proper subspaces of a topological vector space have empty interior

Theorem - Let V be a topological vector spaceMathworldPlanetmath. Every proper subspaceMathworldPlanetmathPlanetmath SV has empty interior.

Proof : Let S be a subspace of V. Suppose there is a non-empty open set AS.

Fix a point a0AS. Since the vector sum operation is continuous, translationsMathworldPlanetmathPlanetmath of open sets are again open sets. In particular, the set A-a0:={x-a0:xA} is an open set of V that contains the origin 0.

As S is a vector subspace and AS, we see that the translation A-a0 is still contained in S.

Since the scalar multiplication operation is continuous it follows easily that, for every xV, the function fx:𝕂V given by


is also continuous.

Consider now any vector vV. The set fv-1(A-a0) is an open set that contains 0. Thus, taking a value λfv-1(A-a0) we see that


i.e. we can multiply v by a sufficiently small λ such that λv belongs to the open set A-a0.

Since the set A-a0 is contained in S, we see that λvS, and therefore vS.

This proves that V=S, i.e. S is not proper.

We conclude that if S is proper then S has empty interior.

Title proper subspaces of a topological vector space have empty interior
Canonical name ProperSubspacesOfATopologicalVectorSpaceHaveEmptyInterior
Date of creation 2013-03-22 17:34:23
Last modified on 2013-03-22 17:34:23
Owner asteroid (17536)
Last modified by asteroid (17536)
Numerical id 6
Author asteroid (17536)
Entry type Theorem
Classification msc 46A99