# If $cf(\alpha)<cf(\beta)$, how to show that every increasing $h:\alpha \to \beta$ has a range that is bounded in $\beta$?

The problem
Let $\alpha$ and $\beta$ be two limit ordinals. Show that $1$. $\implies$ $2$., where

$\qquad 1$. $cf( \alpha)$ < $cf(\beta )$

$\qquad 2$. Every increasing $h: \alpha \to \beta$ has a range that is bounded in $\beta$ .

Note that we do not restrict $2$. to strictly increasing functions.
Does the implication hold if we do not even demand $h$ to be increasing?

My attempt at a solution
Let $\alpha$ and $\beta$ be two limit ordinals, and let $cf(\beta) = \beta_1$.

It seems to me that we must have $\alpha < \beta_1 = cf(\beta)$, because if $\alpha \geq \beta_1$ we would have $cf(\alpha) \geq cf(\beta_1) = \beta_1 = cf(\beta)$, contrary to our assumption $1$. So, since $\beta_1$ is the least ordinal such that there exist an $f:\beta_1 \to \beta$ that has a range that is unbounded in $\beta$ (i.e. $\forall b \in \beta \ \ \exists \, b_1 \in \beta_1$ s.t. $f(b_1) > b$), doesn't it follow that any function from any smaller ordinal than $\beta_1$ would have a range that would be bounded in $\beta$ (in particular including the $h$ in $2$.)?

I feel like I must have overlooked something since I don't use the fact that $h$ is increasing. If anyone could explain where I went wrong (if it is wrong, that is) I would be very grateful!

Some definitions

• If $f:\alpha \to \beta$, $f$ maps $\alpha$ cofinally iff $\text{ran}(f)$ is unbounded in $\beta$.
• The cofinality of $\beta$ ($cf(\beta)$), is the least $\alpha \$ s.t. there is a map from $\alpha$ cofinally into $\beta$.
• $\operatorname{cf}\alpha<\operatorname{cf}\beta$ does not imply $\alpha < \operatorname{cf}\beta$. For example, consider $\alpha=\omega_\omega=\sup_{n<\omega}\omega_n$ and $\beta=\omega_1$. If we assume the choice, then $\operatorname{cf}\omega_1=\omega_1$ but $\operatorname{cf}\omega_\omega=\omega$. Commented Mar 19, 2014 at 11:14
• @tetori: You are, of course, absolutely right. Cofinality still confuses me, thanks for the example :) Commented Mar 20, 2014 at 15:54

consider any $\beta$ with $cf(\beta)>\omega$. Set $\alpha=\beta+\omega$. We have $cf(\alpha)=\omega<cf(\beta)$ by construction. But you can get $h:\alpha\to\beta$ unbounded by mapping the "tail" $\omega$ to zero.
As for $1\Rightarrow 2$ if there is $h:\alpha\to\beta$ unbounded, since there is an unbounded $f:cf(\alpha)\to\alpha$, we get an unbounded $h\circ f:cf(\alpha)\to\beta$. Which forces $cf(\beta)\leq cf(\alpha)$.
• b.t.w. your mistake is concluding that $\alpha<cf(\beta)$ as the example $\alpha=\beta+\omega$ shows Commented Mar 19, 2014 at 11:14