Talk:Dual system

Changes as of 18th Feb 21

@Quondum: Especially, if $K\in \{\mathbb {R} ,\mathbb {C} \}$ we denote it by $\mathbb {K}$ or $\mathbb {k}$ or 𝕂, so that we notice that it is NOT a variable. –Nomen4Omen (talk) 19:17, 18 February 2021 (UTC)

Perhaps not the kind of variable as most people think of it, but here it is a variable that takes values that are sets (or fields). There are numerous articles in mathematics in WP that denote an unspecified field as K or F, for example the articles Field (mathematics), Finite field, Algebra over a field, Vector space, Matrix (mathematics) § Matrices with more general entries, Exterior algebra. —Quondum 20:14, 18 February 2021 (UTC)

Is black-board K a neologism of some kind? I can see why it is appealing, but I've never seen in any math book I've ever read, nor have I seen it in Wikipedia, up until today. 67.198.37.16 (talk) 05:09, 26 November 2023 (UTC)

Making the separation axioms "type check"

In (both) the separation axioms (copied below) we use equations like $b(x,\,\cdot )=0$. Here we are setting a function (or a functional) equal to 0, but it could be ambiguous if the 0 is the constant zero function (i.e., the zero functional) or the zero vector in the primal space. If it were interpreted as the zero vector in the primal space, then a reader might interpret this as $\exists x,\exists y,\,b(x,\,y)=0$, rather than $\exists x,\forall y,\,b(x,\,y)=0$, as intended.

A plain reading seems to imply that $\langle x,0\rangle =0\implies x=0$ when $\langle \cdot ,\cdot \rangle$ is a dual pairing.

Perhaps the alternative definition provided after the first is sufficient to ignore the ambiguity.

1. $Y$ separates (distinguishes) points of $X$: if $x\in X$ is such that $b(x,\,\cdot \,)=0$ then $x=0$; or equivalently, for all non-zero $x\in X$, the map $b(x,\,\cdot \,):Y\to \mathbb {K}$ is not identically $0$ (i.e. there exists a $y\in Y$ such that $b(x,y)\neq 0$ for each $x\in X$);

Tbardwell (talk) 15:32, 5 November 2025 (UTC)