Why is the smallest factor of 1 considered to be $\infty$ in analytic number theory?

Question

In some works and textbooks, the function $P^{-}:\mathbb{N}\to\mathbb{N}\cup {\infty}$ is denoted as the function that returns the smallest prime factor of a positive integer. Furthermore, it is stated that $P^{-}(1)=\infty$ by convention.

Shouldn't $P^{-}(1) = 1$?

In the analysis of numbers free of prime factors smaller than a bound $y$, we have the function $$\Phi(x,y)= |\{ 1\leq n \leq x: P^{-}(n)>y \}| $$ with $P^{-}(1)=\infty$, wouldn't this means that 1 is always counted by $\Phi$?

Other works where I have seen $P^{-}$ with that conventional value for The Distribution of Prime Numbers by Dimitris KouKoulopoulos, Multiplivcative functions in large arithmetic progressions and applications by Fouvry and Tenenbaum

Answer 1

$P^{-}(1) = 1$ doesn't make sense for two reasons: first, $1$ isn't prime, and second, $P^{-}(n)$ has the property that if $n \mid m$ then $P^{(-)}(n) \ge P^{(-)}(m)$ (note that the order is reversed), and setting $P^{-}(1) = 1$ causes this property to not hold. In fact it can only hold if $P^{-}(1) = \infty$, because $P^{-}(1)$ needs to be greater than every prime.

This is an example of the funny behavior that sometimes happens with "empty operations" like the empty sum and empty product. $P^{-}(n)$ can be defined as the infimum over all the prime factors of $n$, so $P^{-}(1)$ is the empty infimum. In general the infimum of a subset $S$ of a partially ordered set $X$, if it exists, is its greatest lower bound; if $S$ is empty, every element is a lower bound, so the empty infimum is the greatest element of $X$.

Now there is no greatest prime number, so you would be justified in arguing that $P^{-}(1)$ is undefined. But in some places it may be convenient to assign it a value, and you can do that by just adjoining a greatest element; that's $\infty$.

It may seem counterintuitive that we get the greatest element here. The idea is that infimums "start at infinity and count down," in the same way that sums "start at zero" and products "start at one" which is why the empty sum and product take those values respectively.

Answer 2

When $n$ is some integer, $n$ is not divisible by primes $p<x$ indicates that $P^-(n)\ge x$. For $n=1$, this $x$ can be arbitrarily large, so $P^-(1)=+\infty$.