# Cardinal arithmetic questions

I have problem to solve:

Let $a,b$ and $c$ be cardinal numbers. Prove that $a+b=b$, $b \le c$ implies $a+c=c$.

And trying to prove this I got couple questions:

1. For infinite cardinal $c$, is $c+c=c$ always? And how to prove it? (Is $c \cdot c=c$?)
2. From $a+b=b$ can I conclude that $a \le b$ since $a \ge b$ and 1. will imply $a+b=a$? I know that statement that every two cardinals are comparable leads to AC (we can use it when solving problems)

(I apologize if my questions are too simple, I tried to google it, but Google is not math-friendly tonight for me =D)

P.S. Is there solution without using AC?

-
To make it clear: you are in fact allowed to use the Axiom of Choice? (1) is implied by AC but won't always hold when Choice doesn't hold. – Steven Stadnicki Jun 18 '13 at 20:48
(And if you are allowed to use Choice, try writing $a+b+c$ in two different ways using associativity...) – Steven Stadnicki Jun 18 '13 at 20:49
@StevenStadnicki Note that the question says "cardinal numbers", i.e. alephs. For those, no choice is needed. – Lord_Farin Jun 18 '13 at 20:49
I am allowed to use Axiom of Choice. But I still don't see how that will help in proving 1.? – Meow Jun 18 '13 at 20:50

Since $b\le c$, then there is a $d$ such that $c=b+d$. Then $$a+c=a+(b+d)=(a+b)+d=b+d=c.$$ Note that this proof does not use the axiom of choice.

If you assume the axiom of choice, then $c+c=c$ for all infinite cardinals $c$. Without choice, it is consistent that this fails in general (see the comments here), though of course it holds in particular cases (for any well-ordered $c$, or for $c=\mathfrak c:=|\mathbb R|$, for example).

That $c\cdot c=c$ for all infinite $c$ is equivalent to choice. To prove that choice gives us this, it suffices to argue that $c\cdot c=c$ whenever $c$ is well-orderable. Typically, one uses some kind of pairing function to achieve this. See for example here. The point is that pairing functions give us explicit bijections between $\kappa\times \kappa$ and $\kappa$ for any infinite well-ordered cardinal $\kappa$. Since $\kappa\le\kappa+\kappa\le \kappa\cdot \kappa$, it follows from this that $\kappa+\kappa=\kappa$ as well.

To see that $c\cdot c=c$ for all infinite $c$ gives us choice, see here.

And yes, if $a+b=b$, then $a\le b$. On the other hand, without assuming choice, just knowing that $a\le b$ does not suffice to conclude that $a+b=b$. For example, if $b$ is infinite and Dedekind finite, then $1<b<b+1$.

-

1. If $c$ is infinite, assuming the axiom of choice, it is possible to show that $c+c=c\cdot c=c$. This is true, without the axiom of choice, for $\aleph$ cardinals. In the presence of choice all cardinals are $\aleph$ cardinals so it is simply true for all sets.

However without the axiom of choice it is consistent that $c+c=c$ for all cardinals, and the axiom of countable choice fails (this is the Ph.D. work of Sageev). On the other hand, if $c\cdot c=c$ for every infinite cardinal then the axiom of choice holds. That is a theorem of Tarski.

2. Note that for every cardinals $a,b$ we have $a\leq a+b$. In particular if $a+b=b$ then $a\leq b$. This does not require the axiom of choice at all. This is simply true because we can compose the obvious injection from $a$ into $a+b$ with the bijection of $a+b$ and $b$.

Lastly, there is a solution for the general question which does not use the axiom of choice. This is because if $b\leq c$ then we can write $c=b+x$ for some $x$, regardless to its size, then we have $$a+c=a+b+x=b+x=c.$$

-

You asked how to prove (1). Assuming the axiom of choice, every cardinal is an aleph.

Let $\kappa=\aleph_\gamma$ for some $\gamma$. To prove that $\kappa+\kappa=\kappa$, note that every ordinal $\alpha$ can be written uniquely in the form $\alpha=\eta+n$, where $\eta$ is $0$ or a limit ordinal, and $n\in\omega$. Say that $\alpha$ is even if $n$ is even and odd if $n$ is odd. Let $K_0=\{\alpha\in\kappa:\alpha\text{ is even}\}$ and $K_1=\{\alpha\in\kappa:\alpha\text{ is odd}\}$. Then

$$\varphi_0:\kappa\to K_0:\alpha=\eta+n\mapsto\eta+2n$$

and

$$\varphi_1:\kappa\to K_0:\alpha=\eta+n\mapsto\eta+2n+1$$

are bijections, and $\kappa=K_0\sqcup K_1$, so

$$\varphi:2\times\kappa\to\kappa:\langle k,\alpha\rangle\mapsto\varphi_k(\alpha)$$

is a bijection from $2\times\kappa$ onto $\kappa$.

-