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consider the space $ω_1$ (the first uncountable ordinal) and $ω_1+1$ together with the order topology. is it true that

  1. $ω_1$ is $G_\delta$-dense subspace of $ω_1+1$?
  2. $\omega_1$ is $C^*$-embedded in $\omega_1+1$?

If both proposition are correct, then we conclude that $\omega_1$ is $C$-embedded in $\omega_1+1$.


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up vote 4 down vote accepted

Every continuous $f:\omega_1\to\Bbb R$ is eventually constant, so $\omega_1$ is indeed $C$-embedded in $\omega_1+1$; you’ll find a proof of the first statement here in Dan Ma’s Topology Blog.

  1. Yes, this is true, since any $G_\delta$ in $\omega_1+1$ containing the point $\omega_1$ contains an interval $(\alpha,\omega_1]$ for some $\alpha<\omega_1$.

  2. This is also true; in fact, $\omega_1+1=\beta\omega_1$.

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eventually constant means there exist a $a\in \omega_1$ such that $f$ is constant on $(a,\omega_1)$ ? – TXC Feb 12 '13 at 7:35
@TXC: Yes, exactly. And the extension that assigns $\omega_1$ that constant value is clearly continuous. – Brian M. Scott Feb 12 '13 at 7:36
with use this property "every continuous $f:ω_1→R$ is eventually constant" i cant show that $f$ can be countinuously extended to $\omega_1+1$. for example, $g(\omega_1)=1$ and $g(x)=c$ for $x\in \omega_1$ ($c$ is constant value), can be continuous extension of $f$? – TXC Feb 12 '13 at 13:24
@TXC: If $f: \omega_1 \rightarrow \mathbb{R}$ is continuous, there exists $\alpha < \omega_1$ and a constant $c \in \mathbb{R}$ such that $f(\beta) = c$ for all $\beta >= \alpha$. Now define $\bar{f}: \rightarrow \mathbb{R}$ by $\bar{f}(\alpha) = f(\alpha)$ for $\alpha < \omega_1$ and $\bar{f}(\omega_1) = c$. Then $\bar{f}$ extends $f$ and is continuous at $\omega_1$ too, as every neighbourhood of it contains one that is contained in $[\alpha,\rightarrow)$. So the extension is of course unique and uses the same $c$ as promised in the eventually constant fact. – Henno Brandsma Feb 12 '13 at 14:47
Is it true that $\omega_1$ is only proper dense subspace of $\omega_1+1$? – TXC Feb 13 '13 at 6:13

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