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

• If [ilmath]Y[/ilmath] is a proper vector subspace of [ilmath]X[/ilmath] then [ilmath]\text{Int} [/ilmath][ilmath](Y)\eq\emptyset[/ilmath] - i.e. it has no interior, or empty interior.
  • Symbolically: [ilmath](Y\subset X)\implies \text{Int}(Y;X)\eq \emptyset[/ilmath]

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]

     • See subset and negation of subset for information on what these mean.
   • 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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