# What would be the consequence of restricting multiplication by Zero to only Finite Cardinals? [closed]

What would be the consequence of restricting multiplication by Zero to only Finite Cardinals?

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## closed as not a real question by Asaf Karagila, tomasz, rschwieb, Emily, BenjaLim Sep 23 '12 at 23:37

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Restricting is usually done when the definition is not "uniform", in the case of cardinal multiplication it is irrelevant whether or not the cardinals are finite or infinite. The result is that restriction of multiplication by zero to finite cardinals makes little to no sense. – Asaf Karagila Jan 20 '12 at 19:14
Obviously, you couldn't get by without zero in ordinary arithmetic. You absolutely need Zero. But, because the arithmetic of infinite cardinals is as it is, it seems to me that you don't need it. It's something extra. You can't get to zero by subtracting cardinals. – mathNotebook Jan 20 '12 at 19:20
This is because often the subtraction is undefined. Multiplication, however, is defined always and in particular if one set is empty and the second is infinite. – Asaf Karagila Jan 20 '12 at 19:22
Leaving multiplication by zero undefined when the other factor is infinite would unnecessarily complicate what is now a very simple definition. – Brian M. Scott Jan 20 '12 at 19:24
If you restrict multiplication and adjust the statements of theorems appropriately (so that the statements are only taken "if multiplication is defined"), then you cannot possibly introduce new contradictions into your theory. – Arturo Magidin Jan 20 '12 at 19:25

Fact: If $|A|=0$ then $A=\varnothing$.

Cardinal arithmetics is just a definition allowing us to observe what is the cardinality of sets created by unions, or by products of sets. If we disallow $\kappa\cdot 0$ for infinite $\kappa$, consider this:

$$\mathbb N\times\varnothing = \varnothing\Rightarrow |\mathbb N\times\varnothing|=0\Rightarrow |\mathbb N|\times0=0$$

We have that cardinality no longer behave nicely. This means that what was simple to define and very natural to begin with will now require elaborate tricks to overcome.

Cardinality, in such case, cannot be defined using bijections, since from one end of the spectrum there exists a bijection from $\mathbb N\times\varnothing$ to $\varnothing$; however the cardinality of the former is "undefined".

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I'm having trouble following this. Why isn't it behaving nicely there? – mathNotebook Jan 20 '12 at 19:44
@mathNotebook: Because on one side you have $|\varnothing|$ which is fine, and on the other you have $\aleph_0\cdot 0$ which is not defined anymore; if something undefined is equal to something else which is defined then there is a problem with your definition. – Asaf Karagila Jan 20 '12 at 19:45
I see. Thank you very much. – mathNotebook Jan 20 '12 at 19:53