# 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

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## See also

- TODO: Do this

## References

Categories:

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