Example:
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)
Using our notation for the cardinality of the natural numbers we can rephrase
this equation (recall that
= aleph null = 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)
Since we can not have two possible answer, we would guess that
- card(N) - card(N) is undefined.
Or, in our 'aleph null' notation we would say:
-
-
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)
Using our notation for the cardinalities of the natural numbers and the continuum we can
rephrase this equation as:
- c + = c
We would actually need to show that the cardinality of
A B can not be strictly
larger that the cardinality of A to establish this. That, however, is
left as an exercise.
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)
In our 'special cardinality' notation we could rephrase this as
To establish this, we also need to show that
card(A B)
card(A)
This is true, and left as an exercise.
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