Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I consider as follows, but i could not proceed it. The topological dimension 2 of a set means that there is a base for the open sets of the set consisting of sets U with topological dimension of boundary of U is 1. My question is: if there exists such a base for the open sets of the set, does it mean that it contains a set homeomorphic to open ball?

share|cite|improve this question
IIRC every subspace of $\mathbb{R}^n$ of full dimension $n$ contains a ball. I cannot recall where I read or heard this, though; I'll look (if someone doesn't give the answer first...) – Henno Brandsma Jun 4 '13 at 11:29
Do you mean with full dimension, topological dimension or vector space dimension? – mtm Jun 4 '13 at 11:50
topological dimension of course! – Henno Brandsma Jun 4 '13 at 11:56
up vote 6 down vote accepted

If we are talking about Lebesgue covering dimension, there is the Menger-Urysohn theorem:

If $M$ is a subset of $\mathbb{R}^{n}$ with empty interior, then $\dim M\leq n-1$.

In brief, the complement $W=\mathbb{R}^{n}\backslash M$ is dense in $\mathbb{R}^{n}$ and we may find sufficiently small coverings $\omega$ of $\mathbb{R}^{n}$ such that the intersection of any $n+1$ elements of $\omega$ lies in $W$. But then $\omega\cap M$ is a small covering of $M$ such that the intersection of any $n+1$ elements is empty. Thus $\dim M\leq n-1$. So the answer to your question is negative.

share|cite|improve this answer
Thanks. I used the topological dimension for small inductive dimension, but i was wondering if there exists a compact set has topological dimension 2 with empty interior. Already, these dimensions are equal to each other on compact metric spaces. – mtm Jun 4 '13 at 19:41
@mtm In fact, small and large inductive dimension and Lebesgue covering dimension all coincide for all separable metrisable spaces . So it does not matter which we use. I just call it the "topological dimension" in that case, and use whatever definition is convenient. Large inductive dimension and Lebesgue covering dimension coincide for all metric spaces, not just the separable ones. – Henno Brandsma Jun 5 '13 at 11:22
@HennoBrandsma Thanks. – mtm Jun 5 '13 at 12:38

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.