Question about nowhere dense sets

  • #1
AxiomOfChoice
533
1
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]?
 

Answers and Replies

  • #2
HallsofIvy
Science Advisor
Homework Helper
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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.
 
  • #3
AxiomOfChoice
533
1
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.
 
  • #4
Landau
Science Advisor
905
0
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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