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.
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10It is not an error; it is a correct inference. By definition, any element of a well-ordered set is either a maximum or a successor. However every element of the reals is neither a maximum nor a successor. Therefore... the reals are not a well-ordered set. $\blacksquare$ The reals are ordered; just not well-ordered. – Graham Kemp Jul 21 '16 at 03:14
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4Where when @Graham says "the reals" he means not the set of real numbers, but the ordered set of real numbers equipped with the usual ordering. – Jul 21 '16 at 06:56
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1The usual ordering is not a well-ordering. No-one has defined a well-ordering of the Reals yet. – fleablood Jul 21 '16 at 08:37
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4@fleablood: I don't believe that's true: I'm pretty sure that people have defined well-orderings of the reals. What happens is that additional hypotheses are required (e.g. the axiom of constructibility) to prove that the definition gives a total ordering. – Jul 21 '16 at 08:44
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@Hurkyl You are quite right and http://math.stackexchange.com/a/6504/176997 gives a great exposition. Favorite points on comments should really count towards something. Good comments are often as important as answers. – DRF Jul 21 '16 at 08:51
2 Answers
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.
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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 $\mathbb Z\subset\mathbb R$.
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