View source for 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

: This is a [[precursor theorem]] to "''[[a proper vector subspace of a topological vector space has no interior]]''".
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==Statement==
Let {{M|(X,\mathcal{J},}}[[Doctrine:K (field)|{{M|\mathbb{K} }}]]{{M|)}} be a [[topological vector space]] and let {{M|(Y,\mathbb{K})}} be a [[vector subspace]] of {{M|(X,\mathbb{K})}}, then{{rFAVIDMH}}:
* {{M|(\exists U\in(\mathcal{J}-\{\emptyset\})[U\subseteq Y])\implies X\eq Y}}
** In words: if there exists a ''[[non-empty]]'' [[open set]] of {{M|(X,\mathcal{J})}}, say {{M|U}}
==Proof==
{{Requires proof|grade=D|msg=Get a picture!
* Got a picture - [[User:Alec|Alec]] ([[User talk:Alec|talk]]) 17:51, 16 February 2017 (UTC)
<gallery>
File:ProofAProperVectorSubspaceHasNoInterior.JPG | The proof here does have precursors though
</gallery>
}}
==See also==
* {{XXX|Do this}}
==References==
<references/>
{{Theorem Of|Functional Analysis|Topology|Linear Algebra|Abstract Algebra}}