Bourbaki

Definition 4

Tag 0788, permanent. Cite it as [Bourbaki, Tag 0788].

Topological Vector Spaces, chapter I, § 1. Topological vector spaces

Let K be a non-discrete valued division ring and E be a left vector space on K with two subsets A and B. We say that A absorbs B if there exists α > 0 such that λA ⊃ B for every λ ∈ K with |λ| ≥ α (or equivalently if μB ⊂ A for μ ≠ 0 and |μ| ≤ α⁻¹). A set A of E is called absorbent if it absorbs every set consisting of a single point.

Let A be a balanced set of E; for it to absorb a set B of E it is sufficient that there exists λ ≠ 0 such that λA ⊃ B; in fact, for |μ| ≥ |λ|, we have λA = (λμ⁻¹) μA, and as μA is balanced and |λμ⁻¹| ≤ 1, it follows that λA ⊂ μA, and thus B ⊂ μA. In particular for a balanced set A of E to be absorbent, it is necessary and sufficient that for every x ∈ E, there exists λ ≠ 0 in K such that λx ∈ A. Every absorbent set of E generates the vector space E. Every finite intersection of absorbent sets is an absorbent set.

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

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Cites

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Transcribed from the printed volume. Not translated.