Weak topology on a normed space

Let $X$ be a normed space over $\mathbb{K}=\mathbb{R}$ or $\mathbb{C}$, and let $X^{\ast }$ be its bounded $\mathbb{K}$-linear dual, with the operator norm, as in The dual space X^* of a normed space and its dual norm. The weak topology $\sigma (X,X^{\ast })$ is the initial topology induced by all maps $f:X\to \mathbb{K}$, $f\in X^{\ast }$, where the scalars have their usual topology. Explicitly it is generated by the subbasis

$\{f^{-1}(V):f\in X^{\ast },V\subseteq \mathbb{K}\text{ open}\}\mathrm{.}$

The initial-topology construction in The initial topology of a family of maps into spaces and the final topology of a family of maps out of spaces, and the subspace topology as the model initial topology establishes existence and says that finite intersections of these sets form a basis. The empty intersection is $X$. This is the coarsest topology making every bounded scalar-linear functional continuous, and it is contained in the norm topology. This definition uses no choice principle and does not assert separation of points without a norming hypothesis. For $X=\{0\}$ it is the unique topology on that singleton.

In the complex case functionals are complex-linear; convexity in subsequent results means convexity for real coefficients, and real-valued separation inequalities use real parts. The topology is a topology on the entire space, not a sequence-convergence prescription.