"Hyper-pigeonhole principle": size of tail-avoiding sequence families I'm working on a problem in complex analysis and I wanted to check my logic on a specific point.
Let $A$ be an infinite set with cardinality $\aleph^A := |A|$ and initial ordinal $\omega^A$.
By abuse of notation, let there be given some bijection $A \leftrightarrow \omega^A$ so that $A$ is considered to have the same ordering as $\omega^A$.
I want to verify my proof of the

Hyper-pigeonhole principle. Let $A^A$ be the set of functions from $A$ to itself, and let $B \subseteq A^A$ have cardinality strictly greater than $A$. Then there exist $f, g \in B$ so that the set $$\{ a \in A: f(a) = g(a) \}$$ is cofinal in $A$.

Proof attempt. We argue by contraposition that, if there are no such $f, g$ in $B$, that $|B| \leq |A|$. Suppose $F \subseteq A^A$ has the property that, for any $f, g \in A^A$, there exists $m \in A$ s.t. $f(x) \neq g(x)$ whenever $x > m$. By assumption, $$F := \cup_{m \in A} F_m,$$ where $$F_m := \{ f \in F: f(x) \neq g(x) \forall x > m, \forall g \neq f \in F \}.$$ But if we let $x = m+1$, then $f(x)$ is an injection of $F_m$ into $A$, by assumption, and so $|F_m| \leq |A|$. Therefore $F$ is a union of $|A|$-many sets, each with cardinality at most $|A|$, which means $|F| = |A|$ since for infinite sets $A$, $A^2 \cong A$ by the axiom of choice.
 A: Never mind, this is totally wrong.
Easy counterexample: let $\Bbb{N}$ be the set of all natural numbers starting with $0$, let $\Bbb{N}^\Bbb{N}$ be the set of all functions from $\Bbb{N}$ to itself, and let $\mathbf{2}^\Bbb{N}$ be the set of all countably infinite binary sequences.
Then I can actually get an injection $$ɪ:\mathbf{2}^\Bbb{N} \hookrightarrow \Bbb{N}^\Bbb{N}$$ such that $ɪ(\mathbf{x})(n) \neq ɪ(\mathbf{y})(n)$ for all but finitely many $n \in \Bbb{N}$ and all $\mathbf{x}, \mathbf{y} \in \mathbf{2}^\Bbb{N}$ such that $\mathbf{x} \neq \mathbf{y}$.
For any sequence $\mathbf{x} = \{ x_k \}_{k \geq 0} \in \mathbf{2}^\Bbb{N}$, we may define $$ɪ(\mathbf{x}) := \{x_0, x_0 + 2x_1, x_0 + 2x_1 + 4x_2, ...\} = \left\{ \sum_{j=0}^k x_j 2^j \right\}_{k \geq 0} \in \Bbb{N}^\Bbb{N};$$ then clearly if $\mathbf{y} = \{ y_k \}_{k \geq 0} \in \mathbf{2}^\Bbb{N}$ is any other binary sequence, $ɪ(\mathbf{x})(n) \neq ɪ(\mathbf{y})(n)$ for any $n \geq d$, where $d$ is the first element in the sequence for which $x_d \neq y_d$, because one of $ɪ(\mathbf{x})(n), ɪ(\mathbf{y})(n)$ will have $2^d$ in its binary expansion for all $n \geq d$ and the other won't.
So we even have $\mathfrak{c} = 2^{\aleph_0}$-many functions from $\Bbb{N}$ to itself so that any two distinct functions only have finitely many elements in common, and every input only has finitely many possible outputs!
