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So I understand that there are infinitely many non-prime natural numbers, and infinitely many prime natural numbers. I am able to show that there are an infinite number of non-prime natural numbers, but I don't know how to show that the entire set of non-prime natural numbers is infinite.

Consider the set $S=\{{p\in\mathbb{N}\space |\space p\space is\space prime}\}$

Assuming we already know that this set is infinite, consider the set $T=\{q\in\mathbb{N}\space|\space q=2p, where\space p\space is\space prime\}$

$T$ is then also an infinite set since it is contained in the set of all non-prime natural numbers

Thus there are an infinite number of non-prime naturals.

But how do I show that the entire set of non-prime naturals is infinite? Is it as easy as saying that it is a subset of the naturals and since the set of naturals is infinite, this set must also be infinite?

Mather Guy
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    Yes. Knowing that a set contains an infinite subset tells us that the set is itself infinite. – lulu Mar 19 '20 at 23:22
  • It shoud not be hard to prove that a finite set cannot contain infinite subset. – user Mar 19 '20 at 23:24
  • Your final paragraph is off base. The important part is that the set of non-primes has an infinite subset (which you show above), not that it is a subset of the naturals. ${5}$ is a subset of the naturals and the naturals are infinite, but ${5}$ is finite. – Ross Millikan Mar 19 '20 at 23:33
  • "Is it as easy as saying that it is a subset of the naturals and since the set of naturals is infinite, this set must also be infinite?" That's backwards. ${2,7,11}$ is a subset of the naturals but is not infinite. But if $A \subset B \subset \mathbb N$ and $A$ is infinite then $B$ is infinite. And it is as easy as saying that. – fleablood Mar 19 '20 at 23:41

3 Answers3

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It looks like there's some confusion over the basic definitions here.

I am able to show that there are an infinite number of non-prime natural numbers, but I don't know how to show that the entire set of non-prime natural numbers is infinite.

These are the same thing: "has infinitely many elements" and "is infinite" are just two ways of phrasing the same property.

Is it as easy as saying that it is a subset of the naturals and since the set of naturals is infinite, this set must also be infinite?

This makes me think that you might be conflating "is infinite" with "is countable." It's certainly not true that every subset of the naturals is infinite, but it is true that every subset of the naturals is countable (or finite - some texts define "countable" as "in bijection with $\mathbb{N}$" as opposed to "in bijection with some subset of $\mathbb{N}$).

So if the problem you're trying to solve is "Show that the set of composite numbers is countable and infinite, then you've more-or-less outlined it:

  • It's a subset of $\mathbb{N}$, so trivially countable (in the broader sense of allowing finiteness).

  • The argument you give in the OP shows that it's infinite.

Noah Schweber
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You have a few issues, first, you say that $T$ is infinite because it is contained in the set of all composite (non-prime) numbers. But this is the set you wanted to prove was infinite! What you really mean to say is that because there are infinitely many distinct prime numbers, the set $T$ is infinite. Because $T$ is contained in the set of composite numbers (every number in $T$ is divisible by $2$ and some prime, so the numbers in $T$ cannot be prime), then it must be that the set of composite numbers is infinite [so in a sense, you have proved what you wanted, modulo phrasing errors]. As for your last sentence, I think you mean how do you prove the set of primes is infinite, for that, see this answer.

But you do not need to invoke the infinitude of primes at all! Consider the set $T= \{ 2n \colon n \in \mathbb{N}, n>1\}$, i.e. the even numbers bigger than $2$. This set is clearly infinite because $\mathbb{N}$ is. Moreover, every number in $T$ is divisible by $2$ and some other integer, so every number in $T$ is composite. Therefore, $T$ is a subset of the composite numbers that is infinite, so that this set is infinite! You do not need to even consider this $T$; for example, $T= \{ 2^n \colon n >1\}$, $T= \{ 2 \cdot 3^b \colon n \geq 1\}$, etc all work as well!

Perhaps what you're worried about is that your (or even my) $T$ do not contain all composite numbers. But you do not need to find them all, just an infinite amount of them to prove that the set of all of them is infinite. Because the set of composite numbers (other than 1) is the complement of the set of prime numbers in $\mathbb{N}$, finding one set is the 'same' as finding the other. But finding all the primes is really hard! So it will be equally hard to find all the composite numbers. Again, luckily, this is not what is required for your proof.

mathematics2x2life
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I think the proposition/Corollary will help you that:

Proposition: Every subset of a finite set is finite.

Corolary: If $A$ is an infinite subset of $B$ then $B$ is also infinite

will help you.

Then as the even numbers (other than $2$) are composite, or the set of $2p$ for all prime $p$ are infinite, or even the fact that ever pair of consecutive primes equal or higher, have a gap of at least one between the, and subsets of the composites, the set of composites is, of course, infinite.

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fleablood
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