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On the Wikipedia page on Cardinal Numbers, Cardinal Arithmetic including multiplication is defined. For finite cardinals there is multiplication by zero, but for infinite cardinals only defines multiplication for nonzero cardinals. Is multiplication of an infinite cardinal by zero undefined? If so, why is it?

Also does $\kappa\cdot\mu= \max\{\kappa,\mu\}$ simply means that the multiplication of the two is simply the cardinality of the higher cardinal? Why is this?

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Of course any infinite cardinal is nonzero... if you want to multiply an infinite cardinal by a finite cardinal, the result is equal to the infinite cardinal if the finite one is nonzero and zero otherwise. – Qiaochu Yuan Jan 20 '12 at 18:38
But what is Zero times a Cardinal? Is it undefined? Why? – mathNotebook Jan 20 '12 at 18:40
Your claims are not true. The article in question defines multiplication for any pair of cardinals. It even notes explicitly that multiplication by $0$ gives $0$. The statement you've picked out is not a definition. It is a theorem, and the reason the cardinals are required to be nonzero is that the result wouldn't be true otherwise. – Chris Eagle Jan 20 '12 at 18:46
Chris Eagle’s link in fact tells you that $\kappa\cdot 0=0\cdot\kappa=0$ for all cardinals. The statement that you quote in your last comment restricts $\kappa$ and $\lambda$ to non-zero cardinals precisely because it would be false if one were $0$: the product would then be $0$, not $\max\{\kappa,\mu\}$. – Brian M. Scott Jan 20 '12 at 18:54
The only way to thank users of the forum who have been of help to you is by upvoting their answers and accepting top answers. I understand that you cannot upvote because you need atleast 15 reputations to do so. But, certainly, you can select an answer by clicking the tick mark on the side of the answer. – user21436 Jan 20 '12 at 20:15

For any cardinal $\kappa$ whatsoever, $0\cdot\kappa=\kappa\cdot 0=0$. This is an immediate consequence of the definition and the fact that for any set $X$, $\varnothing\times X=\varnothing$.

Yes, if one assumes the axiom of choice, the product of two infinite cardinals is simply the larger of them; so is their sum. The product of a non-zero finite cardinal and an infinite cardinal is that infinite cardinal, so it’s also simply the larger of the two. This fails when the finite cardinal is $0$, because then the product is $0$.

Even without the axiom of choice it’s true that if $\kappa$ and $\mu$ are well-orderable cardinals, $\kappa\cdot\mu=\max\{\kappa,\mu\}$. This is proved by constructing a bijection between $\kappa\times\mu$ and $\max\{\kappa,\mu\}$.

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Thanks for these well articulated responses. They were very helpful. I'm a little embarrassed I asked it now because that is basically how I interpreted it in the past. But I didn't understand that the multiplication for infinite cardinals wasn't a definition. Now everything is much clearer to me. – mathNotebook Jan 20 '12 at 19:10

Suppose $A,B$ are sets, we have a good sense about $|A|,|B|$ which are the cardinality of these sets.

The arithmetics of cardinals is quite simple:

  • $|A|+|B| = |(\{0\}\times A)\cup(\{1\}\times B)|$ (That is the disjoint union of $A$ and $B$)
  • $|A|\cdot|B| = |A\times B|$
  • $|A|^{|B|}=|\{f:B\to A\mid f\text{ is a function}\}|$

It is easy to see that these operations are well defined (that is, $|A|=|A'|$ and $|B|=|B'|$ then $|A|+|B|=|A'|+|B'|$ etc.) and it is also quite simple to see that the usual addition, multiplication and exponentiation correspond to the same operations (with the "exception" that $0^0=1$).

If $|B|=0$ then $B=\varnothing$ and we know that $A\times\varnothing=\varnothing$, for every set $A$. Therefore $\kappa\cdot 0=0$.

As for your second question, assuming the axiom of choice holds, every two cardinalities are comparable and the result is that multiplication and addition is the same thing as taking the maximum. This is a result of the well ordering theorem, which is equivalent to the axiom of choice.

Such assumption is widely accepted today and it's quite rare to find people working in contexts where it is false (like yours truly). In situations like that it is usually the case that $|A|\cdot|B|\neq\max\{|A|,|B|\}$.

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Thank You!!!!!!!!!!!!!!!! – mathNotebook Jan 20 '12 at 19:48

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