2.2. Uncountable Infinity

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 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: Since we can not have two possible answer, we would guess that Or, in our 'aleph null' notation we would say:

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 Using our notation for the cardinalities of the natural numbers and the continuum we can rephrase this equation as: 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 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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