Continuity of $\{f(x)\}$ 
Does there exist a function $f(x)$ with a range including all real numbers so that $\{f(x)\}$ has no discontinuities?

So far I have shown that if so, such a function must be decreasing whenever it approaches an integer value, and that it must not ever cross any line defined by $y=n$, where $n$ is an integer, though it can touch such a line. Also, I have shown that the function need not be continuous as long as its "jumps" are integers.
My intuition tells me that no such function exists. How can I prove it?
 A: Such functions exist. To show this, let $g$ be your favorite continuous function $[0,1]\rightarrow [0,1]$ such that $g(0)=g(1)=0$ and $g(x)=1$ for some $x$. Also choose your favorite function $h:\mathbb Z \rightarrow\mathbb Z$ with the property that for every $m\in \mathbb Z$ the set $\{n\in\mathbb Z:h(n)=m\}$ is unbounded above. Lastly, choose your favorite function $i:\mathbb Z\rightarrow[0,1)$ such that $\lim_{n\rightarrow\infty}i(n)=1$.
Now, define a function $f$ as follows, where $n\in \mathbb N$ and $x\in [0,1)$:
$$f(n+x)=h(n)+i(n)g(x).$$
The fractional part of this is just $i(n)g(x)$, which is obviously continuous at non-integers and is continuous at integers because $g(x)$ vanishes near integers. This is surjective since if you choose $y\in \mathbb R$ and write it as $y=n+x$ for $x\in [0,1)$, then you can find arbitrarily large $m$ such that $h(m)=n$. Then, just choose such an $m$ large enough that $i(m)>x$, in which case $f$ will necessarily equal $y$ in the range $[m,m+1)$ by continuity of $f$ on this range.

In case your favorite functions do not have the desired properties, here's some possible choices for the functions mentioned:
$$g(x)=\begin{cases}2x&\text{if }x\leq 1/2 \\ 2-2x&\text{if }x\geq 1/2\end{cases}$$
Define $h(n)$ by first choosing $m^2$ be the nearest (integer) square to $n$ and setting $h(n)=n-m^2$.
Define $i(n)=\frac{1}{1+e^{-n}}$.
