# Is finiteness of Assouad dimension a topological invariant for compact metric spaces?

A metric space is called doubling if there is some $$C>0$$ such that for any $$r>0$$ any ball of radius $$r$$ can be covered by $$C$$ balls of radius $$r/2$$. This is equivalent to having finite so-called Assouad dimension.

My question: If $$(X,d)$$ is a compact doubling metric space, is any metric space homeomorphic to it doubling?

If not, then what about when $$(X,d)$$ is a compact Riemannian manifold?

• This seems like a very nice question for this site, but I am afraid it may end up "on hold" because it is too terse. This site is almost overrun with "PSQ" posts - "problem statement questions" - which have nothing but a question. We tend to look for questions with more background and context. Can you expand to post a little to explain the background, motivation, interest, or anything like that? It would make the question more useful for readers, and would help avoid people downvoting it for lack of context. Sep 30 '18 at 2:11
• Every compact metric space is bounded. Maybe you mean 'doubling' instead of bounded? Sep 30 '18 at 3:20
• Yes, should have said "doubling" instead of bounded! Thanks! (Edited to reflect this). Sep 30 '18 at 3:36
• OK, then this is false already for curves: The doubling condition is not a topological invariant. Oct 2 '18 at 3:37
• Can you give an example/reference? Oct 2 '18 at 6:47