For a vector subspace of a topological vector space if there exists a non-empty open set contained in the subspace then the spaces are equal
From Maths
- This is a precursor theorem to "a proper vector subspace of a topological vector space has no interior".
Contents
Statement
Let [ilmath](X,\mathcal{J},[/ilmath][ilmath]\mathbb{K} [/ilmath][ilmath])[/ilmath] be a topological vector space and let [ilmath](Y,\mathbb{K})[/ilmath] be a vector subspace of [ilmath](X,\mathbb{K})[/ilmath], then[1]:
- [ilmath](\exists U\in(\mathcal{J}-\{\emptyset\})[U\subseteq Y])\implies X\eq Y[/ilmath]
Proof
Grade: D
This page requires one or more proofs to be filled in, it is on a to-do list for being expanded with them.
Please note that this does not mean the content is unreliable. Unless there are any caveats mentioned below the statement comes from a reliable source. As always, Warnings and limitations will be clearly shown and possibly highlighted if very important (see template:Caution et al).
The message provided is:
The message provided is:
See also
- TODO: Do this
References
Categories:
- Pages requiring proofs
- XXX Todo
- Theorems
- Theorems, lemmas and corollaries
- Functional Analysis Theorems
- Functional Analysis Theorems, lemmas and corollaries
- Functional Analysis
- Topology Theorems
- Topology Theorems, lemmas and corollaries
- Topology
- Linear Algebra Theorems
- Linear Algebra Theorems, lemmas and corollaries
- Linear Algebra
- Abstract Algebra Theorems
- Abstract Algebra Theorems, lemmas and corollaries
- Abstract Algebra