A tower of size at most the continuum exists

By [F2], $[\omega {]}^{\omega }$ is well-orderable. Let $\delta$ be its initial cardinal $\mid [\omega {]}^{\omega }\mid$ and choose a bijection $\alpha \mapsto X_{\alpha }$ from $\delta$ onto $[\omega {]}^{\omega }$. This induces a particular well-order of the latter set, so "least" below refers to this enumeration. Since $[\omega {]}^{\omega }\subseteq \mathcal{P}(\omega )$, [F4] gives $\delta \le \mid \mathcal{P}(\omega )\mid =2^{{\mathrm{\aleph }}_0}=\mathfrak{c}$. An arbitrary well-order of $[\omega {]}^{\omega }$ could have order type larger than $\delta$ and would not justify this bound.

For $Y\in [\omega {]}^{\omega }$ let $y_0<y_1<y_2<\cdots$ be the increasing enumeration of $Y$, which exists by [F5] applied to the well-ordered set $Y$; put $E(Y)=\{y_{2n}:n\in \mathbb{N}\}$ and $O(Y)=\{y_{2n+1}:n\in \mathbb{N}\}$. Then $E(Y)$ and $O(Y)$ are infinite, disjoint, $E(Y)\cup O(Y)=Y$, and $E(Y)\subseteq Y$, $O(Y)\subseteq Y$. Consequently for every $X\in [\omega {]}^{\omega }$ at least one of $X\mathrm{̸}{\subseteq }^{\ast }E(Y)$, $X\mathrm{̸}{\subseteq }^{\ast }O(Y)$ holds: if both $X\setminus E(Y)$ and $X\setminus O(Y)$ were finite, then $X\setminus (E(Y)\cap O(Y))=X\setminus \mathrm{\varnothing }=X$ would be finite, contradicting $X\in [\omega {]}^{\omega }$.

Define, by transfinite recursion on $\delta$ ([F3]), values $A_{\alpha },Y_{\alpha }\subseteq \omega$ for $\alpha <\delta$, using the sentinel value $\mathrm{\varnothing }$ for $Y$. Say that $\alpha <\delta$ is free when $Y_{\beta }\mathrm{\ne }\mathrm{\varnothing }$ for every $\beta <\alpha$. For a free $\alpha$ let ${\mathcal{P}}_{\alpha }=\{Y\in [\omega {]}^{\omega }:Y{\subseteq }^{\ast }{A}_{\beta }$ for every $\beta <\alpha \}$. If ${\mathcal{P}}_{\alpha }=\mathrm{\varnothing }$, set $A_{\alpha }=\omega$ and $Y_{\alpha }=\mathrm{\varnothing }$; if $Y_{\alpha }=\mathrm{\varnothing }$ we set $A_{\gamma }=\omega$ and $Y_{\gamma }=\mathrm{\varnothing }$ at every later $\gamma$, so that the recursion is total on $\delta$. If ${\mathcal{P}}_{\alpha }\mathrm{\ne }\mathrm{\varnothing }$, let $Y_{\alpha }$ be the well-order-least member of ${\mathcal{P}}_{\alpha }$ and split it as in step 1.2, setting $A_{\alpha }=E(Y_{\alpha })$ if $X_{\alpha }\mathrm{̸}{\subseteq }^{\ast }E(Y_{\alpha })$, and $A_{\alpha }=O(Y_{\alpha })$ otherwise; by step 1.2 one of the two cases applies, so $A_{\alpha }$ is a well-defined infinite subset of $Y_{\alpha }$ with $X_{\alpha }\mathrm{̸}{\subseteq }^{\ast }{A}_{\alpha }$.

Let $\lambda$ be the least $\alpha \le \delta$ such that either $\alpha =\delta$, or $\alpha <\delta$ and $Y_{\alpha }=\mathrm{\varnothing }$. Such a $\lambda$ exists because $\delta$ is an ordinal and the second alternative is decided for each $\alpha <\delta$; and $\lambda >0$, since ${\mathcal{P}}_0=[\omega {]}^{\omega }\mathrm{\ne }\mathrm{\varnothing }$ and hence $Y_0\mathrm{\ne }\mathrm{\varnothing }$ by step 2.1 and the definition of ${\mathcal{P}}_0$.

For every $\alpha <\lambda$ the stage $\alpha$ was free and ${\mathcal{P}}_{\alpha }\mathrm{\ne }\mathrm{\varnothing }$, so $A_{\alpha }\in [\omega {]}^{\omega }$, $Y_{\alpha }\in [\omega {]}^{\omega }$, $A_{\alpha }\subseteq Y_{\alpha }$, and $Y_{\alpha }{\subseteq }^{\ast }{A}_{\beta }$ for every $\beta <\alpha$; hence $A_{\alpha }{\subseteq }^{\ast }{A}_{\beta }$ for every $\beta <\alpha$, and the family $⟨A_{\beta }:\beta <\lambda ⟩$ is decreasing in the sense of [F1]. Moreover $X_{\alpha }\mathrm{̸}{\subseteq }^{\ast }{A}_{\alpha }$ for every $\alpha <\lambda$, and the family is a set, being the image of the ordinal $\lambda$ under a definable function.

If $\lambda <\delta$, then $Y_{\lambda }=\mathrm{\varnothing }$, which by step 2.1 and the minimality of $\lambda$ happened because ${\mathcal{P}}_{\lambda }=\mathrm{\varnothing }$: no $Y\in [\omega {]}^{\omega }$ satisfies $Y{\subseteq }^{\ast }{A}_{\beta }$ for all $\beta <\lambda$. By step 4.1 the family $⟨A_{\beta }:\beta <\lambda ⟩$ is a decreasing family of infinite sets with no pseudointersection, that is a tower, of length $\lambda <\delta \le \mathfrak{c}$.

If $\lambda =\delta$ and some $X\in [\omega {]}^{\omega }$ were a pseudointersection of $⟨A_{\alpha }:\alpha <\delta ⟩$, then $X=X_{\alpha }$ for some $\alpha <\delta$ by step 1.1, so $X_{\alpha }{\subseteq }^{\ast }{A}_{\alpha }$; but step 4.1 gives $X_{\alpha }\mathrm{̸}{\subseteq }^{\ast }{A}_{\alpha }$, a contradiction. Hence $⟨A_{\alpha }:\alpha <\delta ⟩$ is a tower of length $\delta \le \mathfrak{c}$, again by step 4.1.

In the case $\lambda <\delta$ step 5.1 exhibits a tower of length $\lambda <\delta \le \mathfrak{c}$, and in the case $\lambda =\delta$ step 5.2 exhibits a tower of length $\delta \le \mathfrak{c}$; in both cases the length is at most $\mathfrak{c}=2^{{\mathrm{\aleph }}_0}$. This is the statement. ∎