A proper vector subspace of a topological vector space has no interior

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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] (so [ilmath]Y\subseteq X[/ilmath]), then[1]:

Proof

Let [ilmath](X,\mathcal{J},\mathbb{K})[/ilmath] and let [ilmath](Y,\mathbb{K})[/ilmath] be a vector subspace of [ilmath](X,\mathbb{K})[/ilmath], so [ilmath]Y\subseteq X[/ilmath]. We have 2 cases:

  1. Suppose [ilmath]Y\eq X[/ilmath] - by the nature of logical implication we do not care about the truth or falsity of the RHS and we're done
  2. Suppose [ilmath]Y\subset X[/ilmath] - by the nature of logical implication we must show the RHS holds.
    • Recall: "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"
      • Symbolically: [ilmath](\exists U\in(\mathcal{J}-\{\emptyset\})[U\subseteq Y])\implies X\eq Y[/ilmath]
    • Recall also that the contrapositive of an implication is logically equivalent to the original statement, that is:
      • [ilmath](A\implies B)\iff(\neg B\implies\neg A)[/ilmath]
    • So we use the earlier statement to see:
      • [ilmath]\big[(\exists U\in(\mathcal{J}-\{\emptyset\})[U\subseteq Y])\implies X\eq Y\big][/ilmath]
        [ilmath]\iff[/ilmath]
      [ilmath]\big[X\neq Y\implies(\forall U\in(\mathcal{J}-\{\emptyset\})[U\nsubseteq Y])\big][/ilmath]
    • We have [ilmath]X\neq Y[/ilmath] (as [ilmath]Y\subset X[/ilmath] is a proper subset there must be something in [ilmath]X[/ilmath] that is not in [ilmath]Y[/ilmath]
      • So we see [ilmath]X\neq Y[/ilmath]
        • [ilmath]\implies \forall U\in(\mathcal{J}-\{\emptyset\})[U\nsubseteq Y][/ilmath] - in words, for all non-empty open sets of [ilmath]X[/ilmath], that open set is not contained in [ilmath]Y[/ilmath].
        • Recall the definition of interior:
          • [math]\text{Int}(Y):\eq\bigcup_{U\in\{V\in\mathcal{J}\ \vert\ V\subseteq Y\} } U[/math]
            • blah blah blah
              • We see [ilmath]\text{Int}(Y)\eq\emptyset[/ilmath]
Grade: D
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References

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