If this is well-defined, I would honestly be surprised. But hey, if it is, new biggest number, I guess!
I use \(\mathbb{N}^{\mathbb{N}}\) to denote the set of functions \(\mathbb{N} \to \mathbb{N}\).
For a function \(f\), I define \(\text{dod}(f) := \{x: f(x) \text{ is defined}\}\). I define \(\text{img}(f) = \{f(x): x \in \text{dod}(f)\}\). Let \(f \leq^* g \leftrightarrow \exists N \forall m (m \geq n \Rightarrow f(m) \leq g(m))\). Furthermore, \(f \preceq g \leftrightarrow f \leq^* g \lor \min\{x: f(x) < g(x)\} > \min\{x: g(x) < f(x)\}\).
Define \(\mathit{L} \subseteq \mathbb{N}^{\mathbb{N}}\) like so:
A function \(f\) is an element of \(\mathit{L}\) iff the following hold:
- \(\text{dod}(f) \subseteq \mathbb{N}\).
- \(\text{img}(f) \subseteq \mathbb{N}\).
- For all \((x, y) \in \text{dod}(f)^2\), \(x < y \Rightarrow f(x) < f(y)\).
- For all \(x \in \text{dod}(f)\), \(x \leq f(x)\).
Then, \(F^{\mathit{L}}\) is a function \(\mathit{L} \to \mathbb{R}\) so that:
- For all \((x, y) \in \mathit{L}^2\), \(\neg (x \leq^* y \lor y \leq^* x) \Rightarrow F^{\mathit{L}}(x) = F^{\mathit{L}}(y)\).
- For all \((x, y) \in \mathit{L}^2\), \((x \leq^* y \land x \neq y) \Rightarrow F^{\mathit{L}}(x) < F^{\mathit{L}}(y)\).
Obviously, the \(x = y \Rightarrow F^{\mathit{L}}(x) = F^{\mathit{L}}(y)\) is implicit.
I define the function \(F^{\mathbb{R}}: \mathbb{R} \to \mathit{L}\), where \(F^{\mathbb{R}}(x)\) is the \(\preceq\)-minimal function so that \(F^{\mathit{L}}(F^{\mathbb{R}}(x)) = x\).
Then, \(\Lambda(n) := F^{\mathbb{R}}(n)(n)\).
If this is well-defined, then I'm pretty sure that \(\Lambda\) eventually dominates itself.
My number is \(\Lambda^{10}(10^{100})\).