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Assuming $ZF$ itself is consistent, it is consistent that there are sets $D$ which are infinite but cannot be placed in bijection with any of their proper subsets; such sets are called "strictly Dedekind-finite." Consistently, there is even a Dedekind-finite set of reals.

My question is, is it consistent to be able to partition $\mathbb{R}$ into strictly Dedekind-finite sets?

The simplest way to produce strictly Dedekind-finite sets of reals is to use Cohen forcing and take a symmetric submodel of the resulting forcing extension. We can also create $\kappa$-many disjoint strictly Dedekind-finite sets in a similar fashion, for any $\kappa$, without collapsing $\kappa$ (of course, the continuum is bumped above $\kappa$). However, as far as I can see, there is no simple way to adapt this to provide a partition of $\mathbb{R}$ into such sets.

One simple thing I've been able to figure out: suppose $\kappa$ is a (well-ordered) cardinal, $\{D_i: i\in I\}$ is a partition of $\mathbb{R}$ into strictly Dedekind-finite sets, and $\kappa^+$ injects into $\mathbb{R}$. Then there cannot be an injection $I\rightarrow\kappa$. However, letting $\Psi$ be the least ordinal not injectible into $\mathbb{R}$, it is not even clear to me that we cannot partition $\mathbb{R}$ into $\Psi$-many strictly Dedekind-finite pieces; and of course this says nothing about the case when the index set $I$ itself is non-well-orderable.

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  • $\begingroup$ I'm pretty sure what I'm calling "$\Psi$" actually has a name, but I can't seem to remember/find it right now. $\endgroup$ – Noah Schweber Feb 9 '14 at 4:36
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    $\begingroup$ Nice question! Your cardinal $\Psi$ is called the Hartog number $\aleph(\mathbb{R})$, see en.wikipedia.org/wiki/Hartogs_number. $\endgroup$ – Joel David Hamkins Feb 9 '14 at 5:01
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    $\begingroup$ @Joel, Hartogs, he was French. $\endgroup$ – Asaf Karagila Feb 9 '14 at 5:17
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    $\begingroup$ @Joel: Yeah, although I prefer "Hartogs' number". $\endgroup$ – Asaf Karagila Feb 9 '14 at 5:39
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    $\begingroup$ @AsafKaragila I think he was German (though born in Brusseles). $\endgroup$ – user9072 Feb 9 '14 at 10:49
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YES WE CAN!

Suppose that there is an infinite Dedekind-finite set of real numbers $A$ (e.g. Cohen's first model). Simple cardinal arithmetic shows that, $$|\Bbb R|\leq|\Bbb R\times A|\leq|\Bbb{R\times R}|=|\Bbb R|.$$

Clearly $\Bbb R\times A$ can be partitioned into infinite Dedekind-finite sets, simply consider $\{\{r\}\times A\mid r\in\Bbb R\}$. Now use a bijection of $\Bbb R\times A$ with $\Bbb R$ to transport this partition to a partition of $\Bbb R$ without changing the cardinality of its parts.

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  • $\begingroup$ Nice! A followup question: can we partition $\mathbb{R}$ into fewer than continuum-many amorphous sets? $\endgroup$ – Noah Schweber Feb 9 '14 at 9:35
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    $\begingroup$ No, not even more than continuum many. Any map from an amorphous set to a linearly ordered set has a finite range. So we cannot partition $\Bbb R$ in a way that there is any part which is amorphous. (Well, I'm not counting finite sets, of course.) $\endgroup$ – Asaf Karagila Feb 9 '14 at 10:18
  • $\begingroup$ Sorry, I meant "strictly Dedekind-finite," not amorphous. :P $\endgroup$ – Noah Schweber Feb 9 '14 at 18:27
  • $\begingroup$ Noah, that's a much more difficult task. I'll have to think about it for a bit. $\endgroup$ – Asaf Karagila Feb 9 '14 at 21:09
  • $\begingroup$ I've turned my followup into a real question (mathoverflow.net/questions/157204/…). $\endgroup$ – Noah Schweber Feb 10 '14 at 9:19

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