2
$\begingroup$

I could use some help proving the following:

Let $A$ be a Dedekind Finite set of pairwise disjoint Dedekind finite sets $\left(\mbox{i.e each}\, a\in A\,\mbox{is a Dedekind Finite set}\right)$ show that $\bigcup A$ is Dedekind Finite.

Thanks in advance!

2 Answers 2

4

HINT: Let $S=\bigcup A$, and suppose that there is an injection $\omega\to S$. For $a\in A$ let $N_a=\{k\in\omega:f(k)\in a\}\;.$ Show that $N_a$ is finite for each $a\in A$ and use that result to construct an injection from $\omega$ to $A$, contradicting the Dedekind-finiteness of $A$.

  • 0
    What a good proof and such a good discussion. Wasn't able to understand proof first but ultimately got any single hint from here only2015-06-25
2

If $B\subseteq\bigcup A$ is a countably infinite set, then $B_a=B\cap a$ is a Dedekind-finite set for every $a\in A$.

Show that every partition of a countably infinite set into Dedekind-finite sets is finite, and one of the parts has to be countably infinite.

Deduce that there is some $a\in A$ such that $B_a$ is countably infinite which is a contradiction to Dedekind-finiteness of $a$.


To prove the fact in the middle note that we may assume that the infinite set is $\mathbb N$ and that a partition is merely a surjection. Show, if so, that every surjection from $\mathbb N$ onto a Dedekind-finite set implies that the range is finite. (Hint: If $f\colon\mathbb N\to X$ is a surjection then there is an injection $g\colon X\to\mathbb N$.)

  • 0
    Not$e$ that both proofs are actually the same. It just might be slightly easier to use one over the other depending on how you defined a Dedekind-finite set.2012-12-23