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I've been doing some work with cardinality of sets, and ran into an example I thought was interesting. In proving that the set of prime numbers is a countably infinite set, I've started that showing that the set of prime numbers (integers) $\mathbb{P}$ is a subset of $\mathbb{N}$. Obviously the natural numbers $\mathbb{N}$ can be mapped one-to-one to itself ($1$ to $1$, $2$ to $2$, etc.), so it is a countably infinite set. Following from this, since $\mathbb{P}$ is a subset of a countably infinite set $\mathbb{N}$, then $\mathbb{P}$ must be a countably infinite set as well.

Is this enough information to show $\mathbb{P}$ is a countably infinite set, or must I show a concrete mapping for $\mathbb{P}$?

swolo
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  • Best to show a bijection. ${1,5}$ is a subset of a countably infinite set but it isn't countable infinite. – Integrand Apr 14 '20 at 01:33
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    The fact that the set of primes is a subset of $\Bbb N$ immediately ensures that it is at most countably infinite. To finish the job you must show that it is not finite. – Brian M. Scott Apr 14 '20 at 01:34
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    FYI: the notation $\mathbb{Z}_p$ is not the most common for the set of primes (to the best of my knowledge). Probably better to use something like $\mathbb{P}$ since $\mathbb{Z}_p$ more commonly refers to the set of integers modulo a prime $p$ or the set of $p$-adic integers. – Luke Poeppel Apr 14 '20 at 01:40
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    $\mathbb Z_p$ means other things in mathematics, so consider using $\mathbb P$ instead (I now see @Luke Poeppel commented similarly) – J. W. Tanner Apr 14 '20 at 01:40
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    Thank you @LukePoeppel and J.W.Tanner I have edited my post to better reflect the set. – swolo Apr 14 '20 at 01:54
  • If youy show it is not finite, and a subset of $\mathbb{N}$, then you show it is countably infinite. But right now you have only shown it is countable, but not necessarily infinite. – Arturo Magidin Apr 14 '20 at 01:57

1 Answers1

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One way of proving that a subset of $A \subset \Bbb N$ is countably infinite is by demonstrating that it satisfies the following property,

$\tag 1 (\forall F \subset A) \, \bigr[ \, \text{IF } F \text{ if finite THEN } (\exists x \in \Bbb N) \, [x \notin F \text{ and } x \in A] \, \bigr ]$

See also

$\quad$ Subset of a countable set is itself countable

Now let $\mathbb{P}$ denote the set of prime numbers and let $F$ be any finite subset of $\mathbb{P}$.

If $|F| = 0$ then $2 \notin F$ but $2 \in \Bbb P$.

if $|F| = n \gt 0$ then $F = \{p_1,p_2,\dots\,p_n\}$. Set

$\quad \displaystyle {a = (\prod_{1 \le i \le n} p_i) + 1} $

Using Euclidean division one readily shows that $p_i \nmid a$ for any subscript $i$. So if $q$ is any prime factor of $a$ then $q \ne p_i$ for any subscript $i$. So $q \notin F$ but $q \in \Bbb P$.

The above proof 'gears' Euclid's (direct) proof to the OP's requested set theoretic setting.

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