Ordering on the reals for which every element has a successor and a predecessor 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.
 A: 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 $\langle\{x\},[n]\rangle$.
A: Elements of the reals do not have a predecessor or successor in the usual order, which is $\gt$. The author has defined a different order. Under that order elements have a predecessor and successor. This order is not compatible with addition/multiplication like the usual one, but it is a fine total order on the reals
A: There is nothing wrong with that. Probably the sentence you read was about $(\mathbb{R},<)$ (the usual ordering)
