \eps$ for all $z \in (x, y') \cap [0, 1]$. Therefore $Tf$ is continuous at $x$. Since this holds for all $x \in U$ and $x \in A$, $Tf \in C([0, 1]; E)$.$\square$

Proof. Let $B$ be the closed unit ball of $E^{*}$, equipped with the weak* topology. By the Hahn-Banach Theorem, the linear mapping

is an isometric embedding. By Proposition 14.4.1, $B$ is a compact metric space. The Alexandroff-Hausdorff Theorem then provides a continuous surjection $f: 2^{\natp}\to B$. Thus the composition map

is a linear isometric embedding. Let $\mathcal{C}\subset [0, 1]$ be the Cantor set, then $\mathcal{C}$ is homeomorphic to $2^{\natp}$ through Proposition 10.3.4. Hence $C(2^{\natp}; K)$ is isometrically isomorphic to $C(\mathcal{C}; K)$.

Finally, Lemma 14.4.2 provides yet another linear isometric embedding $C(\mathcal{C}; K) \to C([0, 1]; K)$. Composing the above maps as follows

yields the desired embedding.$\square$

Proof. (1) $\Leftrightarrow$ (2) $\Leftrightarrow$ (3): By Proposition 22.2.6.

(4) $\subset$ (1): Every weakly open set is strongly open.

(2) $\subset$ (4): By Proposition 13.9.7, $\norm{\cdot}_{E}: E \to [0, \infty)$ 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.$\square$