1. Cardinality

Since so much of cardinal numbers involve creating in/bijections I'm not sure if that's what I should be doing here instead...

Prove that the addition and multiplication is well defined for cardinal numbers.

My answer was... (roughly, more to see if I'm on the right track)

Let A and B be two sets with cardinality $\displaystyle \kappa$ and $\displaystyle \lambda$ respectively.

Want to show that if card(A) = card(A') and card(B) = card(B') then...

card$\displaystyle (A \cup B)$ = card$\displaystyle (A' \cup B')$

So we have card$\displaystyle (A' \cup B')$ = card(A') + card(B') = $\displaystyle \kappa + \lambda$ = card(A) + card(B) = card$\displaystyle (A \cup B)$.

Hence addition is well defined... Seems wrong looking back on it. Do I even need to the $\displaystyle \kappa, \lambda$ bit?

2. maybe...

$\displaystyle dim(A \cup B) = dim(A) + dim(B) - dim(A \cap B)$

dont know for multiplication...

do we look $\displaystyle AXB = \{(a,b)| a \in A, b \in B \}$ ?

3. card = card(A') + card(B') = = card(A) + card(B) = card
Doing that you assume the addition for cardinal numbers is well-defined.

Moreover, $\displaystyle cardA=cardA'$ and $\displaystyle cardB=cardB'$ does not mean $\displaystyle card(A\cup B)=card(A'\cup B')$ ; you must assume $\displaystyle A\cap B=A'\cap B'=\emptyset$ to be sure of that (and that's the way cardinal addition is defined).

What you have to prove is: If $\displaystyle cardA=cardA'$ and $\displaystyle cardB=cardB',$ then $\displaystyle card(A\oplus B)=card(A'\oplus B')$ , where $\displaystyle X\oplus Y := X\times\{0\}\cup Y\times\{1\}$ for any two sets $\displaystyle X$ and $\displaystyle Y$.

$\displaystyle cardA=cardA'$ means there exists a bijection $\displaystyle f:A\rightarrow A'$
$\displaystyle cardB=cardB'$ means there exists a bijection $\displaystyle g:B\rightarrow B'$

Question: Can you define a bijection between $\displaystyle A\oplus B$ and $\displaystyle A'\oplus B'$ ?

For the product, the hypotheses and the question are the same with the cartesian product $\displaystyle \times$ instead of $\displaystyle \oplus$.

@josipive: The question is about sets in general, you cannot apply such result about vector subspaces of finite dimension.