Question about nowhere dense sets

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Discussion Overview

The discussion revolves around the properties of nowhere dense sets in the context of metric and topological vector spaces, specifically addressing the implications of scaling a set by a constant and its relationship to the interior of the set.

Discussion Character

  • Exploratory, Technical explanation, Debate/contested

Main Points Raised

  • One participant proposes that if k_0A has nonempty interior, then A may also have nonempty interior, and questions whether kA has nonempty interior for any constant k.
  • Multiple participants express confusion regarding the notation "k_0A," stating that multiplication is not defined in a general metric space.
  • One participant clarifies that the discussion assumes a vector space where a metric is defined.
  • Another participant suggests a potential result in topological vector spaces, asserting that scaling a set by a non-zero scalar preserves the interior, though they have not worked out the details.
  • This participant concludes that if their assertion holds, then both questions posed earlier would have affirmative answers.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the initial questions regarding the properties of the set A and its scaled versions. There is disagreement on the interpretation of "k_0A" and its implications in different mathematical contexts.

Contextual Notes

The discussion is limited by the initial ambiguity surrounding the notation "k_0A" and the assumptions about the underlying space being a vector space with a defined metric.

AxiomOfChoice
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Suppose you know [itex]k_0A[/itex], for some set [itex]A \subset X[/itex] (where [itex]X[/itex] is a metric space) and some constant [itex]k_0[/itex], has nonempty interior. Do you then know that [itex]A[/itex] has nonempty interior, and/or that [itex]k A[/itex] has nonempty interior for any constant [itex]k[/itex]?
 
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I don't know what you mean by "[itex]k_0A[/itex]". Multiplication isn't defined in a general metric space.
 
HallsofIvy said:
I don't know what you mean by "[itex]k_0A[/itex]". Multiplication isn't defined in a general metric space.

Point taken. Suppose we are in a vector space on which a metric has been defined.
 
I have strong feeling the following result holds in any topological vector space X:

Let A be subset of X, and k a non-zero scalar. Then [itex]k\cdot\text{int}A=\text{int}kA[/itex].

(Use that multiplication with k is a homeomorphism. Basically, this means the topology in a TVS is scale invariant. Haven't worked out the details.)

Assuming this is true, the answer to both your questions is then 'yes'.
 

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