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In mathematics, a set is called finite if there exists a one to one correspondence, called a bijective mapping, between it and a set of the form {1, 2, 3, ..., n} for some natural number n. (This includes the empty set for n = 0). A set is called countably infinite if there exists a bijective mapping between it and the set N of all natural numbers. A countable set is a set which is either finite or countably infinite. A set which is not countable is called uncountable. Not all uncountable sets have the same size. The different sizes of infinite sets are investigated in the theory of cardinal numbers. Below you can find a gentle introduction.

The elements of a finite set can be listed, say {a1,a2,...,an}. However, not all sets are finite: for instance, the set of all integers or the set of all real numbers. It might then seem natural to divide the sets into different classes: put all the sets containing one element together; all the sets containing two elements together;...;finally, put together all infinite sets and consider them as having the same size. This view is not tenable, however, under the natural definition of size.

To elaborate this we need the concept of a bijection. Do the sets {1,2,3} and {a,b,c} have the same size?

"Obviously, yes."

"How do you know?"

"Well it's obvious. Look, they've both got 3 elements".

"What's a 3?"

This may seem a strange situation but, although a "bijection" seems a more advanced concept than a "number", the usual development of mathematics in terms of set theory defines functions before numbers, as they are based on much simpler sets. This is where the concept of a bijection comes in: define the correspondence

a ↔ 1, b ↔ 2, c ↔ 3

Since every element of {a,b,c} is paired with precisely one element of {1,2,3} (and vice versa) this defines a bijection.

We now generalise this situation and define two sets to be of the same size precisely when there is a bijection between them. For all finite sets this gives us the usual definition of "the same size". What does it tell us about the size of infinite sets?

Consider the sets A = {1,2,3,...}, the set of positive integers and B = {2,4,6,...}, the set of even positive integers. We claim that, under our definition, these sets have the same size, and that therefore B is countably infinite. Recall that to prove this we need to exhibit a bijection between them. But this is easy: 1 ↔ 2, 2 ↔ 4, 3 ↔ 6, 4 ↔ 8, ...

As in the earlier example every element of A has been paired off with precisely one element of B, and vice versa. Hence they have the same size. This gives an example of a set which is of the same size as one of its proper subsets, a situation which is impossible for finite sets.

Likewise, the set of all pairs of natural numbers is countably infinite through the following "triangular" mapping:

  • (0,0) maps to 0
  • (0,1) maps to 1
  • (1,0) maps to 2
  • (0,2) maps to 3
  • (1,1) maps to 4
  • (2,0) maps to 5
  • ...
  • (m,n) maps to 0.5*k*(k+1)+m, where k=m+n

THEOREM: The Cartesian product of finitely many countable sets is countable.

This triangular mapping recursively generalizes to vectors of finitely many natural numbers by repeatedly mapping the first two elements to a natural number. For example, (2,0,3) maps to (5,3) which maps to 41.

Sometimes it is useful to use more than one mapping. This is where you map the set you want to show as countably infinite to another set. You then map this other set to the natural numbers. For example, the positive rational numbers can easily be mapped to (a subset of) the pairs of natural numbers through p/q maps to (p,q).

What about subsets of infinite countable sets? Are those smaller than N?

THEOREM: Every subset of a countable set is countable.

(Proof omitted)

For example, the set of prime numbers is countable, by mapping the nth prime number to n:

  • 2 maps to 1
  • 3 maps to 2
  • 5 maps to 3
  • 7 maps to 4
  • 11 maps to 5
  • 13 maps to 6
  • 17 maps to 7
  • 19 maps to 8
  • 23 maps to 9
  • etc.

What about sets being "larger than" N? An obvious place to look would be Q, the set of all rational numbers, which is "clearly" much bigger than N. But looks can be deceiving, for we assert

THEOREM: Q (the set of all rational numbers) is countable.

Q can be defined as the set of all fractions a/b where a and b are integers and b > 0. This can be mapped onto the subset of ordered triples of natural numbers (a, b, c) such that b > 0, a and b are relatively prime, and c ∈ {0, 1}, and if a = 0 then c = 0.

  • 0 maps to (0,1,0)
  • 1 maps to (1,1,0)
  • -1 maps to (1,1,1)
  • 1/2 maps to (1,2,0)
  • -1/2 maps to (1,2,1)
  • 2 maps to (2,1,0)
  • -2 maps to (2,1,1)
  • 1/3 maps to (1,3,0)
  • -1/3 maps to (1,3,1)
  • 3 maps to (3,1,0)
  • -3 maps to (3,1,1)

  • 1/4 maps to (1,4,0)
  • -1/4 maps to (1,4,1)
  • 2/3 maps to (2,3,0)
  • -2/3 maps to (2,3,1)
  • 3/2 maps to (3,2,0)
  • -3/2 maps to (3,2,1)
  • 4 maps to (4,1,0)
  • -4 maps to (4,1,1)
  • ...

By a similar development, the set of algebraic numbers is countable, and so is the set of definable numbers.

THEOREM: The union of countably many countable sets is countable.

For example, given countable sets a, b, c ...

Using a variant of the triangular enumeration we saw above:

  • a0 maps to 0

  • a1 maps to 1
  • b0 maps to 2

  • a2 maps to 3
  • b1 maps to 4
  • c0 maps to 5

  • a3 maps to 6
  • b2 maps to 7
  • c1 maps to 8
  • d0 maps to 9

  • a4 maps to 10
  • ...

Note that this only works if the sets a, b, c,... are disjoint. If not, then the union is even smaller and is therefore also countable by a previous theorem.

THEOREM: The set of all finite-length sequences of natural numbers is countable.

This set is the union of the length-1 sequences, the length-2 sequences, the length-3 sequences, each of which is a countable set (finite Cartesian product). So we are talking about a countable union of countable sets, which is countable by the previous theorem.

THEOREM: The set of all finite subsets of the natural numbers is countable.

If you have a finite subset, you can order the elements to get a finite sequence. There are only countably many finite sequences, so there are also only countably many finite subsets.

Further theorems about uncountable sets:

See also:

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