# 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 $k$?

## Answers and Replies

Science Advisor
Homework Helper
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.

Science Advisor
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'.