Examples 2.2.8:
We want to add or subtract the following cardinal numbers:
- card(N) + card(N) = card(N)
- card(N) - card(N) = undefined
- card(R) + card(N) = card(R)
- card(R) + card(R) = card(R)
1. card(N) + card(N) = card(N)
According to the definition, this is the same as the cardinality of A B, where A and B are both countable, disjoint sets . But the countable union of countable sets is again countable. Hence, card(A B) = card(N), so that- card(N) + card(N) = card(N)
- + =
2. card(N) - card(N) = undefined
Although this has not been properly defined, one could say that this should be the same as the cardinality of A \ B, where A and B are both countable sets and B is a subset of A. This creates problems, however, as the following examples show:- A = B = N. Then card(A \ B) = card(0) = 0
- A = all integers, B = even integers. Then card(A \ B) = card(odd integers) = card(N)
- card(N) - card(N) is undefined.
- - is undefined
3. card(R) + card(N) = card(R)
According to the definition, this is the same as the cardinality of A B, where A is uncountable and B is countable and A and B are disjoint. We know that every subset of a countable set is countable or finite. Since A is a subset of A B, the set A B can not be countable. Hence, it must be uncountable. We would therefore guess that- card(R) + card(N) = card(R)
- c + = c
4. card(R) + card(R) = card(R)
This should be the same as the cardinality of A B, where both A and B are uncountable and disjoint. It is easy to find a one-to-one function from A to A B, so that card(A) card(A B). But then card(A B) is again uncountable, so that we would guess that- card(R) + card(R) = card(R)
- c + c = c