-1

Let C be the set of all finite sets of prime numbers. show C is countable.

tried to do this by creating a bijection f : C -> N where X is an element of C and f(X)=the product of all elements in X.

figured it’s injective by looking at sets A and A’ which are elements of C and defining A’ as A-{p} where p is some prime number. from there A≠A’ and f(A)≠f(A’) because p|f(A) but p does not |f(A’). so this function is injective.

next, surjective: tried to argue this by saying let f(A)=b=the product of all elements in A. since b is some natural number and any natural number can be expressed as a product of n many prime numbers it’s also surjective. realized there is an issue with this though since we are looking at sets, not sequences. two of the same prime would not appear in the same set so numbers like 4 (2*2) cause it to fail.

theoretically i would have confirmed it is injective and surjective, therefore there exists a bijection between C and N so C is countable. however, this proof clearly does not work.

how else could i create a bijection or otherwise prove this? thanks.

confusedmathstudent
  • 1
  • 1
    Try $$f(S)=\sum_{p\in S} 2^{\pi(p)-1}$$ where $\pi(x)$ is the prime counting function. https://en.wikipedia.org/wiki/Prime-counting_function – Thomas Andrews May 03 '24 at 18:01
  • To show that a set $C$ is countable, you just need an injective function $f:C\to\Bbb N$. Your argument about injectivity doesn't quite make sense (you can't "define" $A'$ to be something, it has to be arbitrary), but your $f$ is injective by the fundamental theorem of arithmetic. – Karl May 03 '24 at 18:09
  • Also, the role of the prime numbers here is a little misleading since it has nothing to do with countability. Every collection of finite sets of natural numbers is countable (but you'd need a different injective function to show this). – Karl May 03 '24 at 18:11
  • Thomas Andrews' solution implies converting a set into binary digits where you put a 1 in the nth decimal place if it contains the nth prime number, 0 otherwise. A finite set necessarily terminates in a string of zeros. There's at least 1 set for each prime number, so there are infinitely many elements. Since the member sets are all finite, they all terminate in a sequence of 0s. These are all rational numbers interpreted as a binary decimal. Since there are only countably many rationals, it follows that the set in question is also countable. – TurlocTheRed May 03 '24 at 18:58

1 Answers1

1

Let $Q$ denote the squarefree numbers. Consider $$C\overset{f}{\to} Q\overset{g}{\to} \mathbb{N}$$ where $$f(S)=\prod_{s\in S}s$$ and $$g(q)=\vert\{r\in Q:r\leq q\}\vert.$$ Now $g\circ f$ is the wished for bijection.

Here we consider $0\notin\mathbb{N}$ and $\emptyset\in C$.

  • I do think that this is the proof that whoever posed this question, was looking for – Mike May 03 '24 at 18:20