This question was motivated by an answer to this question of Dominic van der Zypen.

It relates to the following classical theorem of Sierpiński.

Theorem (Sierpiński, 1921). For any countable partition of the unit interval $[0,1]$ into closed subsets exactly one set of the partition is non-empty.

Motivated by this Sierpiński Theorem we can ask about the smallest infinite cardinality $\acute{\mathfrak n}$ of a partition of the unit interval into closed non-empty subsets. It is clear that $\acute{\mathfrak n}\le\mathfrak c$. The Sierpinski Theorem guarantees that $\omega_1\le\acute{\mathfrak n}$. So, $\acute{\mathfrak n}$ is a typical small uncountable cardinal living in the segment $[\omega_1,\mathfrak c]$.

Problem 1. Is $\acute{\mathfrak n}$ equal to some other known small uncountable cardinal?

Problem 2. Is $\acute{\mathfrak n}$ equal to $\mathfrak c$ under Martin's Axiom?

We can also can consider the measure modification of the cardinal $\acute{\mathfrak n}$.

Namely, let $\acute{\mathfrak m}$ be the smallest cardinality of a cover of $[0,1]$ by pairwise disjoint closed subsets of Lebesgue measure zero.

The definitions imply that $\acute{\mathfrak n}\cdot\mathrm{cov}({\mathcal N})\le\acute{\mathfrak m}\le\mathfrak c$.

According to Theorem 4 of Miller, the strict inequality $\acute{\mathfrak m}<\mathfrak c$ is consistent. So, $\acute{\mathfrak m}$ is a non-trivial small uncountable cardinal.

Problem 3. Is it consistent that $\acute{\mathfrak n}<\acute{\mathfrak m}$?

Problem 4. Is $\acute{\mathfrak m}$ equal to some known small uncountable cardinal?

Added after analyzing comments to these problems: As was observed by @Ashutosh, the answer to Problem 2 is affirmative. In his paper Miller writes that this was done by Both (1968, unpublished) and Weiss (1972, unpublished). The MA equality $\acute{\mathfrak n}=\mathfrak c$ can be also derived from the ZFC inequality $$\mathfrak d\le\acute{\mathfrak n},$$ which can be proved as follows: given a partition $\mathcal P$ of $[0,1]$ into pairwise disjoint closed sets with $|\mathcal P|=\acute{\mathfrak n}$, we can choose a countable subfamily $\mathcal P'\subset\mathcal P$ such that the space $X=[0,1]\setminus\bigcup\mathcal P'$ is nowhere locally compact and hence is homeomorphic to $\omega^\omega$. Then $\mathcal P\setminus\mathcal P'$ is a cover of $X\cong\omega^\omega$ by compact subsets, which implies that $\acute{\mathfrak n}=|\mathcal P\setminus\mathcal P|'\ge\mathfrak d$ by the definition of the cardinal $\mathfrak d$.

Miller proved the consistency of the strict inequality $\acute{n}<\mathfrak c$. Looking at the diagram of small uncountable cardinals in Vaughan, I found only three small uncountable cardinals above $\mathfrak d$: $\mathfrak i$, $cof(\mathcal M)$ and $cof(\mathcal L)$.

Problem 5. Is $\acute{\mathfrak n}$ equal to one of the cardinals $\mathfrak d$, $\mathfrak i$, $cof(\mathcal M)$ or $cof(\mathcal L)$ in ZFC?

Summing up the progress made sofar. The cardinals $\acute{\mathfrak n}$ and $\acute{\mathfrak m}$ satisfy the following ZFC-inequalities:

$$\mathfrak d\le \acute{\mathfrak n}\le\acute{\mathfrak m}=\acute{\mathfrak n}\cdot\mathrm{cov}_{\sqcup}(\mathcal N)\le\mathfrak c.$$

Here by $\mathrm{cov}_{\sqcup}(\mathcal N)$ we denote the smallest cardinality of a disjoint cover of $[0,1]$ by Borel Lebesgue null sets.

It is clear that $\mathrm{cov}(\mathcal N)\le\mathrm{cov}_\sqcup(\mathcal N)$ and $\mathrm{cov}(\mathcal N)=\aleph_1$ implies $\mathrm{cov}_\sqcup(\mathcal N)=\aleph_1$.

So, $\mathrm{cov}(\mathcal N)=\aleph_1$ implies the equality $\acute{\mathfrak n}=\acute{\mathfrak m}$.

Below I collect two consistency results observed by

Will Brian: $\acute{\mathfrak m}<\mathrm{non}(\mathcal N)$ is consistent;

Ashutosh: $\acute{\mathfrak n}=\aleph_1$ in the random real model (in this model $\acute{\mathfrak m}=\mathrm{cov}(\mathcal N)=\mathrm{non}(\mathcal M)=\mathfrak i=\mathfrak r=\mathfrak c$), so $\acute{\mathfrak n}<\acute{\mathfrak m}$ is consistent.

The above results suggest the following questions:

Problem 6. Is $\mathfrak d<\acute{\mathfrak n}$ consistent?

Problem 7. Is $\acute{\mathfrak n}$ upper bounded by some known small uncountable cardinal different from $\mathfrak c$? For example, is $\acute{\mathfrak n}\le \mathrm{cof}(\mathcal N)$ true in ZFC?

Problem 8. Is $\acute{\mathfrak m}=\acute{\mathfrak n}\cdot\mathrm{cov}(\mathcal N)$?

Problem 9. Is $\mathrm{cov}_\sqcup(\mathcal N)=\mathrm{cov}(\mathcal N)$?

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    $\begingroup$ The answer to 2. is yes. See the last paragraph in math.wisc.edu/~miller/res/cov.pdf This is also theorem 3 here: Stern J. (1978) Partitions of the real line into $\aleph_1$ closed sets. In: Müller G.H., Scott D.S. (eds) Higher Set Theory. Lecture Notes in Mathematics, vol 669. Springer, Berlin, Heidelberg $\endgroup$ – Ashutosh Nov 11 '17 at 12:04
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    $\begingroup$ Theorem 5 says that $\acute{\mathfrak n} = \aleph_1$ in the random real model. In particular $cov(Null) > \acute{\mathfrak n}$ is consistent. I don't know if $\mathfrak{p} \leq \acute{\mathfrak n}$. You should check the proof of MA implies $\acute{\mathfrak n} = \mathfrak{c}$ in Stern's paper - I cannot access it. Maybe the forcing used is sigma-centered? $\endgroup$ – Ashutosh Nov 11 '17 at 13:11
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    $\begingroup$ I just looked at the proof of MA implies $\acute{\mathfrak n} = \mathfrak{c}$ in Stern's paper. He uses a countable poset (Cohen forcing) so $ \mathfrak{p} \leq \acute{\mathfrak n}$ does hold. $\endgroup$ – Ashutosh Nov 11 '17 at 14:26
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    $\begingroup$ My answer at mathoverflow.net/q/48991 shows that $\acute{\mathfrak n}\geq\text{cov}(B)$ (where $B$ means Baire category, also known as $\mathcal M$ and $\mathcal K$). $\endgroup$ – Andreas Blass Nov 11 '17 at 16:34
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    $\begingroup$ The answer to 3. is yes. In the random real model, $\acute{\mathfrak n} = \aleph_1$ while $\acute{\mathfrak m} \geq cov(Null) = \aleph_2 = \mathfrak{c}$. $\endgroup$ – Ashutosh Nov 13 '17 at 21:05

A lot of very good observations have already been put into the comments. I'll add one more observation that's too long for a comment:

It is consistent that $\acute{\mathfrak{m}} < \mathrm{non}(\mathcal L)$.

The basic idea is to start with a model of MA + $\neg$CH (where $\mathrm{non}(\mathcal L)$ is already big), and then to force over this model to make $\acute{\mathfrak{m}}$ smaller while leaving $\mathrm{non}(\mathcal L)$ large. The proof uses two facts:

Fact 1: There is a $\sigma$-centered notion of forcing $P$ that does not change the value of $\mathfrak{c}$ and that forces $\acute{\mathfrak{m}} = \aleph_1$.

Fact 2: Suppose $V$ is a model of MA, and $P$ is a $\sigma$-centered notion of forcing (in $V$). Then $P$ does not lower the value of $\mathrm{non}(\mathcal L)$. In particular, if $V \not\models$ CH and if $P$ does not change the value of $\mathfrak{c}$, then $\mathrm{non}(\mathcal L) = \mathfrak{c} > \aleph_1$ in the extension.

My claim follows easily from these two facts. To get a model of $\acute{\mathfrak{m}} < \mathrm{non}(\mathcal L)$, begin with a model of MA + $\mathfrak{c} > \aleph_1$ and force with the notion of forcing $P$ described in Fact 1. The resulting model has $\acute{\mathfrak{m}} = \aleph_1$ (because $P$ makes this true) and $\mathrm{non}(\mathcal L) = \mathfrak{c} > \aleph_1$ (by the "in particular" part of Fact 2). QED

Fact 2 is possibly "well known" but I don't know the standard reference. It was first explained to me by Andreas Blass, who exposits it nicely in the proof of Corollary 49 in this paper.

For Fact 1, the notion of forcing from Theorem 4 in this paper of Arnie Miller does the job. This forcing -- let us call it $P$ -- is $\sigma$-centered and does not change the value of $\mathfrak c$.$^{(*)}$ $P$ is designed to add a partition of $2^\omega$ into $\aleph_1$ closed sets, and it is easy to show (using a genericity argument) that each set in this partition has measure $0$; Miller even points this out in a comment after the proof of his Theorem 4.

There may be a small issue here about partitioning $2^\omega$ instead of $[0,1]$, but we can get around it. Given our partition of $2^\omega$ into $\aleph_1$ closed measure-zero sets, first observe that none of them has interior in $2^\omega$ (because then it would fail to have measure $0$). Thus there is a countable, dense $D \subset 2^\omega$ that contains no more than one point of any member of our partition. Recall that if $C$ is a closed measure-zero subset of $2^\omega$, then $C$ minus one point is homeomorphic to a countable disjoint union of Cantor sets, so it can be partitioned into countably many closed measure-zero sets. Thus, we may modify our partition by adding in the sets of the form $\{d\}$ with $d \in D$, and then dividing some other sets into countably many pieces. In this way we obtain a partition of $2^\omega$ into $\aleph_1$ closed measure-zero sets, including all sets of the form $\{d\}$ for $d \in D$. Once we have done this, we observe that there is a measure-preserving homeomorphism from $2^\omega \setminus D$ onto $[0,1] \setminus \mathbb Q$. When we push our partition through this homeomorphism, we obtain a partition of $[0,1]$ into $\aleph_1$ closed measure-zero sets.

$(*)$ Neither of these facts is stated explicitly in the linked paper, but neither is difficult to prove either. (Using Miller's notation from the linked paper:) Each forcing of the form $P(X)$ is $\sigma$-centered, because if two conditions agree on the part that asserts sentences of the form "$[s] \cap C_n = \emptyset$'' then they are compatible (take the union of the other part). It is clear that no $P(X)$ increases $\mathfrak c$, because it is too small ($|P(X)| \leq \mathfrak c$ and it has the c.c.c., so there are only $\mathfrak c$ nice names for reals). Thus $P$, which is a length-$\omega_1$, finite support iteration of forcings of the form $P(X)$, also is $\sigma$-centered and does not increase $\mathfrak c$.

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  • $\begingroup$ Thank you for this comment. Maybe it yields more, namely, that $\acute{\mathfrak m}<non(\mathcal L)$ is consistent? By the way, writing about the Baire space, maybe you had in mind $[0,1]\setminus\mathbb Q$ (not $\mathbb Q\cap[0,1]$)? Usually $\mathbb Q$ denotes the set of rational numbers. $\endgroup$ – Taras Banakh Nov 13 '17 at 19:39
  • $\begingroup$ Oops! Yes, I did mean $[0,1] \setminus \mathbb Q$, and I fixed the typo now. $\endgroup$ – Will Brian Nov 13 '17 at 20:00
  • $\begingroup$ As for $\acute{\mathfrak{m}}$, it would follow from this argument that $\acute{\mathfrak{m}} < \mathrm{non}(\mathcal L)$ if you could show that the forcing $P$ mentioned in my answer (or some other $\sigma$-centered forcing) adds an $\aleph_1$-sized partition of $[0,1]$ into closed measure-zero sets. I did consider this, but don't know whether it's true. The partition it adds for $2^\omega$ is a partition into closed measure-zero sets, but it's not obvious (to me, anyway) how to get from there to a partition on $[0,1]$. I will think about it some more, though. $\endgroup$ – Will Brian Nov 13 '17 at 20:03
  • $\begingroup$ @TarasBanakh: OK, I think I can see how to get $\acute{\mathfrak{m}} < \mathrm{non}(\mathcal L)$ now. I have to go teach now, but will edit my answer when I have the time (tomorrow at the latest). $\endgroup$ – Will Brian Nov 13 '17 at 20:20
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    $\begingroup$ @TarasBanakh: $\mathrm{non}(\mathcal M)$ is $\aleph_1$ in this extension because Cohen reals are added at every stage of the iteration (a set $\{r_\alpha : \alpha \in \omega_1\}$ of reals where $r_\alpha$ is Cohen over $V^{P_\alpha}$ will be non-meager in the extension). I'm not sure about $\mathfrak{i}$, but I'll think about it. $\endgroup$ – Will Brian Nov 13 '17 at 21:56

None of the cardinals in Cichoń's diagram is an upper bound for $\acute{\mathfrak{n}}$ (or $\acute{\mathfrak{m}}$). This follows from a theorem of Spinas:

Theorem: there is a proper forcing $P$ with the Sacks property such that forcing with $P$ over a model of $\mathsf{CH}$ makes $\acute{\mathfrak{n}} = \aleph_2 = \mathfrak{c}$. Hence $\mathrm{cof}(\mathcal L) < \acute{\mathfrak{n}},\acute{\mathfrak{m}}$ is consistent.

Of course, $\mathrm{cof}(\mathcal L) = \aleph_1$ in the extension implies that all the cardinals in Cichoń's diagram are $\aleph_1$ in the extension.

It has already been pointed out that $\mathfrak{d} \leq \acute{\mathfrak{n}}$, that $\acute{\mathfrak{n}} < \mathrm{non}(\mathcal L)$ is consistent, and that $\acute{\mathfrak{n}}$ is $\aleph_1$ in the random model. Putting all these facts together, we see

$\mathfrak{d} \leq \acute{\mathfrak{n}}$, but if $\kappa$ is any of the cardinals in Cichoń's diagram not bounded above by $\mathfrak{d}$, then $\acute{\mathfrak{n}}$ is independent of $\kappa$, in the sense that both $\kappa < \acute{\mathfrak{n}}$ and $\acute{\mathfrak{n}} < \kappa$ are consistent.

Spinas proves the above theorem in

O. Spinas, "Partition numbers,'' Annals of Pure and Applied Logic 90 (1997), pp. 243-262. (link)

See Theorem 2.8. The notion of forcing $P$ used in Spinas' proof was actually introduced earlier, in this paper of Arnie Miller (the one Ashutosh links to in the comments). Miller proves that $P$ has the Laver property (in modern terminology -- he doesn't phrase it that way in the paper) and concludes that $\mathrm{cov}(\mathcal M) < \acute{\mathfrak{n}}$ is consistent. Spinas shows that $P$ is $\omega^\omega$-bounding which, combined with Miller's result, means it has the Sacks property, which implies $\mathrm{cof}(\mathcal L) < \acute{\mathfrak{n}}$ is consistent. The forcing $P$ is an iteration of tree-like forcings, each of which kills an $\aleph_1$-sized partition in the ground model. The forcing used to kill a partition $\mathcal X$ is the set of all perfect trees $T \subseteq 2^{<\omega}$ having the property that every member of $\mathcal X$ is relatively meager in $[T]$ (the set of branches through $T$).

Finally, let me note that $P$ having the Sacks property does not suffice to conclude that $\mathfrak{i} = \aleph_1$ in the extension. For example, a countable support iteration of Silver forcings has the Sacks property, but $\mathfrak{r} = \mathfrak{i} = \aleph_2$ in the Silver model. (This is proved in Halbeisen's book.) So we cannot conclude that forcing with $P$ makes $\mathfrak{i} = \aleph_1$, and in fact I don't know what is the value of $\mathfrak{i}$ after forcing with $P$. So your question about whether $\mathfrak{i}$ could be an upper bound for $\acute{\mathfrak{n}}$ or $\acute{\mathfrak{m}}$ remains open, although I suspect the answer is negative.

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  • $\begingroup$ Thank you for another nice answer! $\endgroup$ – Taras Banakh Jun 17 at 21:03

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