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Two simple questions: (Of course ${\frak m}$ denotes a cardinal in the weak sense in the claims below.)

  1. Can we prove in ZF that $\aleph_0\le{\frak m\Rightarrow m+m=m}$?
  2. If not, what is the relationship of this statement for all cardinals ${\frak m}$ to AC?

The weaker principle $\aleph_0\le{\frak m\Rightarrow m}+1={\frak m}$ can be derived in ZF, and it is a well-known result of Tarski that $\forall{\frak m}\ge\aleph_0,{\frak m\times m=m}$ implies the axiom of choice, but I am not sure about the intermediate result with ${\frak m+m=m}$.

Edit: A related question: is it consistent with ZF that there exists a cardinal ${\frak m}\ge\aleph_0$ such that ${\frak m=n+p}$ with ${\frak n<m}$ and ${\frak p<m}$? It's hard to say whether ${\frak m+m=m}$ precludes this scenario.

Asaf Karagila
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Mario Carneiro
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  • Well, if $A$ is infinite and Dedekind-finite then $\mathfrak m=|A\cup\mathbb N|$ is a counterexample to the first question. – Andrés E. Caicedo May 16 '13 at 06:16
  • So this principle would eliminate the possibility of Dedekind-finite sets, then. (How did you do the math on that one? I lack intuition on sets of the form ${\frak m}=|A|+\aleph_0$ where $A$ is Dedekind-finite.) – Mario Carneiro May 16 '13 at 06:20
  • Also, a clarification: ${\frak m}+1={\frak m}$ implies $\aleph_0\le{\frak m}$, right? Because we can just iterate the bijection $\aleph_0$ times to get a countable subset. So Dedekind-finite sets are those infinite sets that satisfy the pigeonhole principle. (Wait, if Dedekind-finite sets don't exist, doesn't that imply CC?) – Mario Carneiro May 16 '13 at 06:25
  • Similar question for $\mathfrak m\cdot\mathfrak m=\mathfrak m$ is here. – Martin Sleziak May 16 '13 at 08:22
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    @Martin: That's actually the other direction, and the more correct link is this one. – Asaf Karagila May 19 '13 at 16:33

2 Answers2

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Suppose that $A$ is infinite and Dedekind finite. Then $\mathfrak m=|A\cup\mathbb N|$ satisfies that $|A|<\mathfrak m$, $\aleph_0<\mathfrak m$, and $\mathfrak m+\mathfrak m>\mathfrak m$.

To see the last inequality, note that if $\mathfrak m+\mathfrak m=\mathfrak m$ then $A\times 2$ embeds into $A\cup\mathbb N$, say via $f$, but only a finite subset of it embeds into $\mathbb N$, so a set strictly larger than $A$ must embed into $A$.

Note that being Dedekind infinite is the same as embedding $\omega$, so if we require $\mathfrak m+\mathfrak m=\mathfrak m$ for all infinite cardinals $\mathfrak m$, or even for all Dedekind-infinite cardinals, then there are no Dedekind-finite sets. But no, Countable Choice is strictly stronger than the lack of infinite Dedekind finite sets. This is due to Pincus, see this MO question.

As for whether $\mathfrak m+\mathfrak m=\mathfrak m$ (the idemmultiple hypothesis) gives us Countable Choice, the answer is again no, as proved by

Gershon Sageev. An independence result concerning the axiom of choice, Ann. Math. Logic 8, (1975), 1–184. MR0366668 (51 #2915).

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Andrés E. Caicedo
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To add slightly on Andres' answer, it is not only that $\frak m+m=m$ does not imply $\sf AC_\omega$, but in the other direction we can have $\sf DC_\kappa$ (for an arbitrary $\kappa$) but $\frak m+m=m$ fails.

We say that $A$ is a $\kappa$-amorphous set if every subset of $A$ has cardinality $<\kappa$, or its complement has such cardinality - but not both.

Note that if $A$ is $\lambda$-amorphous, then for $\kappa>\lambda$, $A$ is $\kappa$-amorphous as well. We say that $A$ is a properly $\kappa$-amorphous set if it is $\kappa$-amorphous, and $\aleph(A)=\kappa$. In other words, if every ordinal smaller than $\kappa$ can be injected into $A$.

For every regular $\kappa$ it is consistent with $\sf ZF+DC_{<\kappa}$ that there exists a [properly] $\kappa$-amorphous set.

Now, if $A$ is a $\kappa$-amorphous set, then $|A|<|A|+|A|$. This is a trivial observation since $A\times2$ can be written as the union of two sets, neither has complement of size $<\kappa$. So there cannot be an injection from $A\times2$ into $A$, but there is an obvious injection in the other direction.

Therefore $\sf ZF+DC_\kappa$ cannot prove $\frak\forall m. m+m=m$.

(On a slightly more confusing note, I am switching between $\sf DC_\kappa$ and $\sf DC_{<\kappa}$, but the former can be thought as $\sf DC_{<\kappa^+}$.)

To the added question, by the way, this is a simple result. If I recall correctly it is by Tarski.

The axiom of choice is equivalent to the assertion that whenever $\frak p,m,n$ are cardinals and $\frak p+m=n$ then $\frak p=n$ or $\frak m=n$.

The proof is simple, let $\frak a$ be a cardinal, and let $\kappa=\aleph(\frak a)$ the Hartogs number of a set of size $\frak a$. Then $\frak a+\kappa=a$ or $\frak a+\kappa=\kappa$. But $\kappa\nleq\frak a$ so the first option is impossible, therefore $\frak a\leq\kappa$, and can be well-ordered. $\square$

In particular if the axiom of choice fails, let $\frak a$ be a non-well ordered cardinal and $\kappa$ its Hartogs number, then $\frak a+\kappa$ is a set which can be decomposed into two strictly smaller cardinals.

It is also consistent with $\sf ZF$ that every cardinal which is not well-orderdable can be written in such way:

Monro, G.P. Decomposable Cardinals. Fund. Math. vol. 80 (1973), no. 2, 101–104.

Asaf Karagila
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    For the definition of $\kappa$-amorphous, do you want $A$ itself not to have size $\mathord{<}\kappa$? – Trevor Wilson May 16 '13 at 19:00
  • Trevor, thanks. – Asaf Karagila May 16 '13 at 19:21
  • @AsafKaragila - is there any known result of the sort "what is the weakest choice principle needed to prove that $m + m = m$ for every infinite cardinal $m$"? – Nagase Aug 16 '15 at 21:09
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    @Nagase: Not that I know of. Since we only really know two [families of] models where $m+m=m$ without choice (and one of them is using urelements, so not even ZF proper), it's hard to tell really. – Asaf Karagila Aug 16 '15 at 21:17