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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:)

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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
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
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 $\omega_1 \times R$.

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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) – arjafi Jul 1 '12 at 2:46
Now it's clear for me:) – Paul Jul 1 '12 at 10:12

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