Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

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'm well aware that for the sequence $x_n=\frac{1}{n}$, $\text{inf }x_n=0$ but $0 \notin (x_n)$. This made me think about something similar but when we are no longer thinking about existence of a number in a sequence but something a bit different. Consider the definition of exterior measure for a set $E \subset\mathbb{R}^d$,

$$m_*(E)=\text{inf }\sum_{j=1}^\infty |Q_j|$$

where the infimum is taken over all countable coverings $E\subset \bigcup_{j=1}^\infty Q_j$ by closed cubes. Does this necessarily imply there is a cover $\{Q_\alpha^*\}_\alpha$ such that

$$\text{inf }\sum_{j=1}^\infty \left|Q_j\right|=\sum_{j=1}^\infty \left|Q_j^*\right|$$

I suspect this is not always the case. I'm wondering about the following:

$1.$ Can one find examples of a set $E\subset \mathbb{R}^d$ (nonfinite) such there is such a "minimal" cover $\{Q_\alpha^*\}$?

$2$. What conditions - if any - on $E$ force there to always be such a "minimal" cover?

$3$. What about when one removes the restriction of $E \subset \mathbb{R}^d$? Are there cases where one clearly can/can't find such "minimal" covers for general metric spaces?

share|cite|improve this question
Your third point does not really make sense. You need a metric space along with a measure for this to make sense. "Cube" doesn't make sense, either, though you can of course replace it with "ball". – tomasz Dec 20 '13 at 5:32
@tomasz I stated that it would be a metric space, which obviously implies a metric. As for replacing cubes with an $\epsilon$-ball, I did not include change that because I believed it was 'obvious' that it would be required. Perhaps I should make it clear with an edit. – mathematics2x2life Dec 20 '13 at 5:35
I was writing about measure, not metric. Metric spaces usually don't come with a natural measure. And in case of those which are not locally compact, if any nontrivial measures exist, they'll most likely be rather strange. About the cubes vs balls, it might be obvious, but it's not exactly a generalization, unless you take the maximum metric on ${\bf R}^d$. – tomasz Dec 20 '13 at 5:37
@tomasz Indeed, I foolishly hadn't even considered whether the normal definition of measure would hold the 'way one would want' for a general metric space. Indeed, this would have to be carefully thought through. However, I am most focused on the case of $\mathbb{R}^d$ and the definition there is clear so I'm focused on looking for those $E$ stated in question $1$ and $2$, as you point out, $3$ needs work and careful thought. Thank you for making that clearer! – mathematics2x2life Dec 20 '13 at 7:06
up vote 4 down vote accepted

If $X$ is an uncountable null set then it cannot have a minimal cover. A fat Cantor set cannot have a minimal cover either since every non-trivial closed cube will leak out of it.

If $X \subseteq \mathbb{R}^d$ has finite outer measure and $\{Q^{\star}_n : n \geq 1\}$ is a minimal cover of $X$, then $X$ is a full outer measure subset of $Y$ where $Y = \bigcup \{Q^{\star}_n : n \geq 1\}$. Isn't this a characterization of such sets - namely sets with full outer measure in a countable union of cubes? Or sets which are a union of countably many full outer measure subsets of cubes?

On real line this just amounts to being a full outer measure subset of a union of an open set and a countable set.

share|cite|improve this answer

In particular, there is a minimal cover for any open set. This follows since any open set can be partitioned into a countable union of half-open cubes of the form $\bar{x}/2^i + [0,1/2^i)^k$ where $\bar{x} \in \mathbb{Z}^k$. The closures of these cubes would give the minimal covering.

share|cite|improve this answer
  1. If $E$ is the countable union of disjoint cubes of course you have that the minimum is reached. This is also true if the cubes have intersection of zero measure (instead of being really disjoint) so that they can touch on the sides. This remains also true if you remove from $E$ a set of measure zero.

  2. sufficient conditions are explained in 1. I think that they might also be necessary... but I haven't investigated the details.

  3. what are cubes in a general metric space? You should use balls, or refer to the diameter of covering sets, as is done with the definition of Hausdorff measure...

share|cite|improve this answer

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.