If $f:\omega_1 \to \mathbb{R}$ is continuous, then there is $a \in \omega_1$ such that if $x\geq a$, then $f(x) = f(a)$.

Question

Show that if $f:\omega_1 \to \mathbb{R}$ is continuous, then there is $a \in \omega_1$ such that if $x\geq a$, then $f(x) = f(a)$.

I'm stuck on this question for a while, although it seems to me that the proof will be related to the fact that $f$ is bounded (I proved this as folows: $f$ is continuous and $w_1$ has the least element of the set of its limit point, say $a$, then $f(a)$ should be the upper bound of $f:\omega_1 \rightarrow \mathbb{R}$. Is this correct?). Can anyone please help on this problem?

Answer 1

Say that a closed interval $[a, b]$ is recurrent if for cofinally many $\alpha\in\omega_1$, we have $f(\alpha)\in (a, b)$. Then we can show a few things:

• If $[a, b]$ is recurrent, then $f$ eventually stays in $[a, b]$: for some $\alpha\in\omega_1$ we have $\forall \beta>\alpha, f(\beta)\in [a, b]$

• For any $\epsilon>0$, there is some closed interval of size $\epsilon$ which is recurrent.

Once we prove these two facts, we then (roughly) can argue as follows:

• Get a sequence $[a_1, b_1], [a_2, b_2], [a_3, b_3], . . . $ of recurrent intervals, $[a_n, b_n]$ of size $2^{-n}$; along with them, get ordinals $\alpha_1, \alpha_2, \alpha_3, . . . $ such that $f(\beta)\in [a_n, b_n]$ for all $\beta>\alpha_n$.

• Now take the intersection of all those closed intervals - this is some real number, $\{r\}$. Meanwhile, the $\alpha_i$s are bounded . . .