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Countable sets (Posted on 2016-03-15) Difficulty: 3 of 5
A set is countable if and only if each of its elements can be associated with a different positive integer. Every finite set is countable. For example, the set {2, 3, 5, 7, 11} is countable.

1↔2
2↔3
3↔5
4↔7
5↔11

Infinite sets can also be countable. For example, the set of all prime numbers is countable.

1↔2
2↔3
3↔5
4↔7
5↔11
6↔13
7↔17
8↔19
9↔23
10↔29
...

1. Is the set of all integers countable?
2. Is the set of all positive rational numbers countable?
3. Is the set of all rational numbers countable?
4. Is the set of all positive real numbers countable?
5. Is the set of all real numbers countable?

No Solution Yet Submitted by Math Man    
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re(3): Interesting Comment 8 of 8 |
(In reply to re(2): Interesting by broll)

Wolfram Alpha may be biased for regular algebra, not infinite set theories.


I'll try to prove 2^I = I^I:

Lets construct a potential set of the natural numbers N: either 1 is or is not in the set, either 2 is or is not in the set, either 3 is or is not in the set, etc.  For each possible set of choices there is a set reflecting the results.

This basically constructs the powerset of N, called P(N).  The simple way of stating its size is 2^N.  (A bit of a misnomer - P(N) is uncountable, the diagonal argument can be used to prove that.)

Now lets construct a new potential set.  For each natural number n, k_n copies of n can be placed in the new potential set, where k_n is any nonnegative integer.  Call the set of all possible sets like this the "superpowerset" of N, I'll call it S(N) for short.  The simple way of stating its size is I^I.

Let M be a member of S(N).  Create a set T based off of M by taking the first k_1 numbers of the form 2d+1, the first k_2 numbers of the form 4d+2, the first k_3 numbers of the form 8d+4, etc.  Each T is a member of P(N) and each M maps to a unique T.  The set of all T is a subset of P(N).  This establishes an injection from S(N) to P(N).

This implies that the size of S(N) cannot be greater than the size of P(N).  But clearly P(N) is a subset of S(N), so the size of S(N) cannot be less than the size of P(N).  Therefore the size of S(N) is the same as the size of P(N), which then implies 2^I = I^I.

  Posted by Brian Smith on 2016-03-16 12:04:02
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