Let $f:\mathbb{N}\to\mathbb{R}$ such that $f(n)=p$ (where $p$ is the $n$th prime number). My doubt is whether this is a function and hence a sequence. I got this doubt because we don't know all the primes, right? So after a certain stage, we don't know what is the output of the function.
$\endgroup$ 92 Answers
$\begingroup$The fact that we humans don't know all the elements of a sequence doesn't stop a sequence from being a sequence. Yes, the sequence of prime numbers is a sequence, as well-defined as any.
$\endgroup$ $\begingroup$Similarly, $(a_n)_{n\in\mathbb N}$ defined as $a_n = n,$ that is an identity function on the set of natural numbers $a:\mathbb N \to \mathbb N$ defined with $a: n \mapsto n$, would be a non-sequence, because we do not know all the natural numbers. Right?
The fact we do not know apriori some / many / almost all of the terms does not invalidate the definition. As long as each term is well defined, the sequence is defined.
The natural numbers are well-ordered, so their subset of prime numbers are well-ordered, too. Hence 'the next prime number' is well defined at each step, and so is the whole sequence. No matter how hard it might be to find the actual value of the 'next term'.
