Continuous function $f$ such that for every sequence $(x_n) \in [-1,1]^{\mathbb Z}, \exists t \in \mathbb R$ satisfying $f(t+n) = x_n$ for all $n$ Does there exist a continuous function $f : \mathbb R \rightarrow \mathbb R$ such that for every sequence $(x_n) \in [-1,1]^{\mathbb Z}$, there is $t \in \mathbb R$ satisfying $f(t+n) = x_n$ for all $n$?
I'm not sure how to approach this question, any help would be highly appreciated.
 A: Here's a construction of such a function, which uses two tools: the Hahn-Mazurkiewicz Theorem applied to the space $X = [-1,+1]^{\mathbb Z}$; and the shift map for $X$.
Let's equip the space $X = [-1,+1]^{\mathbb Z}$ with the product topology. The Hahn-Mazurkiewicz Theorem for $X$ gives a bunch of hypotheses which imply the existence of a continuous surjective function $\gamma : [0,1] \to X$, i.e. of a "space filling curve" in $X$. Let's check these hypotheses one-by-one:


*

*$[-1,+1]^{\mathbb Z}$ is Hausdorff, because it is a product of Hausdorff spaces. 

*$[-1,+1]^{\mathbb Z}$ is compact by Tychonoff's Theorem, because it is a product of compact spaces. 

*$[-1,+1]^{\mathbb Z}$ is connected, because it is a product of connected spaces. 

*$[-1,+1]^{\mathbb Z}$ is locally connected, because it has a basis of connected subsets, namely those sets of the form $\prod_{n \in \mathbb Z} I_n$ for which there exists a finite subset $A \subset \mathbb Z$ such that if $n \not\in A$ then $I_n = [-1,+1]$, and if $n \in A$ then there exists an open subinterval $(a,b) \subset \mathbb R$ such that $I_n = [-1,+1] \cap (a,b)$. Each such basis element is connected, because it is a product of connected sets. 

*$[-1,+1]^{\mathbb Z}$ has a countable basis, because in item 4, we can require that $a,b$ are rational numbers.


That's all the hypotheses of the Hahn-Mazurkiewicz Theorem, so we can apply its conclusion: there exists a surjective continuous path onto the space $\gamma : [0,1] \to X = [-1,+1]^{\mathbb Z}$. Without much trouble, one can strengthen this slightly: for any two points $x,y \in X$ one can find $\gamma$ so that $\gamma(0)=x$ and $\gamma(1)=y$. I'll pick $x$ and $y$ in a moment.
The second tool is the shift map
$$\Sigma : [-1,+1]^{\mathbb Z}
$$
which, for each $x = (x_n) \in [-1,+1]^{\mathbb Z}$ is given by
$$(\Sigma x)_n = x_{n-1}
$$
The function $\Sigma$ is a self-homeomorphism of $[-1,+1]^{\mathbb Z}$: it is one-to-one, onto, and continuous, with a continuous inverse. In particular the (positive and negative) powers $\Sigma^n$ are defined for each $n \in \mathbb Z$, each is a homomeorphism, and $\Sigma^{m+n}=\Sigma^m \circ \Sigma^n$.
Now let's pick any $x^0 = (x^0_n) \in X$, and let $\Sigma x^0 = x^1 = (x^1_n)$. It follows that $x^1_n = x^0_{n-1}$. Applying the slightly strengthened form of the Hahn-Mazurkiewicz theorem we obtain a surjective, continuous function $\gamma : [0,1] \to [-1,+1]^{\mathbb Z}$ such that $\gamma(0) = x^0$ and $\gamma(1) = x^1$. 
The function $\gamma : [0,1] \to [-1,+1]^{\mathbb Z}$ has a unique continuous extension $\gamma : \mathbb R \to [-1,1]^{\mathbb Z}$ such that $\gamma$ is "$\Sigma$-equivariant", by which I mean that $\gamma(t+n) = \Sigma^n \circ \gamma(t)$ for all $t$. Namely, we've already defined $\gamma$ on $[0,1]$. For any $n \in \mathbb Z$ and any $t \in [0,1]$ can define $\gamma$ on $[n,n+1]$ by the formula $\gamma(t+n) = \Sigma^n \circ \gamma(t)$ for all $t \in [0,1]$. One checks that $\gamma$ is well-defined at the point $n$ where the two intervals $[n-1,n]$ and $[n,n+1]$ intersect; one then uses continuity of $\Sigma^n \circ \gamma$ to deduce continuity of $\gamma$ on $[n,n+1]$; and finally one uses the gluing theorem of topology to deduce that $\gamma$ is continuous.
Now define $f : \mathbb R \to \mathbb R$ as follows. Let's switch to "$x$" notation, namely $x^t = \gamma(t)$, so $(x^t_n) = \gamma(t)(n)$. Now define
$$f(t) = x^t_0 = \gamma(t)(0)
$$
The function $f$ is continuous because it is a composition of two continuous functions: the function $\gamma : \mathbb R \to [-1,+1]^{\mathbb Z}$; and the function $[-1,+1]^{\mathbb Z} \to [-1,+1]$ which projects onto the first coordinate, i.e. $(x_n) \mapsto x_0$.
For any sequence $x = (x_n) \in [-1,+1]^{\mathbb Z}$, find $t \in [0,1]$ so that $x = \gamma(t)$, i.e. $x^t_n = x_n$ for all $n$. It follows that
$$f(t+n) = \gamma(t+n)(0) = \Sigma^n(\gamma(t)(0)) = \gamma(t)(n) = x_n
$$
