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".

Statement

Let $(X,\mathcal{J},$$\mathbb{K}$$)$ be a topological vector space and let $(Y,\mathbb{K})$ be a vector subspace of $(X,\mathbb{K})$, then:

• $(\exists U\in(\mathcal{J}-\{\emptyset\})[U\subseteq Y])\implies X\eq Y$
  • In words: if there exists a non-empty open set of $(X,\mathcal{J})$, say $U$

Proof

Grade: D

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Get a picture!

• Got a picture - Alec (talk) 17:51, 16 February 2017 (UTC)

• 

  The proof here does have precursors though

See also

• TODO: Do this