Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

This is a follow-up to this question (in fact, this is what originally motivated me to ask that one.)

Let's say that a sequence $\{s_i\}$ of positive reals covers a set $X\subset\mathbb R$ if there is a collection if intervals $\{I_i\}$ such that $X\subset\bigcup I_i$ and the length of each $I_i$ equals $s_i$.

Does there exist a sequence $\{s_i\}$ such that $\sum s_i<\infty$ and $\{s_i\}$ covers any set of Lebesgue measure zero?

For example, simple things like geometric progressions do not work: they cannot cover a union of infinitely many copies of a compact set of positive Hausdorff dimension, separated by a distance at least $\max_i \{s_i\}$ from one another.

(Sorry for the strange collection of tags. It is hard to see in advance which area this question really belongs to.)

share|improve this question
add comment

1 Answer

up vote 18 down vote accepted

No. If you can cover every set of measure $0$ by your sequence of intervals, you can certainly scale (shrink all intervals some number of times) and still have covering (just cover the expanded set by the original sequence) . If $\sum s_j<+\infty$, then $\sum H(s_j)<+\infty$ for some measuring function $H$ with $H(x)/x\to+\infty$ as $x\to 0$. Thus, every set of measure $0$ would have the Hausdorff measure associated with $H$ zero, which can be ruled out by the standard Cantor construction.

share|improve this answer
add comment

Your Answer

 
discard

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.