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X is an uncountable infinite well ordered set. I need to show that the set of the limit members in X is uncountable.

A limit member is a member of X that is not a successor of any other member in X and not the first number in X

Here are my thoughts- maybe I can show (I'm not sure that the following statement even true) that between a limit member and the next limit member in X there are only countable many elements so it's impossible to have a countable number of limit members because it will contradict the fact that X is uncountable.

I would appreciate it if I can get some help with that. Thanks.

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HINT: Your intuition is correct, but it’s probably not the easiest way to attack the problem. Let $X$ be an uncountable set well-ordered by $\le$. For each $x\in X$ let $P(x)=\{y\in X:y\le x\}$, the set of non-strict predecessors of $x$. Let $U=\{x\in X:P(x)\text{ is uncountable}\}$; if $U\ne\varnothing$, let $u=\min U$, and let $X_0=P(u)\setminus\{u\}$; otherwise, let $X_0=X$. In either case $X_0$ is an uncountable set well-ordered by $\le$ such that $P(x)$ is countable for each $x\in X_0$. Let $$L=\{x\in X_0:x\text{ is a limit element}\}\;;$$ you’re done if you can show that $L$ is uncountable.

Suppose that $L$ is countable. Let $A=\bigcup_{x\in L}P(x)$; $A$ is countable, $L\subseteq A$, and if $y<x\in A$, then $y\in A$. That is, $A$ is a countable initial segment of $X_0$ containing all of the limit elements of $X_0$. (These statements need to be justified, of course.) Let $b=\min(X_0\setminus A)$, and get a contradiction by finding a limit element greater than $b$.

(If $L$ has no largest element, you can actually get the contradiction by showing that $b$ is a limit element, but if $L$ has a largest element, $b$ will be its successor, and you really will have to find a limit element strictly greater than $b$.)

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Thanks. Could you kindly explain 2 things: 1) how can we know that L isn't empty? 2) I'm not sure what to do if L has a largest element.. – user18217 Jan 15 '13 at 21:50
@user18217: The case $L=\varnothing$ requires no special handling: in that case $b=\min X_0$, and in all cases you simply have to show that there is a limit element greater than $b$. Let $A_0=\{x\in X_0:b<x\}$; $A_0$ is uncountable, so it’s certainly non-empty, and we can let $a_0=\min A_0$. Let $A_1=A_0\setminus\{a_0\}$; the same reasoning ensures that we can let $a_1=\min A_1$. Recursively construct $a_n$ for $n\in\Bbb N$. $A\cup\{a_n:n\in\Bbb N\}$ is countable, so there is a least element $c$ of $X_0$ greater than all elements of $A\cup\{a_n:n\in\Bbb N\}$; show that $c$ is a limit element. – Brian M. Scott Jan 15 '13 at 21:57
@user18217: What I’m doing in that last comment is just taking successors repeatedly: $a_{n+1}$ is the successor of $a_n$ in the well-order. – Brian M. Scott Jan 15 '13 at 21:59
@Brain M.Scott: your solution gave me an idea of a much simpler solution, using the idea that I suggested in the beginning: it's easy to show that if a_0 is a limit element and you add to it an infinity series of elements {a_i}_i (as you did in the construction of a_i as the min of A_(i-1)\{A_(i-1)})then the next one will be a limit element as well and so if there are only countable number of limit elements X is countable as well – user18217 Jan 17 '13 at 7:46
@user18217: Making that rigorous takes quite a bit of work $-$ much more than just using it to show that given any $x\in X_0$ there is a limit element greater than $x$. – Brian M. Scott Jan 17 '13 at 7:52

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