Question about nowhere dense sets

[AxiomOfChoice]

Suppose you know k_0A, for some set A \subset X (where X is a metric space) and some constant k_0, has nonempty interior. Do you then know that A has nonempty interior, and/or that k A has nonempty interior for any constant k?

 

[HallsofIvy]

I don't know what you mean by "k_0A". Multiplication isn't defined in a general metric space.

 

[AxiomOfChoice]

I don't know what you mean by "k_0A". 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.

 

[Landau]

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 k\cdot\text{int}A=\text{int}kA.

(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'.