I was thinking of different definitions of small subsets on $\mathbb{R}$, such as meagre or zero-measure. These are quite well-known, so I was searching for different notions.

Define a set has zero-content if for every $\epsilon > 0$, the set can be covered by a finite number of intervals with total length less than $\epsilon$. A set is unimportant if it is a countable union of zero-content sets.

Slightly different: a subset of $\mathbb{R}$ is called tiny if for every $\epsilon > 0$ there exist a sequence of intervals $I_1, I_2, \ldots$ that cover the set such that for every $i$, $|I_i| \leq \epsilon^i$.

I have a few questions concerning these definitions:

  1. Does there exist an uncountable unimportant set?
  2. I already found every unimportant set is of zero-measure, but is the converse true? I guess not.
  3. What is the relation between tiny and zero-measure?

Thanks in advance!

  • 3
    $\begingroup$ It might be useful to note that a set $A$ has zero content iff $A$ is bounded and its closure has measure zero. $\endgroup$ – Chris Eagle Jul 8 '13 at 8:26
  • $\begingroup$ The standard Cantor set is uncountable and of zero content, hence unimportant. $\endgroup$ – Pete L. Clark Jul 8 '13 at 8:33
  • 2
    $\begingroup$ @Pete: ...thus proving the terminology is bad, because the Cantor set is very important! :-) $\endgroup$ – Asaf Karagila Jul 8 '13 at 8:34
  • 1
    $\begingroup$ @Asaf: Why does Chris's comment imply that every zero measure set is a countable union of zero-content sets? $\endgroup$ – Pete L. Clark Jul 8 '13 at 8:43
  • 2
    $\begingroup$ Also, it seems that since $\epsilon \in (0,1) \implies \sum_{n=1}^{\infty} \epsilon^n = \frac{\epsilon}{1-\epsilon}$, which goes to $0$ with $\epsilon$, tiny sets have zero measure. $\endgroup$ – Pete L. Clark Jul 8 '13 at 8:49

On tiny sets: It was already pointed out that tiny sets are null and the classical Cantor set is not tiny. Also, there are uncountable tiny sets. To see this, construct a Cantor scheme of closed intervals in $[0, 1]$ such that the $n$th level of your tree is a disjoint union of $2^n$ intervals each of length less than $\displaystyle (1/n)^{2^{n}}$. The resulting perfect set $C$ is a size continuum tiny set. A similar notion that leads to deeper problems is that of a strongly null set.

On (2): A null dense $G_{\delta}$ set $X \subseteq \mathbb{R}$ cannot be covered by countably many sets of zero (Jordan) content.

  • $\begingroup$ Regarding your comment about (2), note that your set $X$ is not meager in ${\Bbb R},$ whereas every set of zero Jordan content is meager in ${\Bbb R}.$ Thus, $X$ cannot be covered by countably many zero Jordan content sets simply because $X$ cannot be covered by countably many meager sets. By slightly modifying $X$ so that it is a dense $G_{\delta}$ set relative to a measure dense Cantor set (each nonempty intersection of an open set with the Cantor set has positive measure) we get a measure zero and nowhere dense set that cannot be covered by countably many sets of zero Jordan content. $\endgroup$ – Dave L. Renfro Jul 8 '13 at 20:30
  • $\begingroup$ That is right. You can relativize to a fat Cantor set. $\endgroup$ – hot_queen Jul 9 '13 at 5:12
  • $\begingroup$ I forgot to mention (the hopefully obvious) requirement that the dense $G_{\delta}$ subset of the measure dense Cantor set should also have have measure zero. [Start with a countable dense subset (which has measure zero) of the measure dense Cantor set, then use outer regularity of measure to obtain a $G_{\delta}$ set with the same measure (measure zero) that contains the countable dense subset.] $\endgroup$ – Dave L. Renfro Jul 9 '13 at 16:35
  • $\begingroup$ See also this 30 April 2000 sci.math post that surveys the $\sigma$-ideal generated by the closed sets of Lebesgue measure zero (equals the collection of sets each of which is a subset of an $F_{\sigma}$ measure zero set). $\endgroup$ – Dave L. Renfro Jul 9 '13 at 16:43

Tiny sets have zero Hausdorff dimension. Thus any set with positive Hausdorff dimension (certain cantor sets, etc.) cannot be tiny.

The Hausdorff dimension of a set $A \subset \mathbb R$ may be defined via the following process:

  1. For $s \ge 0$ and $\epsilon > 0$ define $\displaystyle H^s_\epsilon(A) = \inf \left\{ \sum_k |I_k|^s : A \subset \cup I_k,\ |I_k| < \epsilon \right\}$.
  2. Since $H^s_\epsilon(A)$ is increasing as $\epsilon \to 0^+$, the quantity $H^s(A) \displaystyle = \sup_{\epsilon > 0 } H^s_\epsilon(A) = \lim_{\epsilon \to 0^+} H^s_\epsilon(A)$ exists and is termed the Hausdorff $s$-measure of $A$. It turns out that $H^s$ is in fact a Borel measure on $\mathbb R$.
  3. There is a unique value $s_0$ with the property that $H^s(A) = \infty$ for all $0 \le s < s_0$ and $H^s(A) = 0$ for all $s > s_0$. This number is the Hausdorff dimension of $A$.

If $A$ is tiny, then for any $s > 0$ and $\epsilon > 0$ there exist intervals $I_k$ with the property that $A \subset \cup I_k$ and $|I_k| < \epsilon ^k$. Thus $$ H^s_\epsilon(A) \le \sum_k |I_k|^s \le \sum_k \epsilon^{sk} = \frac{\epsilon^s}{1 - \epsilon^s}.$$ Let $\epsilon \to 0^+$ to obtain $H^s(A) = 0$. This is the case for any positive $s$, forcing the dimension of $A$ to be $0$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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