Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
add comment

1 Answer 1

up vote 1 down vote accepted

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 $b$.)

share|improve this answer
    
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
1  
@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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.