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This is similar to What does it take to divide by $2$? about $(A\sqcup A\cong B\sqcup B)\Rightarrow A\cong B$ which is valid in $\textsf{ZFC}$ by using cardinalities and also in $\textsf{ZF}$ by some combinatorial argument, but where it remained unclear whether constructive arguments suffice.

This questions concerns products instead of disjoint union, so:

Question: Does $\textsf{ZF}$ prove the implication $(A\times A\cong B\times B)\Rightarrow A\cong B$?

If yes: is anything known in case of intuitionistic set/type theories?

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Hanno
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    @bof Thank you! I'm aware of the result about the 'injectivity' of $n\times -$, but where can I find the result about $(-)^2$ (or even $(-)^n$)? – Hanno Jan 22 '15 at 08:40
  • I'm sorry, I seem to be befuddled. Never mind! – bof Jan 22 '15 at 08:45
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    @bof: No worries, and thanks for carefully looking up the references! Would you mind copying your answer to the question about divisibility by $2$? Even though it wasn't about intuitionistic reasoning, I think it is valuable for anyone generally interested in the topic. – Hanno Jan 22 '15 at 08:47
  • Why isnt' it enough to project onto the first component to get $A\cong B$ – Denis Jan 22 '15 at 13:00

3 Answers3

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The answer is no. It is known that ZF+'$\aleph_1$ and $2^{\aleph_0}$ are incomparable' is consistent if ZF does. Assume that $\aleph_1$ and $2^{\aleph_0}$ are incomparable (over ZF),then

  1. $\aleph_1$, $2^{\aleph_0}<\aleph_1+2^{\aleph_0}$ and
  2. $\aleph_1+2^{\aleph_0}\neq\aleph_1\cdot 2^{\aleph_0}$.

1 is trivial. To prove 2, we can use following theorem:

Theorem (ZF) If $\mathfrak{p}$ is a (possibly non-well-orderable) cardinal and $\alpha$ is an aleph and they satisfy $\mathfrak{p}+\alpha=\mathfrak{p}\cdot\alpha$, then they are comparable.

You can find this theorem and its proof in Jech's 'Axiom of choice', Lemma 11.16.

We will check that $(\aleph_1+2^{\aleph_0})^2=(\aleph_1\cdot 2^{\aleph_0})^2=\aleph_1\cdot 2^{\aleph_0}$. It is just a calculation: $$ \begin{align}(\aleph_1+2^{\aleph_0})^2 &=\aleph_1+2^{\aleph_0}+2\cdot \aleph_1\cdot2^{\aleph_0}\\ &=\aleph_1+2^{\aleph_0}+\aleph_1\cdot2^{\aleph_0}\\ &=\aleph_1+(\aleph_1+1)\cdot 2^{\aleph_0}\\ &=\aleph_1+\aleph_1\cdot 2^{\aleph_0}\\ &=\aleph_1\cdot(1+2^{\aleph_0})\\ &=\aleph_1\cdot2^{\aleph_0}.\\ \end{align}$$ Also, the square of $\aleph_1\cdot 2^{\aleph_0}$ is $\aleph_1\cdot 2^{\aleph_0}$ itself so $(\aleph_1+2^{\aleph_0})^2=(\aleph_1\cdot 2^{\aleph_0})^2$. However, we already know that $\aleph_1+2^{\aleph_0}$ and $\aleph_1\cdot 2^{\aleph_0}$ are not equal.

Hanul Jeon
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  • $2^{\aleph_0}<\aleph_1+2^{\aleph_0}$: This is false, isn't it? – TonyK Jan 22 '15 at 11:48
  • @Tony: No, it's not false. It is consistent that $\aleph_1$ is incomparable with $2^{\aleph_0}$, in which case the strict inequality is true. – Asaf Karagila Jan 22 '15 at 11:51
  • @AsafKaragila: Ah, OK: in a model containing incomparable cardinals, it can't be true that $\kappa+\mu=\max(\kappa,\mu)$ for infinite cardinals $\kappa,\mu$, because $\max(\kappa,\mu)$ is not well-defined. – TonyK Jan 22 '15 at 11:59
  • @Tony: Yes, and every model of $\lnot\sf AC$ contains incomparable cardinals. – Asaf Karagila Jan 22 '15 at 12:00
  • @AsafKaragila: Therefore, "All cardinals are comparable" implies AC? If so, that's a surprise to me. – TonyK Jan 22 '15 at 12:04
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    @TonyK Yes. If every cardinals are comparable, then every set can embed into some ordinal. To be more exact, Hartogs number of the set has larger cardinality than the set. – Hanul Jeon Jan 22 '15 at 12:09
  • @AsafKaragila and tetori: Thank you both for an enlightening session! – TonyK Jan 22 '15 at 12:15
  • Dear tetori, thank you very much! – Hanno Jan 22 '15 at 13:45
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This is a generalization of tetori's nice answer, which in fact shows that this assumption implies the axiom of choice.

Lemma I. If $a$ and $b$ are infinite cardinals equipollent with their own squares, then $(a+b)^2=a\cdot b$.

Proof. We calculate: $$(a+b)^2=a^2+b^2+2\cdot a\cdot b=a^2+b^2+a\cdot b=a+b+a\cdot b=a\cdot b$$ The second equality comes from the fact that $a\leq a+a=2\cdot a\leq a^2=a$, the last one follows from the same reasoning. $\qquad\square$

Lemma II. If $a+a=a$ then $2^a$ equals to its own square.

Proof. $(2^a)^2=2^{a+a}=2^a.\qquad\square$

Tarski's Lemma. If $a$ is a cardinal, $\lambda$ is an aleph, and $a+\lambda=a\cdot\lambda$, then the two are comparable.

(The proof appears in Jech, The Axiom of Choice, Lemma 11.6; and there are generalizations of this lemma out there.)

Theorem. ($\sf ZF$) If $a^2=b^2\implies a=b$ for infinite cardinals, then the axiom of choice holds.

Proof. We will show that if $\kappa$ is an aleph number, then $2^\kappa$ is an aleph number. This implies the axiom of choice, by a theorem of Rubin and Rubin from 1963.

Let $\lambda=\aleph(2^\kappa)$, the least aleph cardinal such that $\lambda\nleq 2^\kappa$. Then $2^\kappa$ and $\lambda$ satisfy the hypothesis of Lemma I, and therefore $(2^\kappa+\lambda)^2=2^\kappa\cdot\lambda=(2^\kappa)^2\cdot\lambda^2=(2^\kappa\cdot\lambda)^2$. By the uniqueness of the square root, $2^\kappa+\lambda=2^\kappa\cdot\lambda$, and by Tarski's lemma $2^\kappa$ and $\lambda$ are comparable.

Finally $\lambda\nleq2^\kappa$, therefore $2^\kappa<\lambda$, so $2^\kappa$ is an aleph number. $\qquad\square$

(I will remark here that the final proof uses $\sf ZF$ in a substantial way. In $\sf ZFA$, where atoms are allowed, having $2^\kappa$ as an aleph does not imply the axiom of choice.)

Asaf Karagila
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  • Dear Asaf, thanks alot! Just to make sure I get it: how do we know that $2^{\kappa}$ and $\aleph(2^\kappa)$ are equal to their own squares? For $2^{\kappa}$, Lemma II tells that it is sufficient to have $\kappa+\kappa = \kappa$, which holds since $2\cdot\lambda$ (ordinal multiplication) has order type $\lambda$ for every limit ordinal $\lambda$, and for $\aleph(2^\kappa)^2=\aleph(2^\kappa)$ we may use $\alpha\beta=\text{max}{\alpha,\beta}$ for $\aleph$-cardinals $\alpha,\beta$? – Hanno Jan 22 '15 at 13:44
  • Anyway, +1 of course for this very helpful answer. – Hanno Jan 22 '15 at 13:45
  • The usual cardinal arithmetic is provable for aleph numbers without choice, choice is needed to show every cardinal is an aleph. – Asaf Karagila Jan 22 '15 at 13:50
  • Aha, ok, then everything's clear. – Hanno Jan 22 '15 at 13:51
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    It would be interesting to know if the theorem even holds in any elementary topos, i.e. if $X^2\cong Y^2\Rightarrow X\cong Y$ implies that the topos satisfies AC. – Hanno Jan 22 '15 at 13:55
  • Well, I'm definitely the wrong man for that question. I think a good start would be to see if it holds in $\sf ZFA$. – Asaf Karagila Jan 22 '15 at 14:10
  • I just saw in Halbeisen's book the unsurprising fact that this equivalence to AC is due to Tarski. I'm guessing that the proof is probably along the line I present here. – Asaf Karagila Jan 25 '15 at 17:56
  • @Asaf regarding Hanno's question, a start would be to see if the above holds in BZ – theHigherGeometer Aug 25 '15 at 00:07
  • @David: Isn't that KP+Power set? – Asaf Karagila Aug 25 '15 at 00:08
  • @Asaf Bounded Zermelo. I.e. no replacement, and with bounded separation, IIRC – theHigherGeometer Aug 25 '15 at 00:09
  • @David: Yes, I understood that. I was being a smartypants since it's 3AM. :-) – Asaf Karagila Aug 25 '15 at 00:10
  • @Asaf well, it's really the theory of a well-pointed topos with natural number object, but since it's 3am, I didn't want to tax you... ;-) – theHigherGeometer Aug 25 '15 at 00:13
  • @David: Taxes like that causes George Bush Sr. not to get re-elected! :-P – Asaf Karagila Aug 25 '15 at 00:14
  • But seriously, can the above be done in BZ? Maybe sleep on it... – theHigherGeometer Aug 25 '15 at 00:17
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This does not answer the question (which I misread, being in a mental fog), but I am undeleting it at the OP's request.

The following summary is copied from p. 174 of Wacław Sierpiński, Cardinal and Ordinal Numbers, second edition revised, Warszawa, 1965:

If, for a natural number $k$ and cardinal numbers $m$ and $n$, we have $km=kn$, then $m=n$.

This theorem was proved without the aid of the axiom of choice for $k=2$ in 1905 by F. Bernstein, who also outlined a proof for natural $k$ (Bernstein [1], p. 122; also Sierpiński [8], p. 1). A detailed proof for natural $k$ (without the aid of the axiom of choice) was published by A. Tarski [12] (p. 77). who proved also (without the aid of the axiom of choice) that if for a natural number $k$ and cardinal numbers $m$ and $n$ we have $km\le kn$, then $m\le n$ (for $k=2$ see also Sierpiński [31], p. 148).

F. Bernstein, Untersuchungen aus der Mengenlehre, Math. Annalen 61 (1905), p. 117-155.

W. Sierpiński, Sur l'égalité $2m=2n$ pour les nombres cardinaux, Fund. Math. 3 (1922), p. 1-16.

W. Sierpiński, Sur l'implication $(2m\le2n)\rightarrow(m\le n)$ pour les nombres cardinaux, Fund. Math. 34 (1947), p. 148-154.

A. Tarski, Cancellation laws in the arithmetic of cardinals, Fund. Math. 36 (1949), p. 77-92.

bof
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