I'm trying to figure out where I'm going wrong with the following proof that all sets which have a well-ordering must be countable. At the heart of it is that every element in a well-ordered set has a unique predecessor (except the minimum element of the set) and a unique successor.
Below let $(\mathbb{N},\leq)$ be linear order on $\mathbb{N}$ and $(X,\preceq)$ be an infinite well-ordered set. Define the sequence $(a_n)_{n\in\mathbb{N}}$ as follows: $$a_1 := \min X \\ a_{n+1} := \text{the successor of } a_n \in (X,\preceq)$$
Claim: For each $x \in X$ there exists a unique $n$ such that $x = a_n$. Thus $X$ is countable. Proof:
- Uniqueness follows from the $a_n$ being a sequence of successors so no two are equal to each other. For existence, proof by contradiction. Suppose there exists an element $c\in X$ such that $c\neq a_n$ for any $n \in \mathbb{N}$.
- Since $X$ is well-ordered by $\preceq$ either (Case A): There exist $i,j\in\mathbb{N}$ such that $a_i\preceq c\preceq{a_j}$ or (Case B): $a_n \prec c \; \forall n\in\mathbb{N}$. Assume Case A first:
- Since each of the $a_i$ is the predecessor of $a_{i+1}$ and $j$ is finite, there are only finitely many $x$ such that $a_i\preceq x \preceq{a_j}$ (i.e. if $a_k\preceq x\preceq a_{k+1}$ then by the construction of $(a_n)$, either $x=a_k$ or $x=a_{k+1}$). Furthermore, each of these $x$ must be an $a_k$ for some $k,\; i\leq k\leq j$ because the construction of $(a_n)$ includes the whole sequence of successors $a_{i+1},\ldots, a_j$. Thus, we must have $c=a_k$ for some $i\leq k\leq j$, a contradiction.
- (Case B). Let $S:=\{ x\;|\; a_n \prec x, \; \forall n\in\mathbb{N} \}$. Without loss of generality let $c := \min S$
- Let $b$ be the predecessor of $c$. Then $b\notin S$ so by the definition of $S$ there exists an $m$ such that $b = a_m$. Then we must have $c=a_{m+1}$, a contradiction.
I've looked it over a few times already and can't find anything. Thanks.
The ordinal $\omega+1 = \{0,1,2,3,\ldots, \omega\}$ is a well-ordered set in which the element $\omega$ (the maximum) does not have a direct predecessor. So your basic assumptions are wrong from the start... Your procedure does not go beyond $\omega$, but limit ordinals exist.