• Real Analysis
• 1. Sets and Relations
• 2. Infinity and Induction
• 2.1. Countable Infinity
• 2.2. Uncountable Infinity
• 2.3. The Principle of Induction
• 2.4. The Real Number System
• 2.5. Pi and e are irrational
• 3. Sequences of Numbers
• 4. Series of Numbers
• 5. Topology
• 6. Limits, Continuity, and Differentiation
• 7. The Integral
• 8. Sequences of Functions
• 9. Historical Tidbits
• Java Tools

Examples 2.2.8:

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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.

Interactive Real Analysis, ver. 2.0.1(b)
(c) 1994-2017, Bert G. Wachsmuth
Page last modified: Mar 3, 2017