I'm going through Problem and Theorems in Classical Set Theory by Komjath and Totik, and the answer to this question is: Let $[x]$ (resp. $\{x\}$) denote the integral (resp. fractional) part of $x$, and set $x ≺ y$ if $\{x\} < \{y\}$ or if $\{x\} = \{y\}$ and $[x] < [y]$.
It is clear that in this ordering $x − 1$ is the predecessor, and $x + 1$ is the successor of $x$.
Everything I had read states that elements of the the reals $\mathbb{R}$ don't have an immediate successors or predecessor. Could someone help explain this? Thanks.
In the usual ordering of the reals no element has an immediate predecessor or successor, but this ordering is very different from the usual one. We can think of a real number as an ordered pair, specifically, as an element of $[0,1)\times\Bbb Z$: the ordered pair $\langle\alpha,n\rangle$ represents the real number $n+\alpha$, where $n$ is the integer part and $\alpha$ the fractional part. It’s easy to see that this is a bijection between $[0,1)\times\Bbb Z$ and $\Bbb R$.
We now let $\preceq$ be the lexicographic order on $[0,1)\times\Bbb Z$: $\langle\alpha,m\rangle\preceq\langle\beta,n\rangle$ if and only if either $\alpha<\beta$, or $\alpha=\beta$ and $m\le n$. I claim that in this ordering the pair $\langle\alpha,n\rangle$ has $\langle\alpha,n-1\rangle$ as immediate predecessor and $\langle\alpha,n+1\rangle$ as immediate successor. Suppose, for instance, that
$$\langle\alpha,n\rangle\preceq\langle\beta,m\rangle\preceq\langle\alpha,n+1\rangle\;.$$
Since $\langle\alpha,n\rangle\preceq\langle\beta,m\rangle$, either $\alpha<\beta$, or $\alpha=\beta$ and $n\le m$, so in particular $\alpha\le\beta$. And since $\langle\beta,m\rangle\preceq\langle\alpha,n+1\rangle$, either $\beta<\alpha$, or $\beta=\alpha$ and $m\le n+1$, so in particular $\beta\le\alpha$. Thus, we must have $\alpha=\beta$, in which case $n\le m\le n+1$. But then clearly either $m=n$ or $m=n+1$, so $\langle\beta,m\rangle$ is equal either to $\langle\alpha,n\rangle$ or to $\langle\alpha,n+1\rangle$. This shows that there is no point of $[0,1)\times\Bbb Z$ strictly between $\langle\alpha,n\rangle$ and $\langle\alpha,n+1\rangle$ in the order $\preceq$, i.e., that $\langle\alpha,n+1\rangle$ is the immediate successor of $\langle\alpha,n\rangle$. The proof that $\langle\alpha,n-1\rangle$ is the immediate predecessor of $\langle\alpha,n\rangle$ is similar.
The ordering of $\Bbb R$ described in the question is exactly the same as this ordering $\preceq$ of $[0,1)\times\Bbb Z$ under the correspondence that matches $\langle\alpha,n\rangle$ with the real number $n+\alpha$; the inverse of this bijection is the one that takes the real number $x$ to the ordered pair $\langle\{x\},[n]\rangle$.