For what follows, I work in ZF+AD+DC. However, the questions below are not obviously trivial in ZFC, so I'm also interested in results in that system.

Suppose I have a set $X\subseteq \mathbb{R}$. Let $\Theta(X)$ be the supremum of the ordinals onto which $X$ surjects. In the presence of choice, this is of course just the successor of the cardinality of $X$; in the presence of determinacy, the $\Theta$ function takes on only values $\le\omega_1$ or $\Theta=\Theta(\mathbb{R})$.

However, things get much weirder when we try to cover an ordinal by many sets simultaneously. For $\mathcal{F}$ a family of sets, let $$Spread(\mathcal{F})=\sup\{\alpha: \exists f:\mathbb{R}\rightarrow ON: \forall X\in\mathcal{F}(f(X)=\alpha)\},$$ that is, the sup of all ordinals onto which the elements of $\mathcal{F}$ simultaneously surject. My questions are below. Some initial observations (again, this is in ZF+AD+DC):

  • $Spread(\{$perfect sets$\})$ is $1$. For $\alpha>1$ countable, if $f:\mathbb{R}\rightarrow \alpha$ then some $\eta\in\alpha$ has uncountable preimage; this preimage contains a perfect set, by AD. For $\alpha$ uncountable, let $S\subset \alpha$ be an uncountable proper subset of $\alpha$; then the $f$-preimage of $S$ again contains a perfect subset, for any $f: \mathbb{R}\rightarrow\alpha$.

  • For $n\in\omega$, let $\mathbb{R}=\bigsqcup_{i\in n} P_i$ and let $\mathcal{F}=\{S: \forall i<n(S\cap P_i\not=\emptyset)\}$. Then it's easy to see that $Spread(\mathcal{F})=n$.

  • Similarly, let $\mathbb{R}=\bigsqcup_{i\in\omega}P_i$ and let $\mathcal{F}=\{S: \exists j\in\omega\forall i\in\omega(i>j\rightarrow S\cap P_i\not=\emptyset)\}$. Then $Spread(\mathcal{F})=\omega_1$: consider the map $f(r)=k$ iff $r\in P_{\langle k, m\rangle}$ for some $m$.

  • EDIT: Let $\mathcal{A}=\{S: \forall \alpha<\omega_1\exists x\in S(\omega_1^x\ge\alpha)\}$. Then $Spread(\mathcal{A})=\omega_2$: there is an obvious surjection onto $\omega_1$ (hence any $\alpha<\omega_2$). Conversely, for $f:\mathbb{R}\rightarrow\omega_2$ and $\alpha<\omega_1$ let $S_\alpha$ be the set of reals $x$ with $\omega_1^{CK}\ge\alpha$ which have minimal $f$-value among all such reals; by the regularity of $\omega_2$, $\bigcup_{\alpha<\omega_1} S_\alpha$ is not mapped onto $\omega_2$ by $f$. I suspect that we can get all successor ordinals below $\Theta$ in a similar way, using the Coding Lemma.

OK, so onto the question. First, a couple minor preliminary questions:

  • Has this notion been studied before? I have a strong sensation that I'm (yet again) reinventing the wheel; however, googling around hasn't been helpful here.

  • Can we have $Spread(\mathcal{F})=\omega$ for $\mathcal{F}\subseteq\mathcal{P}(\mathbb{R})$? This is a very minor question, I'm just curious.

My main question is:

What is $Spread(\{$dominating sets$\})$?

More generally, I'm interested in how $Spread$ interacts with cardinal characteristics of the continuum - e.g. it's easy to see that $Spread(\{$unbounding sets$\})\le Spread(\{$dominating sets$\})$ - but this seems a good test case.

  • 2
    $\begingroup$ In the presence of choice, $\Theta(X)$ is the successor of the cardinality of $X$. If $X$ is countable, then in ZF we have $\Theta(X)=\omega_1$. $\endgroup$
    – Goldstern
    Jul 6, 2016 at 20:46
  • $\begingroup$ @Goldstern Derp. This is why post-beer math is a bad idea. Fixed! $\endgroup$ Jul 6, 2016 at 21:01
  • 3
    $\begingroup$ Noah, agree to disagree on that comment. $\endgroup$
    – Asaf Karagila
    Jul 6, 2016 at 21:06
  • $\begingroup$ Note: a previous version of this question had the main question garbled to the point of meaninglessness, although I think what I was trying to ask was clear from context. I've fixed that now. $\endgroup$ Jul 10, 2016 at 22:06

1 Answer 1


I still don't know whether this has been studied before (and so references are still welcome!), but unsurprisingly, $Spread(\{$dominating families$\})=1$.

We begin with an easy lemma. (Note that in an earlier draft of this question, I mistakenly claimed that it wasn't obvious whether this was true. I'm going to go ahead and blame that on the Saison.)

Lemma. The set of ordinals onto which a family of sets spreads, is closed downwards.

Proof. Suppose $\mathcal{A}$ spreads onto $\alpha$ via $f$ and $\beta<\alpha$; let $g: \alpha\rightarrow\beta$ be surjective and consider $g\circ f$.

Observation. The set $\mathcal{D}$ of dominating families does not spread onto $2$.

Proof. Let $f:\omega^\omega\rightarrow 2$ and suppose neither $f^{-1}(0)$ nor $f^{-1}(1)$ is a dominating family. Then there are $x\in f^{-1}(0)$ and $y\in f^{-1}(1)$ such that

  • $x$ escapes every element of $f^{-1}(1)$, and

  • $y$ escapes every element of $f^{-1}(0)$.

But then $x+y$ escapes every real, including itself. $\Box$

Based on this, the only cardinal characteristic-y question left is:

Do any of the usual "large" families of sets of reals have $Spread>1$?

Of course, there are still questions left:

  • What is the range of $Spread$? In particular, can we have $Spread(\mathcal{A})=\omega$? And does the answer here depend on choice, determinacy, etc.?

But this does answer the specific question I asked here.


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