This is a follow up of an interesting [recent question](https://mathoverflow.net/questions/50548/approximating-polynomials-in-rx-using-integer-valued-polynomials) 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](http://books.google.com/books?id=AlAluH5is6AC&printsec=frontcover&dq=integer+valued+polynomials&source=bl&ots=prKsKY2osw&sig=1sxbmty-6sb7AXpA6FtM5pmkVz4&hl=en&ei=CzYbTY--Msfvsga7yNWVBA&sa=X&oi=book_result&ct=result&resnum=2&sqi=2&ved=0CCEQ6AEwAQ#v=onepage&q&f=false) 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. 



The closures in $C^0(\mathbb{R}, \mathbb{R})$ of the set of integer valued polynomials, resp, of polynomials with integer coefficients
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.


