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

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Statement

Let (X,J,K) be a topological vector space and let (Y,K) be a vector subspace of (X,K) (so YX), then[1]:

  • If Y is a proper vector subspace of X then Int(Y)= - i.e. it has no interior, or empty interior.
    • Symbolically: (YX)Int(Y;X)=

Proof

Let (X,J,K) and let (Y,K) be a vector subspace of (X,K), so YX. We have 2 cases:

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

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