Complete set

in a topological vector space $X$ over a field $K$

A set $A$ such that the set of linear combinations of the elements from $A$ is (everywhere) dense in $X$, i.e. the closed subspace generated by the set $A$, i.e. the closed linear hull of $A$, coincides with $X$. For example, in the normed space $C$ of continuous functions on $[0,1]$ with values in $\mathbf C$ the set $\{x^n\}$ is a complete set. If $K$ is a non-discretely normed field, each absorbing set (and in particular each neighbourhood of zero in $X$) is a complete set.

In order for $A=\{a_t\}$, $t\in T$, to be a complete set in the weak topology $\sigma(X,X^*)$ of a space $X$ it is necessary and sufficient that there exist an index $t$ such that $\langle a_t,\xi\rangle\neq0$ for each $\xi\in X^*$; this means that no closed hyperplane contains all the elements $a_t$, i.e. that $A$ is a total set. Moreover, if $X$ is a locally convex space, a complete set in the weak topology will be a complete set in the initial topology also.

Comments

Of course, a complete set in a topological vector space can also mean a set $A$ such that every Cauchy sequence in $A$ converges in $A$; and this is by far the most frequently occurring meaning of the phrase. For the notion of an absorbing set cf. Topological vector space.