# Interactive Real Analysis - part of MathCS.org

Next | Previous | Glossary | Map | Discussion

## Examples 2.2.8:

We want to add or subtract the following cardinal numbers:
1. card(N) + card(N) = card(N)
2. card(N) - card(N) = undefined
3. card(R) + card(N) = card(R)
4. 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
• c + c = c
To establish this, we also need to show that card(A B) card(A) This is true, and left as an exercise.
Next | Previous | Glossary | Map | Discussion