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Recently in the chat, we are doing some studies on properties of infinite Dedekind finite sets (iD-finite sets). We started with the basics by trying to prove that weakly even and weakly odd iD-finite sets have the same properties as when adding even and odd natural numbers

Let $D$ be a iD-finite set. Then $D$ is weakly even if $D$ can be expressed as a disjoint union of pairs $\{a,b\}$ and weakly odd if $D$ can be expressed as a disjoint union of pairs plus a singleton.

In this answer, it is mentioned that $D$ can be weakly even, weakly odd but never both. Intrigued, we tried to prove that via a proof by contradiction:

Attempt proof: Let $D$ be iD-finite. Suppose $D$ is both weakly even and weakly odd. Then there exists a bijection $f : C \mapsto A$ such that:

$$|D|=|A|=|C|$$

where $$A = \bigsqcup_{a,b} \{a,b\}$$ and

$$C = \bigsqcup_{c,d} \{c,d\} + \{e\}$$

Since $A, C$ are iD-finite, $K = f (\bigsqcup_{c,d} \{c,d\})$ and $K \subsetneq A$. Therefore what remains is to show that any bijection between $A-K$ and $f(\{e\})$ will lead to a contradiction. However, I cannot seemed to find any suitable bijection or otherwise to guarentee that

$$f(\{e\}) \subsetneq A - K$$

Because these are Dedekind cardinals, I cannot really trust my intuition that the difference between a collection of pairs must be a pair (and I cannot use the fact weakly even + weakly even = weakly even either because the same strategy is used as in this attempted proof). How can I show a Dedekind cardinal of pairs cannot be surreally decomposed to give an extra singleton?

Bernard
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Secret
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  • First, you want "union of pairs" to be a "disjoint union of pairs", otherwise any set with two elements or more is even. Now, what if it cannot be expressed as a union of pairs, but can be expressed as a union of sets of size $4$? – Asaf Karagila Oct 11 '18 at 08:18
  • hmm... I suspect any finite set of even natural number should be partitionable into pairs so a disjoint union of 4s should be partitionable into a disjoint union of pairs because for each 4s we can split that into two pairs... o wait a minute, then I am making an uncountable number of splittings which it might not exist without choice or without a given bijection, hmmm... – Secret Oct 11 '18 at 08:31

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Introducing new sets is just confusing. You have a set $D$ with a partition into pairs, say $P$, and a partition into pairs + a singleton, call it $O$. Let's call that singleton $\{d\}$.

Now we will define the following function:

$f(d)=a_0$ such that $\{a_0,d\}\in P$. This $a_0$ exists, and it is not $d$ itself, of course, since $P$ is a partition into pairs. Next, $f(a_0)=a_1$ such that $\{a_0,a_1\}\in O$, this $a_1$ exists since the unique singleton in $O$ was $\{d\}$ and $d\neq a_1$.

And so we proceed: $f(a_{2n+1})=a_{2n+2}$ such that $\{a_{2n+1},a_{2n+2}\}\in O$ and $f(a_{2n+2})=a_{2n+3}$ such that $\{a_{2n+2},2_{2n+3}\}\in P$.

But wait, what have we done??? This is really a function from $\Bbb N$ into $D$. And it is injective!

So $D$ is Dedekind-infinite.

Asaf Karagila
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  • P.S. I purposefully left some details out, such as why this function is injective. Let me know if you want me to elaborate on this. – Asaf Karagila Oct 11 '18 at 08:30
  • I think some elaboration will be good, as I felt I don't fully understand how $f$ combined with the presence of the singleton lead to a well ordering without choice and I felt understanding the details of its injectivity may help me. – Secret Oct 11 '18 at 08:52
  • What? Who said well-ordering? – Asaf Karagila Oct 11 '18 at 08:52
  • Uh, I got that impression because of the subscripts of $a$ as $f$ seemly map from one element to the next in $D$ starting from the singleton and how you said we end up with a function from $\Bbb{N}$ into $D$, and if I recall, $\Bbb{N}$ is well ordered? I might be confused on something... – Secret Oct 11 '18 at 08:57
  • Yes, $\Bbb N$ is well-ordered, but what does that have to do with the question? Do you know that a set is Dedekind-infinite if and only if there is an injective function from $\Bbb N$ into the set? – Asaf Karagila Oct 11 '18 at 08:58
  • Ok, I guess my confusion is the following: Why defining $f$ does not invoke the axiom of countable choice, as the construction basically maps one element to the next infinitely? – Secret Oct 11 '18 at 09:03
  • Yes, because there is no choice needed. There exists exactly one option to choose from. (And just as a side remark, that kind of definition usually requires Dependent Choice, not just countable choice.) – Asaf Karagila Oct 11 '18 at 09:04
  • Ah I see, the singleton provided a unique reference point, thus $f$ can map one element to the next without repetition, hence ensuring injectivity and choiceless. – Secret Oct 11 '18 at 09:06
  • Exactly the point. Yes. – Asaf Karagila Oct 11 '18 at 09:06