proper subspaces of a topological vector space have empty interior
Proof : Let be a subspace of . Suppose there is a non-empty open set .
As is a vector subspace and , we see that the translation is still contained in .
Since the scalar multiplication operation is continuous it follows easily that, for every , the function given by
is also continuous.
Consider now any vector . The set is an open set that contains . Thus, taking a value we see that
i.e. we can multiply by a sufficiently small such that belongs to the open set .
Since the set is contained in , we see that , and therefore .
This proves that , i.e. is not proper.
We conclude that if is proper then has empty interior.
|Title||proper subspaces of a topological vector space have empty interior|
|Date of creation||2013-03-22 17:34:23|
|Last modified on||2013-03-22 17:34:23|
|Last modified by||asteroid (17536)|