Dual ball as a closed subset of a product

If $f\in B_{X^{\ast }}$, then $\mathrm{\mid }f(x)\mathrm{\mid }\le \parallel x\parallel$, so $E(f)$ belongs to the displayed product; evaluations separate functionals, so $E$ is injective. By the two initial-topology descriptions, the subspace topology pulled back by $E$ is exactly $\sigma (X^{\ast },X)$ on the ball.

Inside the product let $C$ be the set of all $z=(z_x{)}_{x\in X}$ satisfying $z_{x+y}=z_x+z_y,z_{\lambda x}=\lambda z_x(x,y\in X,\lambda \in \mathbb{K})\mathrm{.}$ Each equality defines a closed set: it is the inverse image of $\{0\}$ under a continuous finite linear combination of coordinate projections. Hence $C$, their intersection, is closed.

Every $E(f)$ lies in $C$. Conversely, if $z\in C$, then $f_z(x)=z_x$ is linear and its coordinate bound gives $\mathrm{\mid }{f}_z(x)\mathrm{\mid }\le \parallel x\parallel$ for every $x$. Thus $f_z$ is bounded with $\parallel f_z\parallel \le 1$, so $z=E(f_z)$. Consequently $E[B_{X^{\ast }}]=C$.

Steps 1.1 and 2.1 show that $E$ is a homeomorphism onto the closed subspace $C$. When $X=\{0\}$, both the ball and the product are one-point spaces and the same argument applies.