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

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This is a precursor theorem to "a proper vector subspace of a topological vector space has no interior".

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]
    • In words: if there exists a non-empty open set of [ilmath](X,\mathcal{J})[/ilmath], say [ilmath]U[/ilmath]

Proof

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

See also

  • TODO: Do this

References

  1. Functional Analysis - Volume 1: A gentle introduction - Dzung Minh Ha