Cardinality

**Deadstar** · October 7th 2009, 08:30 AM

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 $\kappa$ and $\lambda$ respectively.

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

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

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

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

**josipive** · October 7th 2009, 11:12 AM

maybe...

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

dont know for multiplication...

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

**clic-clac** · October 7th 2009, 12:34 PM

card

= card(A') + card(B') =

= card(A) + card(B) = card

Doing that you assume the addition for cardinal numbers is well-defined.

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

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

$cardA=cardA'$ means there exists a bijection $f:A\rightarrow A'<br />$
$cardB=cardB'$ means there exists a bijection $g:B\rightarrow B'$

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

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

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

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