Multiplying Cardinal Numbers

I was just reading a proof of the dimension theorem in Steven Roman's Advanced Linear Algebra. In addressing the cases of infinite bases, Roman proceeds to show that if $\mathcal{B}$ and $\mathcal{C}$ are bases of a space $V$, then $|\mathcal{B}|\leq |\mathcal{C}|$, working up to an application of the BSC-Theorem. Anyway, he uses the string

$$|\mathcal{B}|\leq\aleph_0|\mathcal{C}|=|\mathcal{C}|.$$

Sorry if it's an elementary question, but why does the equality follow? Here $\mathcal{C}$ is any infinite basis. Is it definition? I tried looking up multiplication of ordinals, but didn't find anything useful. Thanks.

By hypothesis $\mathcal C$ is infinite so $\aleph_0|\mathcal{C}|= \max(\aleph_0,|\mathcal C|) = |\mathcal{C}|$ by basic cardinal arithmetic.

• Thanks, I see what I needed now that you pointed out what to look for. – MSE Dec 3 '10 at 3:30

No, it's not a definition. A theorem of Tarski says that if $A$ is an infinite set, then $|A| = |A \times A|$. An acessible proof of this can be found in Hungerford's "Algebra". So using it and the Schröder-Bernstein theorem, it's easy to prove that if $A$ is an infinite set and $B$ is a non-empty set such that $|B| \le |A|$, then $|A||B| = |A|$.

• I'll try to get my hands on a copy of that book, thanks! – MSE Dec 3 '10 at 3:30
• @MSE: It's on page 20. – Nuno Dec 3 '10 at 3:44
• Actually Tarski's theorem is that $|A|=|A\times A|$ implies the axiom of choice; it was Zermelo who proved that $|A|=|A\times A|$ for infinite sets (assuming, of course, the axiom of choice). – Asaf Karagila Sep 30 '11 at 4:59

Seeing how the essential question was answered, I want to stress something else in your post which needs to be pointed out:

It is true that cardinals (namely, Aleph numbers) are usually treated as ordinals, however the multiplications and addition of cardinals and ordinals are very different, and most of all - exponentiation is different as well.

For ordinals $\alpha$ and $\beta$ we define the sum to be:

• $\alpha + 0 = \alpha$
• $\alpha + (\beta + 1) = (\alpha + \beta) + 1$ (where $+1$ is the successor ordinal)
• $\alpha + \beta$ for a limit ordinal $\beta$ is the limit of $\alpha+\gamma$ for $\gamma<\beta$

One can notice that it is usually non-commutative as $2+\omega = \sup\{2+n\colon n<\omega\} = \omega \neq \omega+2$.

The ordinal multiplication is defined in a similar way, as well exponentiation. (Namely a simple rule for zero, and successor and a limit for limits) and an interesting result is that $\omega^\omega$ is countable when dealing with ordinal exponentiation.

In contrast, if $\lambda$ and $\mu$ are cardinals then $\lambda + \mu = \mu + \lambda = \lambda \cdot \mu = \mu \cdot \lambda = \max \{\lambda, \mu\}$, and the exponentiation is defined as $\lambda^\mu = |\{f | f\colon\mu\to\lambda\}|$ - that is the cardinality of the collection of functions from $\mu$ into $\lambda$. For further information and definitions you can see this wikipedia link

So once you were dealing with the cardinality of the basis you were looking for cardinal arithmetics and not ordinal arithmetics. Which are two different things.