19
votes

Accepted

### Is $[0,1]$ a disjoint union of $\aleph_1$ compact subsets with empty interior?

It is independent of ZFC. As you mention, CH implies the answer is yes. A different axiom, MA+$\neg$CH, implies the answer is no.
A little more precisely, there is a cardinal number denoted $\mathrm{...

16
votes

### Brief history of cardinal characteristics of the continuum

There is a historical note to §3 (“Six cardinals”) in chapter 3 (“The Integers and Topology”, by Erik K. van Douwen) in the 1984 Handbook of Set-Theoretical Topology edited by Kunen and Vaughan. For ...

15
votes

Accepted

### How “disconnected” can a continuum be?

The answer to all three of your questions is yes.
The cardinal $\mathrm{disc}([0,1])$ is discussed in this MO question of Taras Banakh. He calls this cardinal the Sierpiński cardinal and denotes it $\...

12
votes

### Diagonalizing against $\omega_1$-sequences of functions mod finite

This isn't an answer, as you're working in ZFC. But it seems worth noting.
Assume ZFC + AD$^{L(\mathbb{R})}$. Then
$L(\mathbb{R})$ satisfies ZF + AD + DC + "the statement is false".
Proof: ...

12
votes

Accepted

### $\operatorname{cof}(\mathcal{L}) = \aleph_1$ and $2^{\aleph_0} = \aleph_3$

Yes, this is consistent. Probably the simplest way to get a model of this theory is to begin with a model of GCH, and then force with a countable support product of $\aleph_3$ copies of the Sacks ...

11
votes

Accepted

### Is there a condensation from $\aleph_1^{\aleph_0}$ onto a metrizable compact space?

If $|D| < \aleph_\omega$, then there is a condensation from $D^\omega$ onto $\omega^\omega$ (the Baire space) if and only if there is a partition of $\omega^\omega$ into exactly $|D|$ Borel sets.
...

11
votes

Accepted

### Can we force $\mathfrak{r}<\mathfrak{s}$?

The inequality $\mathfrak{r} \leq \mathfrak{u}$ is provable in ZFC (because every base for an ultrafilter is a reaping family). Blass and Shelah proved the consistency of $\mathfrak{u} < \mathfrak{...

10
votes

### Brief history of cardinal characteristics of the continuum

Juris Steprans's article in the Handbook of the History of Logic: "History of the Continuum in the 20th century".
http://www.math.yorku.ca/~steprans/Research/PDFSOfArticles/hoc2INDEXED.pdf

10
votes

### Cofinal well-founded subset in mod finite order

The answer is yes, because indeed every partial order admits a well founded cofinal (i.e. dominating) subset. There is no need to consider cardinal characteristics.
Theorem. Every partial order admits ...

9
votes

Accepted

### Effect of adding one Hechler real versus adding two on the meager ideal

If $c$ is Cohen over $V$, and $d$ is dominating over $V[c]$ (not necessarily Hechler-generic), then in $V[c][d]$ there is a meager set covering all meager sets from $V$. Hence 2 successive Hechler ...

9
votes

Accepted

### A simple cardinal characteristic associated with $\omega^\omega$

I claim that $\mathfrak{c}\mathfrak{p}= \mathfrak{d}$.
For any $x\in \omega^{*\omega}$ define its "inverse" $x'$ by $x'(n) = \min \{k\mid\forall j\ge k : x(j)>n\}$. If $x$ grows very fast, then $...

9
votes

### Is there a condensation from $\aleph_1^{\aleph_0}$ onto a metrizable compact space?

The answer is affirmative and can be derived from
Theorem (Banakh, Plichko). The Hilbert space $\ell_2(\aleph_1)$ condenses onto the Hilbert cube.
By the way, this theorem is related to Problem 1 ...

9
votes

Accepted

### Is it true that $\mathit{MA}(\omega_1)$ iff $\omega_1<\mathfrak{p}$?

Q1: No, see Between Martin's Axiom and Souslin's Hypothesis by Kunen and Tall. Note: Bell proved in The combinatorial principle $P(\mathfrak{c})$ that $\mathfrak{p}>\aleph_1$ is equivalent to $\...

9
votes

Accepted

### Cofinal rectangles in poset

Let $P$ be the set of all countable subsets of $\omega_2$ ordered by set inclusion. If $2^{\aleph_0}\le\aleph_2$ then $P$ has cofinality $\aleph_2$ and additivity $\aleph_1$ and contains no chain of ...

8
votes

Accepted

### What if $\mathbb{R}$ is in bijection with the cardinals less than $\frak{c}$?

Yes. Start with a model of $\sf CH$, then take the least fixed point with uncountable cofinality. Call that $\kappa$. Now add $\kappa$ Cohen reals.
Since fixed points form a club of ordinals, you can ...

8
votes

Accepted

### The "strong" measure number

Chapter 8 of "Set Theory On the structure of the real line" by Bartoszynski and Judah compiles a lot of combinatorial results about strong measure zero sets. Theorem 8.1.14 gives an old result of ...

8
votes

### Self-embeddings of uncountable total orders, 2

For the first question, the answer is YES, even if we just assume that $S$ is not rigid. Indeed, fix $f\in\mathrm{Aut}(S)$ and $a\in S$ such that $a<f(a)$. Then $\bigl(\mathrm{id}\restriction(-\...

7
votes

Accepted

### Pseudo-intersections, splitting families, and ultrafilters

The answer is no -- it is consistent that every $U \in \omega^*$ has $\tau(U) < \mathfrak{s}$.
I had an idea for proving this earlier today, using the Mathias model. I couldn't quite make things ...

7
votes

### A new cardinal characteristic (related to partitions)?

This is not an answer, but hopefully it's a helpful observation:
(1) If $U$ is an ultrafilter on $\omega$ and $\mathcal{P}$ is a finitary partition of $\omega$, then there is $A\in U$ such that $A\...

7
votes

Accepted

### Is $\mathfrak p=\omega_1$ equivalent to the existence of a Hausdorff gap without infinite pseudointersection?

Yes. This is a result due to Nyikos and Vaughan from 1983, appearing the paper
Nyikos, Peter J.; Vaughan, Jerry E., On first countable, countably compact spaces. I: ((\omega_ 1,\omega^*_ 1))-gaps, ...

7
votes

### Are the Sierpiński cardinal $\acute{\mathfrak n}$ and its measure modification $\acute{\mathfrak m}$ equal to some known small uncountable cardinals?

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}(\...

7
votes

Accepted

### The cardinal characteristic $\mathfrak r_{(X,f)}$ of a dynamical system

Unfortunately, $\mathfrak r_{(2^\omega,f)}\ge\mathfrak r$. Indeed, let $\mathcal R$ be a family of infinite subsets of $\omega$ such that $|\mathcal R|=\mathfrak r_{(2^\omega,f)}$ and for any $x=(x_n)...

7
votes

Accepted

### "Compactness length" of Baire space

Baire space is the union of $\mathfrak d$ (the dominating number) compact subsets. So, using equivalence relations that collapse those sets one at a time (i.e., one equivalence class is the set to be ...

6
votes

Accepted

### A sequence of cardinal characteristics constructed with hypergraph coloring

The cardinals $\bf k_n$ ($2\le n\lt\omega$) are all equal.
Lemma. Let $\kappa$ be an infinite cardinal. Given a set $A\subseteq[\omega]^\omega$ with $|A|=\kappa$ and $\chi(\omega,A)\gt n$, we can ...

6
votes

### Can we separate the almost-disjointness sunflower numbers?

The answer to Question 1 is NO:
Split $\omega$ into two disjoint infinite sets $X_0$ and $X_1$, fix $\mathcal A_0$ a countable AD family of subsets of $X_0$ such that
$\mathcal A_0$ does not contain ...

6
votes

Accepted

### Diagonalizing against $\omega_1$-sequences of functions mod finite

The arguments in Section 6 of my paper "The nonstationary ideal in the $\mathbb{P}_{\mathrm{max}}$ extension" show that there is a proper forcing adding a function from $\omega_{1}$ to $\...

6
votes

Accepted

### Comparing bornologies for domination/escaping

Note that $\mathfrak{b}=\mathfrak{d}$ is equivalent to the existence of a $<^\ast$-increasing sequence $(f_\alpha)_{\alpha<\mathfrak{d}}$ which is cofinal in $(\mathcal{N},{<^\ast})$, where $...

6
votes

Accepted

### Cofinal trees in $({}^\omega \omega , \leq^\ast )$

Hechler proved that you can force the existence of a cofinal subset of $\omega^\omega$ having any shape you like (subject to one or two necessary restrictions):
Theorem: (Hechler) Suppose that $(P,\...

6
votes

Accepted

### Comparing bornologies for cardinal characteristics via Borel maps

A bornomorphic map $i\colon\mathcal{N}\to\mathcal{N}$ for these bornologies cannot be Borel.
First, I shall use $f<^\infty g$ to mean that $(\forall m\in\mathbb{N})(\exists n>m)f(n)<g(n)$, ...

5
votes

Accepted

### What's the minimal weight of a maximal space?

No, these cardinals are not provably equal.
This follows from a result of El'kin 1:
Let $X$ be a set and let $x \in X$. If $\tau$ is a maximal topology on $X$, then $\{U \setminus \{x\} \,:\, x \...

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