[proofplan]
We prove that every dominating family in $\mathbb{N}^{\mathbb{N}}$ is automatically unbounded. If a dominating family $D$ had a single eventual upper bound $g$, then domination would produce an element $h \in D$ eventually above the shifted function $g+1$. For sufficiently large indices, this would force $g(n)+1 \leq h(n) \leq g(n)$, an impossibility in $\mathbb{N}$. Hence any family witnessing $\mathfrak{d}$ also witnesses unboundedness, so the least size of an unbounded family is at most the least size of a dominating family.
[/proofplan]
[step:Show that a dominating family cannot have a single eventual upper bound]
Let $D \subset \mathbb{N}^{\mathbb{N}}$ be a dominating family. We prove that $D$ is unbounded.
Suppose, for contradiction, that $D$ is bounded in $(\mathbb{N}^{\mathbb{N}}, \leq^*)$. Then there exists a function $g \in \mathbb{N}^{\mathbb{N}}$ such that $h \leq^* g$ for every $h \in D$.
Define the shifted function $s \in \mathbb{N}^{\mathbb{N}}$ by
\begin{align*}
s: \mathbb{N} \to \mathbb{N}, \quad n \mapsto g(n)+1.
\end{align*}
Since $D$ is dominating, there exists $h \in D$ such that $s \leq^* h$. Since $g$ bounds every element of $D$ and $h \in D$, we also have $h \leq^* g$. Therefore there exist $N_1, N_2 \in \mathbb{N}$ such that $s(n) \leq h(n)$ for every $n \geq N_1$, and $h(n) \leq g(n)$ for every $n \geq N_2$.
Let $N = \max\{N_1, N_2\}$. Then for every $n \geq N$,
\begin{align*}
g(n)+1 = s(n) \leq h(n) \leq g(n).
\end{align*}
This is impossible because $g(n)+1 > g(n)$ in $\mathbb{N}$. Hence no such eventual upper bound $g$ exists, and $D$ is unbounded.
[guided]
Let $D \subset \mathbb{N}^{\mathbb{N}}$ be a dominating family. To prove that $D$ is unbounded, we must show that there is no single function $g \in \mathbb{N}^{\mathbb{N}}$ that eventually dominates every element of $D$.
Assume, toward a contradiction, that such a bound exists. Thus there is a function $g \in \mathbb{N}^{\mathbb{N}}$ such that for every $h \in D$, we have $h \leq^* g$. By the definition of $\leq^*$, this means that each $h \in D$ is eventually below $g$.
Now define a new function $s \in \mathbb{N}^{\mathbb{N}}$ by
\begin{align*}
s: \mathbb{N} \to \mathbb{N}, \quad n \mapsto g(n)+1.
\end{align*}
The point of shifting $g$ upward by $1$ is that $s$ is eventually strictly larger than $g$ at every coordinate, so it cannot be squeezed below $g$ for all sufficiently large $n$.
Because $D$ is dominating, it must eventually dominate this particular function $s$. Therefore there exists $h \in D$ such that $s \leq^* h$. By the definition of eventual domination, there exists $N_1 \in \mathbb{N}$ such that $s(n) \leq h(n)$ for every $n \geq N_1$.
On the other hand, since $g$ was assumed to bound every member of $D$ and since $h \in D$, we also have $h \leq^* g$. Thus there exists $N_2 \in \mathbb{N}$ such that $h(n) \leq g(n)$ for every $n \geq N_2$.
Let $N = \max\{N_1, N_2\}$. Then both eventual inequalities hold simultaneously for every $n \geq N$. Hence, for every such $n$,
\begin{align*}
g(n)+1 = s(n) \leq h(n) \leq g(n).
\end{align*}
This contradicts the strict inequality $g(n)+1 > g(n)$ in $\mathbb{N}$. Therefore the assumed bound $g$ cannot exist. So every dominating family $D \subset \mathbb{N}^{\mathbb{N}}$ is unbounded.
[/guided]
[/step]
[step:Compare the least possible cardinalities]
By the previous step, every dominating family in $(\mathbb{N}^{\mathbb{N}}, \leq^*)$ is an unbounded family. Let $D_0 \subset \mathbb{N}^{\mathbb{N}}$ be a dominating family with $|D_0| = \mathfrak{d}$, whose existence is part of the definition of $\mathfrak{d}$ as the least cardinality of a dominating family. Since $D_0$ is unbounded, its cardinality is an admissible cardinality in the definition of $\mathfrak{b}$. Therefore
\begin{align*}
\mathfrak{b} \leq |D_0| = \mathfrak{d}.
\end{align*}
This proves $\mathfrak{b} \leq \mathfrak{d}$.
[/step]
