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?
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.