A proper vector subspace of a topological vector space has no interior
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[hide]Statement
Let (X,J,K) be a topological vector space and let (Y,K) be a vector subspace of (X,K) (so Y⊆X), then[1]:
- If Y is a proper vector subspace of X then Int(Y)=∅ - i.e. it has no interior, or empty interior.
- Symbolically: (Y⊂X)⟹Int(Y;X)=∅
Proof
Let (X,J,K) and let (Y,K) be a vector subspace of (X,K), so Y⊆X. We have 2 cases:
- 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
- Suppose Y⊂X - 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−{∅})[U⊆Y])⟹X=Y
- Recall also that the contrapositive of an implication is logically equivalent to the original statement, that is:
- (A⟹B)⟺(¬B⟹¬A)
- So we use the earlier statement to see:
- [(∃U∈(J−{∅})[U⊆Y])⟹X=Y]
- ⟺
- [X≠Y⟹(∀U∈(J−{∅})[U⊈
- See subset and negation of subset for information on what these mean.
- [(∃U∈(J−{∅})[U⊆Y])⟹X=Y]
- 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
- blah blah blah
- \text{Int}(Y):\eq\bigcup_{U\in\{V\in\mathcal{J}\ \vert\ V\subseteq Y\} } U
- So we see X\neq Y
- 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"
Grade: D
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