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.

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 $\omega_1+1$.

thanks.

share|improve this question

1 Answer 1

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, $\omega_1+1=\beta\omega_1$.

share|improve this answer
    
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
3  
@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

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.