# "Compactness length" of Baire space

Intuitively, my question is: how many times do we have to mod out by an closed equivalence relation with all classes compact in order to collapse Baire space $$\omega^\omega$$ to a singleton?

In more detail, given a space $$\mathcal{X}$$ let $$\mathsf{CoLe}(\mathcal{X})$$ be the least $$\theta$$ such that there is a sequence of spaces $$(\mathcal{X}_\eta)_{\eta<\theta+1}$$ where

• $$\mathcal{X}_0=\mathcal{X}$$,

• each $$\mathcal{X}_{\eta+1}$$ is the quotient of $$\mathcal{X}_\eta$$ by a closed equivalence relation each of whose classes is compact in the sense of $$\mathcal{X}_\eta$$,

• for $$\lambda<\theta$$ a limit ordinal, the space $$\mathcal{X}_\lambda$$ is the colimit of the family of $$\mathcal{X}_\eta$$s with $$\eta<\lambda$$, and

• $$\mathcal{X}_\theta$$ is a singleton.

For example, if we use $$\mathbb{R}$$ instead of $$\omega^\omega$$ then the corresponding ordinal is $$\omega$$: at stage $$n$$ we can collapse $$[-n,n]$$ to a point, and at stage $$\omega$$ this gives us the one-element space. Similarly, by collapsing just a pair of points at a time, we clearly have an upper bound of $$\mathfrak{c}$$, and it's not hard to show that $$\omega_1$$ is a lower bound for $$\mathsf{CoLe}(\omega^\omega)$$ (essentially this is an elaboration on the non-$$\sigma$$-compactness of $$\omega^\omega$$).

However, beyond that things aren't clear to me. In particular:

Is it consistent with $$\mathsf{ZFC}$$ that $$\mathsf{CoLe}(\omega^\omega)<\mathfrak{c}$$?

Baire space is the union of $$\mathfrak d$$ (the dominating number) compact subsets. So, using equivalence relations that collapse those sets one at a time (i.e., one equivalence class is the set to be collapsed and the other equivalence classes are singletons), we can collapse Baire space in at most $$\mathfrak d$$ steps. In particular, this can consistently be $$\aleph_1$$ steps while the continuum is large.