If $(X,<)$ is an uncountable well-ordered set whose sections are countable, $f:X\to X,f(x)For an uncountable set $X$, with a well-ordering $<$ (i.e. every subset of $X$ has a minimum element), and a property that for any $x\in X$, $\{y\in X|y<x\}$ is countable, we have a function $f:X\to X$ such that $f(x)<x$ for all $x\neq\min(X)$. I need to prove that $\exists x\in X$ such that $f^{-1}(x)$ is uncountable.
Here's what I think. 
First, we have the property that every countable subset of $X$ must be bounded. (assuming countable union of countable sets is countable)
Then, for any $x\in X$, there exists a minimum element $x'\in X,x'>x$, by the well-ordering.
Then, the set $X'=\{x\in X|\text{exists a maximum }x'\in X, x'<x\}$ is unbounded, and thus uncountable.
Then, $X',<$ satisfies all hypothesis but we can find an injection from $X'\to X'$ i.e. $f(x)=$ maximum element $x'\in X'$ such that $x'<x$. This contradicts to that $\exists x\in X',f^{-1}(x)$ is uncountable.
Is there anything wrong with this argument? Or what is the correct approach. Any help is appreciated.
 A: Your problem is that even though $X'$ is uncountable, given $x\in X'$ there is no guarantee that the maximum $x'<x$ is also in $X'$. In fact, it is not in many cases. An $x\in X\setminus X'$ is either the least element of $X$ or what we call a limit ordinal. The set of limit ordinals is also uncountable. To be explicit: The successor of any limit ordinal is an element of $X'$, but its predecessor is a limit, so it is not in $X'$. The function $f$ you defined is certainly $1$-$1$, but its range is not $X'$. It is, in fact, all of $X$, so this is not a counterexample. 
The result you state in the first paragraph is correct. The modern approach actually establishes more, namely that for some $x$, $f^{-1}(x)$ is what we call stationary. This is stronger than simply being uncountable. This version is called Fodor's lemma or, sometimes, the "pressing-down" lemma. The version you state was proved earlier, and the argument for it is somewhat simpler. 
See here for some references on the history of the result and its terminology.
A: First note that the condition that each section is countable is unnecessary. For each x consider the set {f(x),f(f(x)), ...} then this set has a least element, in fact it is 0 the least element of X, (and it is also finite). Set $ A_{0}=\{{0}\}; A_{n+1}=f^{-1}(A_{n}) $ .   Consider U$A_{n} $. On the one hand it is countable if $f^{-1}(x)$ is, on the other hand it is all of X a contradiction.  ER
