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I want to prove that the following are equivalent:

  1. CH
  2. There exists a Sierpinski set and $\operatorname{non}(\mathcal{N})=2^{\aleph_0}$
  3. There exists a Luzin set and $\operatorname{non}(\mathcal{M})=2^{\aleph_0}$
  4. There exists a Sierpinski- and a Luzin set and one of them has size continuum.

I can show that 1 implies 2-4.

In Case 2-4 we either have a Luzin or Sierpinski set of size continuum, how can I deduce that it has size $\aleph_1$?

Here $\mathcal{N}$ is the ideal of Lebesgue null sets, $\mathcal{M}$ is the ideal of meager sets.

A Sierpinksi set is an uncountable set $S\subseteq \mathbb{R}$ s.t. $S\cap N$ is countable for all $N\in\mathcal{N}$. A Luzin set is an uncountable set $L\subseteq \mathbb{R}$ s.t. $L\cap M$ is countable for all $M\in\mathcal{M}$.

2 Answers 2

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Here is a partial answer: I will show that 2 implies 1. Let $\{r_\alpha : \alpha<\mathfrak{c}\}$ be an enumeration of $\mathbb{R}$. Since $\mathrm{non}(\mathcal{N})=\mathfrak{c}$, every subset of $\mathbb{R}$ whose cardinality is strictly less than $\mathfrak{c}$ is measurable, and in fact measure 0.

Let $S$ be a Sierpinski set, then $S_\gamma := S\cap \{r_\alpha : \alpha<\gamma\}$ is countable for each $\gamma < \mathfrak{c}$. Moreover if $\beta<\gamma<\kappa$ then $S_\beta \subseteq S_\gamma$. Therefore $\langle S_\gamma : \gamma<\mathfrak{c}\rangle$ forms an increasing sequence of countable sets. If $\mathfrak{c}\ge \omega_2$, then $S_\gamma$ is eventually constant (since it is always countable) and $S=S_\gamma$ for large $\gamma$, which implies $S$ itself is uncountable measure-zero set. Therefore $\mathfrak{c} = \omega_1$.

You can also prove the second implication $(3)\implies (1)$ with little modification, by changing some words in the above proof.

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    How is this not also an answer for $(3)\implies(1)$ (mutatis mutandis)? – 2017-02-17
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    @AsafKaragila After posting an answer I have tried to prove remaining implications and I realize and my proof also gives a proof of $3\to 1$, and so I've concentrated on the last implication. But I gave up to prove it and forgot to write the remaining obvious implication. – 2017-02-17
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    @Asaf so I edited my answer, thanks for your comment. – 2017-02-17
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$(4)\rightarrow (1):$ Let $L$ be a Luzin set, $S$ be a Sierpinski set and $S'$ a subset of $S$ of cardinality $\aleph_1$ Then clearly $S'$ is also a Sierpinksi set. Thus $S'$ in not measure $0$ and by Rothbergers Theorem we have

$$\operatorname{cov}(\mathcal{M})\leq \operatorname{non}(\mathcal{N})\leq |S'|=\aleph_1$$

Take a sequence $(M_\alpha \mid \alpha<\aleph_1)$ of meager sets covering $\mathbb{R}$. Then

$$|L|=|\bigcup_{\alpha<\aleph_1} L\cap M_\alpha|=\aleph_1$$

Similarly $|S|=\aleph_1$. Since either $|S|=\mathfrak{c}$ or $|L|=\mathfrak{c}$, $\mathfrak{c}=\aleph_1$.