Editing Countable set (section)

==Formal overview==

By definition, a set <math>S</math> is ''countable'' if there exists a [[bijection]] between <math>S</math> and a subset of the [[natural numbers]] <math>\N=\{0,1,2,\dots\}</math>. For example, define the correspondence
<math display=block>
a \leftrightarrow 1,\ b \leftrightarrow 2,\ c \leftrightarrow 3
</math>
Since every element of <math>S=\{a,b,c\}</math> is paired with ''precisely one'' element of <math>\{1,2,3\}</math>, ''and'' vice versa, this defines a bijection, and shows that <math>S</math> is countable. Similarly we can show all finite sets are countable.

As for the case of infinite sets, a set <math>S</math> is countably infinite if there is a [[bijection]] between <math>S</math> and all of <math>\N</math>. As examples, consider the sets <math>A=\{1,2,3,\dots\}</math>, the set of positive [[integer]]s, and <math>B=\{0,2,4,6,\dots\}</math>, the set of even integers. We can show these sets are countably infinite by exhibiting a bijection to the natural numbers. This can be achieved using the assignments <math>n \leftrightarrow n+1</math> and <math>n \leftrightarrow 2n</math>, so that
<math display=block>\begin{matrix}
0 \leftrightarrow 1, & 1 \leftrightarrow 2, & 2 \leftrightarrow 3, & 3 \leftrightarrow 4, & 4 \leftrightarrow 5, & \ldots \\[6pt]
0 \leftrightarrow 0, & 1 \leftrightarrow 2, & 2 \leftrightarrow 4, & 3 \leftrightarrow 6, & 4 \leftrightarrow 8, & \ldots
\end{matrix}</math>
Every countably infinite set is countable, and every infinite countable set is countably infinite. Furthermore, any subset of the natural numbers is countable, and more generally:
{{math theorem | math_statement = A subset of a countable set is countable.<ref>{{harvnb|Halmos|1960|page=91}}</ref>}}

The set of all [[ordered pair]]s of natural numbers (the [[Cartesian product]] of two sets of natural numbers, <math>\N\times\N</math> is countably infinite, as can be seen by following a path like the one in the picture: [[File:Pairing natural.svg|thumb|300px|The [[Cantor pairing function]] assigns one natural number to each pair of natural numbers]] The resulting [[Map (mathematics)|mapping]] proceeds as follows:
<math display=block>
0 \leftrightarrow (0, 0), 1 \leftrightarrow (1, 0), 2 \leftrightarrow (0, 1), 3 \leftrightarrow (2, 0), 4 \leftrightarrow (1, 1), 5 \leftrightarrow (0, 2), 6 \leftrightarrow (3, 0), \ldots
</math>
This mapping covers all such ordered pairs.

This form of triangular mapping [[recursion|recursively]] generalizes to <math>n</math>-[[tuple]]s of natural numbers, i.e., <math>(a_1,a_2,a_3,\dots,a_n)</math> where <math>a_i</math> and <math>n</math> are natural numbers, by repeatedly mapping the first two elements of an <math>n</math>-tuple to a natural number. For example, <math>(0, 2, 3)</math> can be written as <math>((0, 2), 3)</math>. Then <math>(0, 2)</math> maps to 5 so <math>((0, 2), 3)</math> maps to <math>(5, 3)</math>, then <math>(5, 3)</math> maps to 39. Since a different 2-tuple, that is a pair such as <math>(a,b)</math>, maps to a different natural number, a difference between two n-tuples by a single element is enough to ensure the n-tuples being mapped to different natural numbers. So, an injection from the set of <math>n</math>-tuples to the set of natural numbers <math>\N</math> is proved. For the set of <math>n</math>-tuples made by the Cartesian product of finitely many different sets, each element in each tuple has the correspondence to a natural number, so every tuple can be written in natural numbers then the same logic is applied to prove the theorem.

{{math theorem | math_statement = The [[Cartesian product]] of finitely many countable sets is countable.<ref>{{Harvard citation no brackets|Halmos|1960|page=92}}</ref>{{efn|'''Proof:''' Observe that <math>\N\times\N</math> is countable as a consequence of the definition because the function <math>f:\N\times\N\to\N</math> given by <math>f(m,n)=2^m\cdot3^n</math> is injective.<ref>{{Harvard citation no brackets|Avelsgaard|1990|page=182}}</ref>  It then follows that the Cartesian product of any two countable sets is countable, because if <math>A</math> and <math>B</math> are two countable sets there are surjections <math>f:\N\to A</math> and <math>g:\N\to B</math>. So <math>f\times g:\N\times\N\to A\times B</math>
is a surjection from the countable set <math>\N\times\N</math> to the set <math>A\times B</math> and the Corollary implies <math>A\times B</math> is countable. This result generalizes to the Cartesian product of any finite collection of countable sets and the proof follows by [[mathematical induction|induction]] on the number of sets in the collection.
}}}}

The set of all [[integer]]s <math>\Z</math> and the set of all [[rational number]]s <math>\Q</math> may intuitively seem much bigger than <math>\N</math>. But looks can be deceiving. If a pair is treated as the [[numerator]] and [[denominator]] of a [[vulgar fraction]] (a fraction in the form of <math>a/b</math> where <math>a</math> and <math>b\neq 0</math> are integers), then for every positive fraction, we can come up with a distinct natural number corresponding to it. This representation also includes the natural numbers, since every natural number <math>n</math> is also a fraction <math>n/1</math>. So we can conclude that there are exactly as many positive rational numbers as there are positive integers. This is also true for all rational numbers, as can be seen below.

{{math theorem | math_statement = <math>\Z</math> (the set of all integers) and <math>\Q</math> (the set of all rational numbers) are countable.{{efn|'''Proof:''' The integers <math>\Z</math> are countable because the function <math>f:\Z\to\N</math> given by <math>f(n)=2^n</math> if <math>n</math> is non-negative and <math>f(n)=3^{-n}</math> if <math>n</math> is negative, is an injective function. The rational numbers <math>\Q</math> are countable because the function <math>g:\Z\times\N\to\Q</math> given by <math>g(m,n)=m/(n+1)</math> is a surjection from the countable set <math>\Z\times\N</math> to the rationals <math>\Q</math>.}}}}

In a similar manner, the set of [[algebraic number]]s is countable.<ref>{{Harvard citation no brackets|Kamke|1950|pages=3–4}}</ref>{{efn|1='''Proof:''' Per definition, every algebraic number (including complex numbers) is a root of a polynomial with integer coefficients. Given an algebraic number <math>\alpha</math>, let <math>a_0x^0 + a_1 x^1 + a_2 x^2 + \cdots + a_n x^n</math> be a polynomial with integer coefficients such that <math>\alpha</math> is the <math>k</math>-th root of the polynomial, where the roots are sorted by absolute value from small to big, then sorted by argument from small to big. We can define an injection (i. e. one-to-one) function <math>f:\mathbb{A}\to\Q</math> given by <math>f(\alpha) = 2^{k-1} \cdot 3^{a_0} \cdot 5^{a_1} \cdot 7^{a_2} \cdots {p_{n+2}}^{a_n}</math>, where <math>p_n</math> is the <math>n</math>-th [[prime number|prime]].}}

Sometimes more than one mapping is useful: a set <math>A</math> to be shown as countable is one-to-one mapped (injection) to another set <math>B</math>, then <math>A</math> is proved as countable if <math>B</math> is one-to-one mapped to the set of natural numbers. For example, the set of positive [[rational number]]s can easily be one-to-one mapped to the set of natural number pairs (2-tuples) because <math>p/q</math> maps to <math>(p,q)</math>. Since the set of natural number pairs is one-to-one mapped (actually one-to-one correspondence or bijection) to the set of natural numbers as shown above, the positive rational number set is proved as countable.

{{math theorem | math_statement = Any finite [[union (set theory)|union]] of countable sets is countable.<ref>{{Harvard citation no brackets|Avelsgaard|1990|page=180}}</ref><ref>{{Harvard citation no brackets|Fletcher|Patty|1988|page=187}}</ref>{{efn|1='''Proof:''' If <math>A_i</math> is a countable set for each <math>i</math> in <math>I=\{1,\dots,n\}</math>, then for each <math>i</math> there is a surjective function <math>g_i:\N\to A_i</math> and hence the function
<math display="block">G : I \times \mathbf{N} \to \bigcup_{i \in I} A_i,</math>
given by <math>G(i,m)=g_i(m)</math> is a surjection.  Since <math>I\times \N</math> is countable, the union <math display="inline">\bigcup_{i \in I} A_i</math> is countable.
}}}}

With the foresight of knowing that there are uncountable sets, we can wonder whether or not this last result can be pushed any further. The answer is "yes" and "no", we can extend it, but we need to assume a new axiom to do so.

{{math theorem | math_statement = (Assuming the [[axiom of countable choice]]) The union of countably many countable sets is countable.{{efn|1='''Proof''': As in the finite case, but <math>I=\N</math> and we use the [[axiom of countable choice]] to pick for each <math>i</math> in <math>\N</math> a surjection <math>g_i</math> from the non-empty collection of surjections from <math>\N</math> to <math>A_i</math>.<ref>{{cite book |last1=Hrbacek |first1=Karel |last2=Jech |first2=Thomas |title=Introduction to Set Theory, Third Edition, Revised and Expanded |date=22 June 1999 |publisher=CRC Press |isbn=978-0-8247-7915-3 |page=141 |url=https://books.google.com/books?id=Er1r0n7VoSEC&pg=PA141 |language=en}}</ref> Note that since we are considering the surjection <math>G : \mathbf{N} \times \mathbf{N} \to \bigcup_{i \in I} A_i</math>, rather than an injection, there is no requirement that the sets be disjoint.}}}}

[[File:Countablepath.svg|thumb|300px|Enumeration for countable number of countable sets]]
For example, given countable sets <math>\textbf{a},\textbf{b},\textbf{c},\dots</math>, we first assign each element of each set a tuple, then we assign each tuple an index using a variant of the triangular enumeration we saw above:
<math display=block>
\begin{array}{ c|c|c } 
\text{Index} & \text{Tuple} & \text {Element} \\ \hline
0 & (0,0) & \textbf{a}_0 \\
1 & (0,1) & \textbf{a}_1 \\
2 & (1,0) & \textbf{b}_0 \\
3 & (0,2) & \textbf{a}_2 \\
4 & (1,1) & \textbf{b}_1 \\
5 & (2,0) & \textbf{c}_0 \\
6 & (0,3) & \textbf{a}_3 \\
7 & (1,2) & \textbf{b}_2 \\
8 & (2,1) & \textbf{c}_1 \\
9 & (3,0) & \textbf{d}_0 \\
10 & (0,4) & \textbf{a}_4 \\
\vdots & &
\end{array}
</math>

We need the [[axiom of countable choice]] to index ''all'' the sets <math>\textbf{a},\textbf{b},\textbf{c},\dots</math> simultaneously.

{{math theorem | math_statement = The set of all finite-length [[sequence]]s of natural numbers is countable.}}

This set is the union of the length-1 sequences, the length-2 sequences, the length-3 sequences, and so on, each of which is a countable set (finite Cartesian product). Thus the set is a countable union of countable sets, which is countable by the previous theorem.

{{math theorem | math_statement = The set of all finite [[subset]]s of the natural numbers is countable.}}

The elements of any finite subset can be ordered into a finite sequence. There are only countably many finite sequences, so also there are only countably many finite subsets.

{{math theorem | math_statement = Let <math>S</math> and <math>T</math> be sets.
# If the function <math>f:S\to T</math> is injective and <math>T</math> is countable then <math>S</math> is countable.
# If the function <math>g:S\to T</math> is surjective and <math>S</math> is countable then <math>T</math> is countable.}}

These follow from the definitions of countable set as injective / surjective functions.{{efn|'''Proof''': For (1) observe that if <math>T</math> is countable there is an injective function  <math>h:T\to\N</math>.  Then if <math>f:S\to T</math> is injective the composition <math>h\circ f:S\to \N</math> is injective, so <math>S</math> is countable.

For (2) observe that if <math>S</math> is countable, either <math>S</math> is empty or there is a surjective function <math>h:\N\to S</math>.  Then if <math>g:S\to T</math> is surjective, either <math>S</math> and <math>T</math> are both empty, or the composition <math>g\circ h:\N\to T</math> is surjective. In either case <math>T</math> is countable.
}}

'''[[Cantor's theorem]]''' asserts that if <math>A</math> is a set and <math>\mathcal{P}(A)</math> is its [[power set]], i.e. the set of all subsets of <math>A</math>, then there is no surjective function from <math>A</math> to <math>\mathcal{P}(A)</math>. A proof is given in the article [[Cantor's theorem]]. As an immediate consequence of this and the Basic Theorem above we have:
{{math theorem | name = Proposition | math_statement = The set <math>\mathcal{P}(\N)</math> is not countable; i.e. it is [[uncountable]].}}

For an elaboration of this result see [[Cantor's diagonal argument]].

The set of [[real number]]s is uncountable,{{efn|See [[Cantor's first uncountability proof]], and also [[Finite intersection property#Applications]] for a topological proof.}} and so is the set of all infinite [[sequence]]s of natural numbers.