How Can You Define Successors Over The reals? I've been going through Introduction To Modern Set Theory by Judith Roitman, and am confused by her exposition of well orderings. She gives the following proof that every element $x$ of a well-ordered set $X$ is a maximum or has a successor: $$\text{If}\ x\ \text{is not maximum, then}\ {S=\{y \in X\ |\ y > x\} \neq \emptyset}\ \\ \text{The minimum of }S \text{ is the successor of }x.$$ 
I hit a cognitive roadblock when thinking about such a minimum element in a dense field like the reals. She defines $x$ to be a limit in a well-order $X$ iff $$\forall y \in X, x \neq S(y)$$that is, $x$ is a limit if it is not the successor of some element $y$. She says that every element of $\mathbb{R}$ is a limit, which agrees with my intuition. What I do not understand, specifically, if why the corresponding condition of $\forall y \in X, y \neq S(x)$ does not hold for $\mathbb{R}$ with the usual ordering. If that held, it seems to me to go against the claim that every element of a well-ordered set is a maximum or has a successor (since every element of $\mathbb{R}$ would be neither maximum nor have a successor). Any help in getting around whatever cognitive error it is that I'm making would be much appreciated. 
 A: It seems like what you're missing is that the usual order on $\mathbb{R}$ is not a well-ordering.  So there is no problem with the fact that no element of $\mathbb{R}$ has a successor.
A: By definition, '$x$ is maximum element' in an ordered set $D$ with ordering relations $<$ iff there is no element in $D$ which is greater than $x$. So if $x$ is not maximal, then there exist elements (at least one) in $D$, which are greater than $x$. They form a subset $S$ of $D$. And that subset inherits the ordering relation $<$ from $D$. If we assume $D$ is well-ordered, so is $S$. And the main property of a 'well-order' is that each non-empty subset has a least element. Name it $q$.
As we chose $S$ to be a set of elements greater than $x$, its least element $q$ is greater than $x$ but it's smaller than all other elements of $S$. So there exists no element between $x$ and $q$, hence $q$ is a successor of $x$.
Note, however, the standard $<$ relations on $\mathbb R$ is not a well-order; there are subsets of $\mathbb R$ which have no least elements. In particular the $\mathbb R$ itself has no least element, similary any open interval $(a,b)$ or a set of integers $\mathbb Z\subset\mathbb R$.
