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I'm reading Kunen's Set Theory, and I'm stuck on his proof that least size of a dominating family of $\omega^\omega$, denoted by $\mathfrak{b}$, is less than or equal to the least size of a maximal almost disjoint family of subsets of $\omega$, denoted by $\mathfrak{a}$.

He starts off by saying that to see $\mathfrak{b}\le \mathfrak{a}$, it suffices to show

\begin{equation} \forall \kappa(\kappa<\mathfrak{b}\rightarrow\kappa<\mathfrak{a})\tag{1} \end{equation}

where $\kappa$ denotes an infinite cardinal. Up to here, there's no significant leap in logic. But then he states:

[...] so fix an infinite $\kappa<\mathfrak{b}$ and fix an almost disjoint family $\mathcal{D}=\{X_\alpha:\alpha<\kappa\}$; we shall show that $\mathcal{D}$ is not maximal.

I don't see why this implies $\kappa <\alpha$. By taking $\kappa<\mathfrak{b}$, he can ensure that any family of $\kappa$-many functions is not unbounded (by minimality of $\mathfrak{b}$), but how can you conclude that the non-existence of a mad family of size $\kappa$ implies $\kappa<\mathfrak{a}$. We have the conditional:

$$\forall \kappa (\kappa<\mathfrak{a}\rightarrow \text{any $\mathcal{E}\subset\mathcal{P}(\omega)$ with $|\mathcal{E}|=\kappa$ is not mad})$$

so it seems he is reverting it; why is this licit? It would suffice to show $\kappa\ge \mathfrak{a}$ implies there is a mad family of size $\kappa$, but I don't know if this is true, or how to prove it.

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    The non existence of a mad family of size $\kappa$ does not imply $\kappa < \mathfrak{a}$, you're right there. But we are showing that for any $\kappa < \mathfrak{b}$ there is no mad family of size $\kappa$, so no $\kappa < \mathfrak{b}$ can be the least size of a mad family, thus $\mathfrak{a} \geq \mathfrak{b}$. – 2017-01-31
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    @Jonathan That should be an answer! – 2017-01-31
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    @Jonathan Aha, got it know, thank you very much! Just to clarify, he's not really showing that if $\kappa<\mathfrak{b}$ then $\kappa<\mathfrak{a}$, but rather applying the contrapositive of the conditional he proves to directly obtain $\mathfrak{b}\le\mathfrak{a}$. Also, I agree with Noah Schweber, if you post your comment as an answer, I'll be happy to accept it. – 2017-01-31

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The non existence of a mad family of size $\kappa$ does not imply $\kappa < \mathfrak{a}$, you're right there, namely you can have holes in the spectrum of sizes of mad families. But we are showing that for any $\kappa<\mathfrak{b}$ there is no mad family of size $\kappa$, so no $\kappa<\mathfrak{b}$ can be the least size of a mad family, thus $\mathfrak{a} \geq \mathfrak{b}$.