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This question already has an answer here:

If $R$ be a Union of zero measure (lebesgue) sets , what can we say about the cardinal of index set? Does this question related to continuum hypothesis? Thanks.

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marked as duplicate by BrianO, Claude Leibovici, user91500, user228113, Eric Wofsey Feb 4 '16 at 7:26

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  • $\begingroup$ What have you done so far to solve it? Where are you stuck? $\endgroup$ – Nate 8 Feb 4 '16 at 3:34
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Suppose $\kappa$ is a cardinal such that $\mathbb{R} = \bigcup_{\alpha < \kappa} A_\alpha$, where each $A_\alpha$ is measure zero. Then $\kappa$ is greater than equal to a cardinal called $\text{cov(null)}$, which is by definition the smallest cardinal $\kappa$ such that there exists a family $\{A_\alpha : \alpha < \kappa\}$ of null sets whose union is $\mathbb{R}$.

It is clear that $\text{cov(null)} > \aleph_0$. Hence if the continuum hypothesis is true then, $\text{cov(null)} = 2^{\aleph_0}$.

However, there are models (where the continuum hypothesis is false) in which $\text{cov(null)} < 2^{\aleph_0}$.

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  • $\begingroup$ Where I can find this result , thanks. $\endgroup$ – Idele Feb 4 '16 at 4:18
  • $\begingroup$ This can be proved using iterated forcing. I believe Combinatorial Set Theory by Halbeisen should have a proof of this result. $\endgroup$ – William Feb 4 '16 at 4:23
  • $\begingroup$ @hctb By the way, one explicit example is to start off the with the constructible universe $L$ as forcing using the $\omega_2$ length countable support iteration of Sacks forcing (perfect trees). $\endgroup$ – William Feb 4 '16 at 4:25
  • $\begingroup$ @hctb Also it is possible for the continuum hypothesis to be false and $\text{cov(null)} = 2^{\aleph_0}$. For this you can use an $\omega_2$ length countable support iteration of Random forcing over $L$. $\endgroup$ – William Feb 4 '16 at 4:26

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