What is $\text{inf} \{ \prod_{i=0}^{n}(x-i) ; n \in \mathbb{N_{0}} \}$ of relation $\sqsubseteq$?

Question

On the set of polynomials with real coefficients we define $\sqsubseteq$ as:

$p \sqsubseteq q \iff p=q\ \vee \exists n \in \mathbb{N_{0}}:(p(n) <q(n)\ \wedge \forall m \in \mathbb{N_{0}}:(m<n \Rightarrow p(m)=q(m)))$

First part of the question was to prove that this is total order, which I had no problem with. Then it also asks what is $\text{inf} \{ \prod_{i=0}^{n}(x-i) ; n \in \mathbb{N_{0}} \}$?

My initial though was to take $\prod_{i=0}^{\infty}(x-i)$, which would have all natural numbers as its zeroes, but if I remember correctly, a polynomial can't have infinitely many zeroes unless it is constant. Now I have two questions:

1. Why can't polynomial have infinitely many zeroes, as does $\prod_{i=0}^{\infty}(x-i)$?
2. If not $\prod_{i=0}^{\infty}(x-i)$, then what is $\text{inf} \{ \prod_{i=0}^{n}(x-i) ; n \in \mathbb{N_{0}} \}$ equal to?

Answer 1

For $n\in\Bbb N_0$ let $p_n(x)=\prod_{i=0}^n(x-i)$. Clearly for each $n\in\Bbb N_0$ we have $p_n(k)=0$ for all $k\in\Bbb N_0$ such that $k\le n$, while $p_n(k)>0$ for $k\in\Bbb N_0$ such that $k>n$. It follows that if $m,n\in\Bbb N_0$, and $m<n$, then $p_m(k)=p_n(k)=0$ for all $k\in\Bbb N_0$ such that $k\le m$, while $p_m(m+1)>0=p_n(m+1)$, so $p_n\sqsubseteq p_m$ (and in fact $p_n\sqsubset p_m$).

Let $p(x)$ be a real polynomial. It is clear that if $p\sqsubseteq p_n$ for all $n\in\Bbb N_0$, then $p(k)\le 0$ for all $k\in\Bbb N_0$: if some $p(k)>0$, then $p_k\sqsubset p$. The simplest real polynomial satisfying this requirement is the constant function $z(x)=0$, and indeed $z\sqsubset p_n$ for each $n\in\Bbb N_0$. It is also clear that if $p$ is any real polynomial such that $p(k)\le 0$ for all $k\in\Bbb N_0$, and there is a $k_0\in\Bbb N_0$ such that $p(k_0)<0$, then $p\sqsubset z$, and hence $p\ne\inf\{p_n:n\in\Bbb N_0\}$. The only possible candidate for $\inf\{p_n:n\in\Bbb N_0\}$ besides $z$ would therefore be some real polynomial $p$ such that $p(k)=0$ for each $k\in\Bbb N_0$. But it is a well-known consequence of the fundamental theorem of algebra that a non-zero real polynomial of degree $n$ has at most $n$ real roots, so there is no such $p$: the zero polynomial $z$ is the only real polynomial taking the value $0$ at each non-negative integer, and it follows that $\inf\{p_n:n\in\Bbb N_0\}=z$.

marty cohen has already pointed out that $\prod_{i\ge 0}(x-i)$ is undefined for all $x\in\Bbb R\setminus\Bbb N_0$: unless $x$ is a non-negative integer, all of the factors are non-zero, exactly two have absolute values less than $1$, and there are factors with arbitrarily large absolute value. It is also not a polynomial, so it is irrelevant here: the infimum is taken in the domain of the order $\sqsubseteq$, i.e., in the set of real polynomials.

Answer 2

$\prod_{i=0}^{\infty}(x-i)$ is undefined for any $ x\not\in \mathbb{N_{0}}$.