Free set

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

A linearly independent system of vectors from $ X $, that is, a set of elements $ A = \{ a _ {t} \} \subset X $, $ t \in T $, such that the relation $ \sum \xi _ {t} a _ {t} = 0 $, where $ \xi _ {t} = 0 $ for all but a finite number of indices $ t $, implies that $ \xi _ {t} = 0 $ for all $ t $. A non-free set is also called dependent.

A free set in a topological vector space $ X $ over a field $ K $( a topologically-free set) is a set $ A = \{ a _ {t} \} \subset X $ such that for any $ s \in T $ the closed subspace generated by the points $ a _ {t} $, $ t \neq s $, does not contain $ a _ {s} $. A topologically-free set is a free set in the vector space; the converse is not true. For example, in the normed space $ C $ of continuous functions on $ [ 0, 1] $, the functions $ \mathop{\rm exp} [ 2 \pi kix] $, $ k \in Z $, form a topologically-free set, in contrast to the functions $ x ^ {k} $( since, e.g., $ x $ is contained in the closed subspace generated by $ \{ x ^ {2k} \} $).

The set of all (topologically-) free sets in $ X $ is, in general, not inductive under inclusion; in addition, it does not necessarily contain a maximal topologically-free set. For example, let $ X $ be the space over $ \mathbf R $ formed by the continuous functions and endowed with the following Hausdorff topology: a fundamental system of neighbourhoods of zero in $ X $ consists of the balanced absorbing sets $ V _ {s, \epsilon } = \{ {x } : {| f ( x) | \leq \delta \textrm{ everywhere outside an open set } \textrm{ (depending on } f \textrm{ ) of measure } \leq \epsilon, 0 \langle \epsilon < 1, \delta \rangle 0 } \} $. Then every continuous linear functional vanishes, and $ X $ does not contain a maximal free set.

For $ A $ to be a (topologically-) free set in the weak topology $ \sigma ( X, X ^ {*} ) $ in $ X $ it is necessary and sufficient that for each $ t $ there is a $ b _ {t} \in X ^ {*} $ such that $ \langle a _ {t} , b _ {t} \rangle \neq 0 $, and $ \langle a _ {s} , b _ {t} \rangle = 0 $ for all $ s \neq t $. For a locally convex space a free set in the weak topology is a free set in the original topology.