How to prove that the space $ \omega_1\times R $ has countable extent? 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:)
 A: 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 $\omega_1 \times R$.
