3

I am self-studying Horst Herrlich, Axiom of Choice (Lecture Notes in Mathematics, Vol. 1876), and I'm getting quite confused about cardinal arithmetic without AC.

Here (Which sets are well-orderable without Axiom of Choice?) OP writes that $\mathbb R$ is not necessarily well-orderable in ZF. At the same time, it seems to me that $\mathcal P(\omega)$ is well-orderable in ZF (attempt of proof follows). Therefore, either my proof is wrong, either the answer to the original question is "Yes". Any help? Thanks.

Claim ZF $\vdash$ "$\mathcal P(\omega)$ is well-orderable"

Proof We know that $2=\{0,1\}$ is a cardinal number, therefore it is well-ordered w.r.t. "$\in$" (of course this could be manually checked too). Same holds for $\omega$. (I'll use "$<$" instead of "$\in$"). Let's consider $2^\omega = \{f \mid f : \omega \to 2\}$. The relation defined by $$f \prec g \iff \exists n \, [f(n) < g(n) \wedge \forall i\!\!<\!\!n \;\; f(i)=g(i)]$$ is a well-order on $2^\omega$ (and this is true in ZF). To conclude, just observe that ZF $\vdash 2^\omega \approx \mathcal P(\omega)$.

Update: The proof is wrong, since the lexicographic order on $2^\omega$ is clearly not a well-order, as pointed out in Asaf Karagila's answer. Asaf claims also that ZF $\not\vdash$ "$2^\omega$ is well-orderable". But then, I have another question: in the book I cite above, there is the following proof (page 51):

enter image description here

Now: If $2^\omega$ is not necessarily well-orderable, why is the author allowed to use $2^{\aleph_0}$ as a cardinal number? Or, equally, why is he allowed to talk about the cardinality of $\prod_{n \in \mathbb N} B_n$?

Solution to update: It was just a matter of definitions. In the text I've studied these topics, cardinal numbers are defined as ordinals for which there is no bijection with smaller ordinals. Then, given a set $A$, $|A|$ is a cardinal number, therefore it is defined iff $A$ is well-orderable. On the contrary, in this book $|A|$ is intended as the equivalence class of $A$ under the equivalence relation $A \sim B \iff $ there exist a bijection $A \to B$. This way, $|A|$ is always well-defined.

Community
  • 1
aerdna91
  • 1,144
  • 6
  • 19
  • 1
    The order you define is called the lexicographic order. It is not a well-ordering. (Try finding an infinite descending chain!) – Zhen Lin Jul 22 '14 at 18:24
  • 2
    While you're right that your question has a positive answer and your proof is wrong, it's perfectly possible that the answer is negative and that the proof is wrong as well. You don't account for all the possible outcomes! :-) – Asaf Karagila Jul 22 '14 at 18:26
  • @AsafKaragila D'OH! – aerdna91 Jul 22 '14 at 19:25

2 Answers2

9

You don't need the axiom of choice to show that $|\Bbb R|=2^{\aleph_0}$. If you observe carefully neither the injection from $\Bbb R$ into $\mathcal P(\omega)$ defined by enumerating the rationals and using Dedekind cuts, nor the injection from $\mathcal P(\omega)$ into $\Bbb R$ defined by creating the Cantor set use the axiom of choice.

Since the proof of the Cantor-Bernstein theorem does not rely on the axiom of choice either, this shows that $\sf ZF\vdash|\Bbb R|=|\mathcal P(\omega)|$.

As for your proof that $2^\omega$ is a well-orderable set, this relation is not well-founded (so it cannot be a well-order). Let $f_n$ be the function which is constant $0$ until $n$ and then $1$ from $n$ onwards. Then $f_n\prec f_k$ if and only if $n>k$ which means this is a decreasing sequence.

To your edit, note that cardinal exponentiation is still well-defined. Just as a product of two cardinals need not be equal to the infinite sum of one by an index of the other, the exponentiation need not be the product of one by itself over the index of the other.

This just gives more weight to the old saying that multiplication is not repeated addition (and exponentiation is not repeated multiplication).

We define the cardinal arithmetic directly: $|A|+|B|$ is the cardinality of $|A\cup B|$ where $A\cap B=\varnothing$; $|A|\cdot|B|=|A\times B|$; and $|A|^{|B|}=|A^B|$, namely the cardinality of all functions from $B$ into $A$.

And note that $\prod_{i\in I}X_i$ is a particular set, so it has a well-defined cardinality. Whether or not it is equal to $\prod_{i\in I}Y_i$ under the assumption that $|X_i|=|Y_i|$ for all $i\in I$ is a whole other story. But particular products have particular cardinalities.

leo
  • 10,433
Asaf Karagila
  • 393,674
  • 1
    Or, alternatively, that ${f_n\mid n\in \mathbb N}\subseteq 2^{\mathbb N}$ is non-empty and does not contain a least element. (Not sure many beginners know the term "well-founded.") – Thomas Andrews Jul 22 '14 at 18:36
  • 1
    @ThomasAndrews I'm not a beginner (despite my SE reputation), I perfectly know what well-order means, and I've seen Asaf's counterexample so many times that I can't even count them (without AC, at least :). Which makes it even worse. The fact is that I am dumb and sometimes when I am tired I rely on memory instead of logic. And each time I miserably fail. This time, I thought that I remembered that lexicographic order was a well-order on $2^\omega$, and I didn't even check. My bad. Thanks Asaf anyway...always the best (and the first too)! Anyway, you should switch 1 and 0 in your definition. – aerdna91 Jul 22 '14 at 19:21
  • @aerdna91 "Well-founded" is not "well-ordered." Well-founded is a generalization to non-order relationships. – Thomas Andrews Jul 22 '14 at 19:30
  • Ok, now I see what you meant. BTW, I knew that too. – aerdna91 Jul 22 '14 at 19:39
  • @aerdna91: You're welcome. – Asaf Karagila Jul 22 '14 at 20:28
  • @leo: Thanks for the edit. – Asaf Karagila Jul 22 '14 at 20:57
  • @AsafKaragila I'm sorry but I still can't fully understand. Are you claiming that if $\lambda$ and $\kappa$ are cardinal number, then $^\lambda \kappa$ (the set of all the functions $\lambda \to \kappa$) is well-orderable (and then $|^\lambda \kappa|$ is well-defined)? – aerdna91 Jul 22 '14 at 21:06
  • 1
    @aerdna91: No, I'm claiming that cardinals are not necessarily ordinals. And that cardinal exponentiation is well-defined (in the sense that if $|A|=|A'|$ and $|B|=|B'|$ then $|A^B|=|A'^{B'}|$). It's a freshmen level exercise in painful composition of bijections, and I'll leave it to you to verify. – Asaf Karagila Jul 22 '14 at 21:07
  • Now I understand why I was totally confused! Here the author is dealing with cardinal numbers in a "naive set theory" way! Well, IMHO in this case it would have been clearer to write $|2^\omega|$ instead of $2^{\aleph_0}$. Thank you again – aerdna91 Jul 22 '14 at 21:17
  • @aerdna91: But $2^{\aleph_0}$ is the correct way of writing this. It is a well-defined cardinal. Perhaps a better way would be to write in the introduction that cardinal exponentiation of two well-ordered cardinals need not be a well-ordered cardinal itself; but then again, have you read the introduction? :-) – Asaf Karagila Jul 22 '14 at 21:19
  • Yeah, I did it, I swear :) I had also looked for "cardinal" by Ctrl+F, because I was starting to suspect that my problem could be a misunderstanding of definitions with the author. To be honest, it seems to me that there is no clear definition of cardinal operations through the book. Still, the book deals with ordinals, therefore I thought that was the approach he was using (i.e. cardinal numbers are ordinals, and $|A|$ is a cardinal number and therefore it is defined iff $A$ is well-orderable). – aerdna91 Jul 22 '14 at 21:39
0

It seems to me that you are somehow mixing well-orderability with cardinality. Two sets $A$ and $B$ have the same cardinality iff there exists a bijection $A\to B$. It doesn't say whether they can or cannot be well-ordered.

According to Carl Mummert's answer here, it is consistent with ZF that there doesn't exist a well-ordering of the reals.

Community
  • 1
Peter Franek
  • 11,522
  • 1
    If there exists a bijection $f: \lambda \to A$, where $\lambda$ is a well-order, then $A$ is well-orderable. It is true that ZF $\not\vdash$ "$\mathbb R$ is well-orderable". If it was true that ZF $\vdash \mathcal "P(\omega)$ is well-orderable" and ZF $\vdash \mathbb R \approx \mathcal P(\omega)$, then we had a serious problem ;) – aerdna91 Jul 22 '14 at 18:44
  • $\mathbb{R}\simeq\mathcal{P}(\omega)$ is provable in $ZF$. The well-orderability of $\mathbb{R}$ (or, equivalently, $\mathcal{P}(\omega)$) is not, as far as I know. But even if it were provable, I don't see the serious problem. – Peter Franek Jul 22 '14 at 18:52
  • 2
    Peter, it seems that the OP wasn't sure that their "proof" is wrong, and therefore wasn't sure whether or not $|\Bbb R|=|\mathcal P(\omega)|$ is provable without the axiom of choice, and hence the question. – Asaf Karagila Jul 22 '14 at 18:54