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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!

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    $\begingroup$ 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
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    $\begingroup$ @Pete: ...thus proving the terminology is bad, because the Cantor set is very important! :-) $\endgroup$ – Asaf Karagila Jul 8 '13 at 8:34
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    $\begingroup$ @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
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    $\begingroup$ 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
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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.

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  • $\begingroup$ 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
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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 $0$.

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