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This is a follow up of an interesting recent question on the topic. The answer given there by fedia shows that the matter is rich and complicated, and I can't resist to submit here a further question.

Q1. What is the closure of the integer valued polynomials in $C^0(\mathbb{R},\mathbb{R})$ w.r.to the uniform convergence on bounded sets?

For instance, a nontrivial element in the closure is, by the Euler infinite product

$$\frac {\sin(\pi x)} \pi x = \lim_{n \to \infty}\, \binom {x-1} n \binom {x+n} n $$

but can one reach all integers-to-integers continuous functions on $\mathbb{R}$?

(After a quick Google search, among the large literature on integer valued polynomials, I was only able to find this book concerning approximation by IVP, that however treats approximation of functions on the p-adic integers rather than on $\mathbb{R}$; I'm not p-adic enough to understand if there are implications to the real case).

[edit] A further question raised by SJR

Q2. What is the closure of the polynomials with integer coefficients in $C^0(\mathbb{R},\mathbb{R})$ w.r.to the uniform convergence on bounded sets?

At a first glance I'd have said $\mathbb{Z}[x]$ is closed in $C^0(\mathbb{R}, \mathbb{R})$ for some elementary reason, but now I don't see it. It is certainly a countable, metrizable commutative group; so it is closed iff it is discrete. It has an additional ring structure and it's a monoid by composition, both structure being compatible with the topology.

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The $p$-adic situation is very different since $\mathbb{Z}$ is dense in $\mathbb{Z}_p$. –  Felipe Voloch Dec 30 '10 at 1:25
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Pietro--I believe it's known that Z[x] is closed in C^0(R,R); see the reference I give in my comment to your answer. –  paul Monsky Dec 31 '10 at 2:49
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up vote 4 down vote accepted

I'd say that the closure of the integer-valued polynomials (IVP) by uniform convergence on bounded sets is the whole set $C^0\big((\mathbb{Z},\mathbb{R}),(\mathbb{Z},\mathbb{R})\big)$ of continuous functions mapping integers to integers.

Let $r\le s\in\ \mathbb{N}$. Then $\big|\binom x s \big| < 1 $ holds for all $x \in ]0,r[$. Consider the IVP $f_n(x):=\frac 1 2 (x + x^n)$, that maps $]-1,1[$ into itself, and converges to $\frac x 2 $ uniformly on compact sets of $]-1,1[$. So th $k$-fold iterate $f_n^k:=f_n\circ\dots\circ f_n$ converges to $\frac x {2^k}$, and for all $p\in\mathbb{Z}$ the sequence of IVP $pf^k _n\big(\binom x s \big)$ converges to $\frac p {2^k} \binom x s $ as $n\to \infty$ on $]0,r[$. By density of the dyadic rationals, and by linearity, we get at least any polynomial multiple of $\binom x r$ as a uniform limit of IVP on compact sets of $]0,r[$, and of course, any polynomial taking integer values on $x=0,1\dots, r$ as well. By density, we also get any function $f\in C^0(]0,r[)$ that takes integer values on the integer points; in particular the restriction to $]-1,r[$ of any $f\in C^0\big((\mathbb{Z},\mathbb{R}),(\mathbb{Z},\mathbb{R})\big)$. By translation invariance, and since $r$ is arbitrary, this is true for any interval in place of $]0,r[$, and we conclude by a diagonal argument.

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I see, but let me say that you should think more before posting a question, and avoid these schyzoid dialogues with yourself. –  Pietro Majer Dec 30 '10 at 8:40
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No, this is very nice - Thanks for being two. What about the closure of $\mathbb{Z}[x]$? –  SJR Dec 30 '10 at 9:16
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Or maybe Z[x] is closed? –  SJR Dec 30 '10 at 11:03
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Pietro, SJR--I believe the answer is yes and that a proof is to be found in Le Baron O. Ferguson's book: "Approximation by polynomials with integral coefficients", Math Surveys and monographs vol. 17. He also had a survey article in the Monthly on the topic; perhaps there's a proof there as well. –  paul Monsky Dec 31 '10 at 2:14
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@Paul: Hi Paul! I wrote a survey of Ferguson's results which you can find at the very bottom of this page umpa.ens-lyon.fr/~lberger/publications.html (it's in French, sorry). The main results are that if $K$ is a compact subset of $R$ then $Z[X]$ is discrete in $C^0(K,R)$ if the capacity of $K$ is $\geq 1$ and otherwise there is a finite set $J(K) \subset K$ such that a function $f$ is a limit of elements of $Z[X]$ if and only if there exists some $P(X) \in Z[X]$ such that $f(a)=P(a)$ for all $a \in J(K)$. For example if $K=[0,1]$ then $J(K)=\{0,1\}$. –  Laurent Berger Dec 31 '10 at 10:25
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