# Thread: Distributivity of Cardinal Numbers

1. ## Distributivity of Cardinal Numbers

If $\kappa, \lambda, \mu$ be three cardinals. Prove that $\kappa(\lambda + \mu) = \kappa \lambda + \kappa \mu$.

Thoughts...
My basic problem is that I have never really understood how to construct in/sur/bijections so all this cardinal number stuff is proving to be a bit hard...
But my thoughts on this are that...
Let $\kappa, \lambda, \mu$ be the cardinal numbers of the sets A,B and C respectively such that B and C are disjoint. Then $A \times B$ and $A \times C$ are disjoint and $A \times (B \cup C) = A \times B \cup A \times C$.

I'm not sure whether I'm allowed to assume the disjoint part and how to construct a bijection between the two...

2. Hi

Let $\lambda$ and $\mu$ be two cardinal numbers, then $\lambda +\mu$ is, by definition, the cardinal of the disjoint union of two sets whose cardinalities are $\lambda$ and $\mu$ respectively.

If in the definition of the cardinal addition the sets weren't assumed to be disjoint, that would not make sense: for instance we would have $1+1=card(\{0\}\cup\{0\})=card(\{0\})=1$ !

If this is true, then you've finished: the identity map will be a bijection between these two sets, which means, under the assumptions and the remark:

Let be the cardinal numbers of the sets A,B and C respectively such that B and C are disjoint. Then and are disjoint
that $\kappa (\lambda +\mu )=\kappa\lambda + \kappa\mu$

Therefore all you have to do is to prove the equality. By the way, this equlity holds even if $B\cap C\neq\emptyset$ (the case $B\cap C\neq\emptyset$ does not interest us for this problem, but we don't have to suppose $B$ and $C$ are disjoint to prove the equality)

The proof is quite simple, it can be done with two inclusions; the only things you have to use are the definitions of union and cartesian product of sets.