0

Here is my proof, given that $|A|=a$ and $\omega=|\mathbb{N}|$(also commonly denoted as $\aleph_0$):

  1. Let $F$ be the set of all bijections of the form from $B^{\omega}$ to $B$ where $B\subset A$ ($F$ is not empty, since $(2^{\omega})^{\omega}=2^{\omega}$);

  2. Order $F$ using extension of function, apply Zorn's Lemma for a maximal function $g:B^{\omega}\to B$;

  3. To show that $|B|=|A|$, notice that otherwise $A\setminus B$ contains a set $C$ with the same cardinality as $B$, and so $g$ can be extended to $(B\cup C)^{\omega}\to (B\cup C)$, which violates the maximality of $B$;

  4. Since $|B|=|A|=a$ and $|B^{\omega}|=|B|$, we have $a^{\omega}=a$.

This looks very fishy, as I failed to discover a theorem that directly implies this. Is the proof above correct?

Ginger88895
  • 184
  • 3
    How do you know that Zorn's lemma applies? The union of a chain of bijections $f$ for which there is a $B\subseteq A$ such that $f!:B^{\aleph_0}\to B$ does not need to have this form. – Andrés E. Caicedo Jul 28 '18 at 13:55
  • Ah I see... taking $\bigcup B_i$ does not work since $B^{\aleph_0}$ is an infinite sequence. Thanks for the help! :D – Ginger88895 Jul 28 '18 at 14:12

1 Answers1

1

The claim is false in the first place. A consequence of Konig's theorem is that for all infinite cardinals $\kappa$, it holds $$\kappa<\kappa^{\operatorname{cof}\kappa}$$ and there is plenty of cardinals $\kappa$ such that $\kappa>2^{\aleph_0}$ and $\operatorname{cof}\kappa=\aleph_0$. Notably, $\beth_\omega$ (see here for the definition of beth-sequence).

  • Thanks... after some further research I am aware that Konig's theorem disproves this. However, I am very curious about where my proof fails :) – Ginger88895 Jul 28 '18 at 13:29
  • @Ginger88895 It is quite unclear how you can extend $g$ in point (3). You seem to be under the false impression that $(B\cup C)^\omega=B^\omega\cup C^\omega$, while actually the LHS is much larger: there are all the functions that send some natural numbers to elements of $B$ and the other ones to elements of $C$. –  Jul 28 '18 at 13:37
  • Actually I picked $C$ so that $|C|=|B|$... then since $B$ is infinite, $|B\cup C|=|B|$, and also $|(B\cup C)^{\omega}\setminus B^{\omega}|=|C|$, which indicates the existence of a bijection from $(B\cup C)^{\omega}\setminus B^{\omega}$ to $C$... – Ginger88895 Jul 28 '18 at 13:40