# A set that can be covered by arbitrarily small intervals

Let $X$ be a subset of the real line and $S=\{s_i\}$ an infinite sequence of positive numbers. Let me say that $X$ is $S$-small if there is a collection $\{I_i\}$ of intervals such that the length of each $I_i$ equals $s_i$ and the union $\bigcup I_i$ contains $X$. And $X$ is said to be small if it is $S$-small for any sequence $S$.

Obviously every countable set is small. Are there uncountable small sets?

Some observations:

• A set of positive Hausdorff dimension cannot be small.

• Moreover, a small set cannot contain an uncountable compact subset.

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For those who somehow feel familiar with this question, it showed already up on mathoverflow as an answer: mathoverflow.net/questions/1924/… – j.p. Dec 6 '10 at 17:05
Why are you interested in these sets? Do they have some special geometric properties and/or characterisations? One such characterisation is that a set X is small iff for every closed measure 0 set C, the sum C+X has measure 0. – mmm Jun 16 '11 at 16:49

The sets you are calling small are commonly referred to in the literature as "strong measure zero sets." The Borel conjecture is the assertion that any strong measure zero set is countable.

This is independent of the usual axioms of set theory. For example, Luzin sets are strong measure zero; if MA (Martin's axiom) holds, then there are strong measure zero sets of size continuum. In fact, both CH and ${\mathfrak b}=\aleph_1$ contradict the Borel conjecture.

However, the Borel conjecture is consistent. This was first shown by Laver, in 1976; in his model the continuum has size $\aleph_2$. Later, it was observed (by Woodin, I think) that adding random reals to a model of the Borel conjecture, preserves the Borel conjecture, so the size of the continuum can be as large as wanted.

All this is described very carefully in Chapter 8 of the very nice book "Set Theory: On the Structure of the Real Line" by Tomek Bartoszynski and Haim Judah, AK Peters (1995).

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@Denis : Thanks for fixing the typo! – Andrés E. Caicedo Dec 6 '10 at 16:50
@Andres: Just to make sure I understand it correctly: assuming CH, one can prove that uncountable strong measure zero sets do exist, but not in general, right? This looks so counter-intuitive to me... – Sergei Ivanov Dec 6 '10 at 17:32
@Sergei : Yes, I imagine that when the results began to appear, they were somewhat surprising, in that it is easy to find models where the conjecture fails (so there are uncountable strong measure zero sets), but it was significantly trickier to find models where the conjecture holds. – Andrés E. Caicedo Dec 6 '10 at 18:28

The problem was also studied by Besicovitch from the geometric measure-theoretic point of view in the 1930s. In particular, Besicovitch was motivated by the problem of determining the sets of reals on which the variation of any continuous monotone function vanishes. He proved that assuming CH there exists an uncountable set of reals which has strong measure zero (see "Concentrated and rarified sets of points", Acta Mathematica (1934), Vol. 62, pp. 289-300).

Besicovitch constructed what he called a concentrated set (an uncountable set of reals $E$ is said to be concentrated on a countable set $H$ iff for any open set $U$ if $H \subset U$, then $E \setminus U$ is countable). The Besicovitch concentrated sets can be viewed as a weaker measure-theoretic analogue of Lusin sets.

The earlier stages of the theory of strong measure zero sets are summarized in Sierpinski's monograph Hypothèse du continu. Sierpinski refers to these objects as sets satisfying Property C (see the definition on p. 37).

[EDIT. Somewhat surprisingly the work of Besicovitch and Sierpinski on strong measure zero sets is not mentioned at all in the book recommended by Andres. The historical development of the theory is discussed in detail in the fundamental survey article "History of the Continuum in the 20th Century" by Steprans.]

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Many thanks for the references, Andrey! – Andrés E. Caicedo Jan 28 '11 at 1:39
@Andres: You are welcome. – Andrey Rekalo Jan 28 '11 at 7:07
Andrey's and Andres's answers are also related since the existence of a Besicovich-concentrated set was proved by Lawrence to be equivalent to $\mathfrak{b}=\aleph_1$. Interestingly, in this case there are uncountable sets with much more pathological properties, e.g., real sets $X$ whose product with every $\gamma$-set is a $\gamma$-set. A real set is a $\gamma$-set iff the function space $C_p(X)$ is Frechet-Urysohn. – Boaz Tsaban Mar 31 at 20:19