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  clickbait: 'Some infinities are bigger than others.  Countable infinities are the smallest infinities.',
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  text: 'The [-3jz set] of *counting numbers*, or of *positive integers*, is the set $\\mathbb{Z}^+ = \\{1, 2, 3, 4, \\ldots\\}$.\n\nA set $S$ is called *countable* or *enumerable* if there exists a [4bg surjection] from the counting numbers onto $S$.\n\n### Example: The rational numbers ###\n\nThe set of *rational numbers*, $\\mathbb Q$, is the set of integer fractions $\\frac{p}{q}$ in reduced form; the greatest common divisor of $p$ and $q$ is one, with $q > 0$.\n\n**Theorem** The rational numbers are countable.\n\nThe proof is, essentially, that $\\mathbb Z^+ \\times \\mathbb Z^+$ is isomorphic to $\\mathbb Z$; we count in a roughly spiral pattern centered at zero.\n\n**Proof**  Define the *height* of $\\frac{a}{b}$ to be $|a| + |b|$.  We may count the rational numbers in order of height, and ordering by $a$, and then $b$, when the heights are the same.  The beginning of this counting is $0 / 1$, $-1 / 1$, $1 / 1$, $-2 / 1$, $-1 / 2$, $1 / 2$, $2 / 1$, $\\ldots$   Since there are at most $(2d+1)^2$ rational numbers of height less than or equal to $d$, a rational number with height $d$ is mapped on to by one of the counting numbers up to $(2d+1)^2$; every rational number is mapped onto by this counting.  Thus, the rational numbers are countable.  $\\square$\n\n*Note*: It is not hard to extend this proof to show that $(\\mathbb Z^+)^n$ is countable for any finite $n$.\n\n**Theorem** If there exists a surjection $f$ from a countable set $A$ to a set $B$, then $B$ is countable.\n**Proof** By definition of countable, there exists an enumeration $E$ of $A$. Now, $E\\circ f$ is an enumeration of $B$, so $B$ is countable.\n\n##Exercises\n\n>Show that the set $\\Sigma^*$ of [ finite words] of an enumerable [ alphabet] is countable.\n\n%%hidden(Show solution):\nFirst, we note that since $\\mathbb N^n$ is countable, the set of words of length $n$ for each $n\\in \\mathbb N$ is countable. \n\nLet $E_n: \\mathbb N  \\to \\mathbb N^n$ stand for an enumeration of $\\mathbb N ^n$, and $(J_1,J_2)(n)$ for an enumeration of $\\mathbb N^2$.\n\nConsider the function $E: \\mathbb N \\to \\Sigma^* , n\\hookrightarrow E_{J_1(n)}(J_2(n))$ which maps every number to a word in $\\Sigma^*$. Then a little thought shows that $E$ is an enumeration of $\\Sigma^*$.\n\n$\\square$\n%%\n\n\n\n>Show that the set $P_\\omega(A)$ of finite subsets of an enumerable set $A$ is countable.\n\n%%hidden(Show solution):\nLet $E$ be an enumeration of $A$.\n\nConsider the function $E': \\mathbb N^* \\to P_\\omega(A)$ which relates a word $n_0 n_1 n_2 ... n_r$ made from natural numbers to the set $\\{a\\in A:\\exists m\\le k E(n_m)=a\\}\\subseteq A$. Clearly $E'$ is an enumeration of $P_\\omega(A)$.\n%%\n\n>Show that the set of [ cofinite subsets] of an enumerable set is countable.\n\n%%hidden(Show solution):\nSimply consider the function which relates each cofinite set with its complementary.\n%%',
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