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.

How to proof that the space $ \omega_1\times R $ has countable extent? The topological space $\omega_1$ is the first uncountable ordinal with order topology.

A space $X$ has countable extent if every uncountable subset of $X$ has a limit point in $X$.

Thanks for any help:)

share|improve this question
    
Do you possibly mean $\omega_1$ is the first uncountable ordinal? $\omega + 1$ is certainly countable. –  Ben Millwood Jun 30 '12 at 11:49
    
Maybe I have made a silly mistake. Yes, because I can't type the $\omega_1$ (now I could); then the proof from Arthur Fischer can't answer the question. –  Paul Jul 1 '12 at 1:52
2  
Please indicate such a substantial edit clearly as such in the main body of your question. Now it looks as if Arthur made this rather silly mistake. (Why didn't you just ask a new question with $\omega+1$ replaced by $\omega_1$? Also: I find it a bit bizarre that you even accepted his answer and notice only much later that you had a completely different ordinal in mind...) –  t.b. Jul 1 '12 at 2:44

1 Answer 1

up vote 4 down vote accepted

By $R$ I assume you are denoting the real line.

(Oh, dear; the question seems to have been substantially altered. Please ignore the now silly sounding striked-out paragraph.)

Suppose that $A \subseteq (\omega + 1 ) \times R$ is uncountable. Note that there must be a $i \leq \omega$ such that $A_i = \{ x \in R : (i,x) \in A \}$ is uncountable. But as $R$ has countable extent it follows that $A_i$ has a limit point $x$ in $R$. It easily follows that $(i,x)$ is a limit point of $A$ in $(\omega +1 ) \times R$.

Let $A \subseteq \omega_1 \times R$ be uncountable. If there is an $\alpha < \omega_1$ such that $A_\alpha = \{ x \in R : (\alpha , x ) \in A \}$ is uncountable, then $A_\alpha$ has a limit point $x$ (as $R$ has countable extent), and it is easy to show that $(\alpha , x )$ is a limit point of $A$.

So assume that $A_\alpha$ is countable for each $\alpha < \omega_1$. We may then recursively construct a sequence $\langle (\alpha_i , x_i ) \rangle_{i \in \omega}$ in $A$ such that:

  • $\alpha _i < \alpha_{i+1}$ for all $i \in \omega$; and
  • $\langle x_i \rangle_{i \in \omega}$ is a convergent sequence in $R$.

Let $\alpha = \sup_{i \in \omega} \alpha_i < \omega_1$ (and note that $\alpha$ is a limit ordinal). Let $x = \lim_{i \in \omega} x_i$. It is easy to show that $( \alpha , x )$ is a limit point of $A$ in $\omega_1 \times R$.

share|improve this answer
    
Maybe I have made a silly mistake. Because I can't type the $\omega_1$ (now I could); In your proof, $\omega+1$ is seen as the successor of $\omega$, however, I mean that it is the first uncountable ordinal $\omega_1$. –  Paul Jul 1 '12 at 1:53
    
@John: Well, I thought that it was too easy.... (gosh darned it) –  Arthur Fischer Jul 1 '12 at 2:46
    
Now it's clear for me:) –  Paul Jul 1 '12 at 10:12

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.