In functional analysis and related areas of mathematics, a complete topological vector space is a topological vector space (TVS) with the property that whenever points get progressively closer to each other, then there exists some point
But unlike metric-completeness, TVS-completeness does not depend on any metric and is defined for all TVSs, including those that are not metrizable or Hausdorff.
Completeness is an extremely important property for a topological vector space to possess.
Prominent examples of complete TVS that are (typically) not metrizable include strict LF-spaces such as the space of test functions
Explicitly, a topological vector spaces (TVS) is complete if every net, or equivalently, every filter, that is Cauchy with respect to the space's canonical uniformity necessarily converges to some point.
This notion of "TVS-completeness" depends only on vector subtraction and the topology of the TVS; consequently, it can be applied to all TVSs, including those whose topologies can not be defined in terms metrics or pseudometrics.
This section summarizes the definition of a complete topological vector space (TVS) in terms of both nets and prefilters.
and similarly their difference is defined to be the image of the product net under the vector subtraction map
is called a complete topological vector space if any of the following equivalent conditions are satisfied: where if in addition
is sequentially complete if any of the following equivalent conditions are satisfied: The existence of the canonical uniformity was demonstrated above by defining it.
This section is dedicated to explaining the precise meanings of the terms involved in this uniqueness statement.
and use the neighborhood definition of "open set" to obtain a topology on
is called a complete uniform space (respectively, a sequentially complete uniform space) if every Cauchy prefilter (respectively, every elementary Cauchy prefilter) on
We review the basic notions related to the general theory of complete pseudometric spaces.
is called a complete pseudometric if any of the following equivalent conditions hold: And if addition
Two norms on a vector space are called equivalent if and only if they induce the same topology.
is a complete TVS that contains a dense vector subspace that is TVS-isomorphic to
For example, the vector space consisting of scalar-valued simple functions
Similarly, the completion of the injective tensor product of the space of scalar-valued
[16] But nevertheless, every Hausdorff TVS still has infinitely many non-isomorphic non-Hausdorff completions.
denote any TVS endowed with the indiscrete topology, which recall makes
[16] But nevertheless, as shown above, every Hausdorff TVS still has infinitely many non-isomorphic non-Hausdorff completions.
is a Hausdorff topological vector space then its completion is linearly isomorphic to the set of all
[37] There exists a sequentially complete locally convex TVS that is not quasi-complete.
[3] This remains true if "TVS" is replaced by "commutative topological group.
contains a closed vector subspace that has no topological complement.
be a separable locally convex metrizable topological vector space and let
[43][proof 2] Thus a closed and totally bounded subset of a complete TVS is compact.
Every relatively compact subset of a Hausdorff TVS is totally bounded.
is not Hausdorff then there exist compact complete sets that are not closed.