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We've been studying various properties of $\omega_1$ (equipped with the order topology), and I recently came across these questions. Can anyone help?

If $f$ maps $\omega_1$ onto a metric space $X$ and $f$ is continuous, why is $X$ compact? Also, why does there exist $\alpha \in \omega_1$ such that $f$ maps the interval $[0, \alpha]$ onto $X$?

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I suppose that you assume that $f$ is continuous - this should be added to the question. The space $\omega_1$ is considered with order topology? – Martin Sleziak May 23 '12 at 11:49
For the first part, note that $\omega_1$ is sequentially compact. – Chris Eagle May 23 '12 at 11:49
Yes, $f$ is continuous, and yes, $\omega_1$ is considered with the order topology. I will add these. – Maria May 23 '12 at 11:53
up vote 4 down vote accepted

As an ordinal space $\omega_1$ is sequentially compact, but not compact. If it maps onto a metric space then the metric space is also sequentially compact, however for metric spaces the notions of compactness are equivalent. So $X$ would have to be compact as well.

Since $X$ is a compact metric space it has to be separable, so we have a countable dense subset $D$. So there is some $\alpha<\omega_1$ such that $D$ is covered by $[0,\alpha]$. Since $\omega_1$ is sequentially compact all the limit points of $D$ must lie in the interval $[0,\alpha]$.

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Suppose that $f:\omega_1\to X$ is a continuous map of $\omega_1$ onto a metric space $X$ with metric $d$. A metric spaces is compact iff every infinite set has a cluster point, so if $X$ is not compact, it contains a closed, discrete set $C=\{x_n:n\in\omega\}$. For $n\in\omega$ let $\alpha_n\in\omega_1$ be such that $f(\alpha_n)=x_n$, and let $A=\{\alpha_n:n\in\omega\}$. $A$ is an infinite subset of $\omega_1$, so there is a strictly increasing sequence $\langle n_k:k\in\omega\rangle$ in $\omega$ such that $\langle \alpha_{n_k}:k\in\omega\rangle$ is strictly increasing in $\omega_1$. Let $\alpha=\sup\{\alpha_{n_k}:k\in\omega\}$; then $\langle \alpha_{n_k}:k\in\omega\rangle$ converges to $\alpha$ in $\omega_1$, so $\langle f(\alpha_{n_k}):k\in\omega\rangle=\langle x_{n_k}:k\in\omega\rangle$ converges to $f(\alpha)$ in $X$, contradicting the choice of $C$. Thus, $X$ must be compact.

Every compact metric space is separable, so let $D$ be a countable dense subset of $X$. Then there is a countable $A\subseteq\omega_1$ such that $f[A]=D$; let $\alpha=\sup A\in\omega_1$. For each $x\in X$ there is a sequence $\langle x_n:n\in\omega\rangle$ in $D$ converging to $x$. For $n\in\omega$ choose $\xi_n\in[0,\alpha]$ such that $f(\xi_n)=x_n$. As before, some subsequence $\langle\xi_{n_k}:k\in\omega\rangle$ must converge to some $\eta\le\alpha$, and hence $$\langle x_{n_k}:k\in\omega\rangle=\langle f(\xi_{n_k}):k\in\omega\rangle\to f(\eta)\;.$$ But $\langle x_{n_k}:k\in\omega\rangle\to x$, so $x=f(\eta)$, and $f$ maps $[0,\alpha]$ onto $X$.

You might like to try to prove the closely related fact that every continuous real-valued function on $\omega_1$ is constant on a tail of $\omega_1$. For each $\alpha\in\omega_1$ let $K_\alpha=f[\omega_1\setminus\alpha]$; the result above shows that $K_\alpha$ must be a compact subset of $\Bbb R$. Let $K=\bigcap_{\alpha\in\omega_1}K_\alpha$. Clearly $K$ is non-empty. Show that $K$ contains exactly one point, say $x$, and that $f(\xi)=x$ for all sufficiently large $\xi\in\omega_1$.

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