Why is cf(α) a cardinal for any limit ordinal α?

Question

In Jech, one of the lemmas state that for every limit ordinal α, cf(α) is a regular cardinal. Every source I tried to search on the Internet claimed that it was obvious to see that cf(α) should be an initial ordinal (a cardinal), but I'm having trouble understanding why this is the case. This is what I have so far: Suppose that $cf(α) = β$ is not a cardinal, then there exists some $γ < β$ such that $|γ|=|β|$, namely $|β|$.

Here, Jech states that using a mapping from $|β|$ onto β, one can construct a cofinal sequence in β of length $\leq |β|$. This is the part that I do not understand, as in my head, knowing that $|β|$ bijects with β means we have a bijection that is not necessarily order preserving, and does not yield any increasing functions from $|β|$ to β that would allow us to construct the cofinal sequence.

Referencing another SE post, Why cofinality is a cardinal., I understand that

From a bijection f between κ=|α| and α take S={β∈κ: f(γ)<f(β) for all γ<β} then if ξ is >the order type of S and g is the isomorphism from ξ to S the function f∘g is a cofinal >sequence in α and ξ<α.

However, why are we certain that there are at least a limit ordinals number of terms in S to allow us to construct the cofinal sequence? Is it not possible, for example, for the bijection between $|β|$ and β to be such that S only contains one element?

How would one begin to go about constructing such a cofinal sequence? Furthermore, does this imply that given any two ordinals α and β such that $α<β$ and $|α|=|β|$, we can construct a cofinal sequence in β of length $\leq α$?

Any help would be greatly appreciated!

Edit: Was confused about how f: |α| -> α maps to elements in α and there would not be f(β) = α for any β. I get the construction now, thank you!

Answer 1

So, you want to prove that $\mathrm{cf}(\alpha)\le |\alpha|$ for any ordinal $\alpha$.

Take a surjective function $f:|\alpha|→\alpha$. Define now, for all ordinals $\eta<|\alpha|$:

$$g(\eta):=\max(f(\eta),\sup\{g(\xi)+1.\ \xi<\eta\}).$$

So basically, $g$ is defined in a recursive way but the important things are:

1. $g(\eta)\ge f(\eta)$ for all $\eta<|\alpha|$;
2. $g$ is strictly increasing.

Now, $g$ is an order isomorphism from $|\alpha|$ into its image. It should be easy now to see that using this $g$ we can construct a cofinal sequence in $\alpha$ of size $|\alpha|$, since $g$ is strictly increasing but also $f$ was surjective on $\alpha$.