Is the output of a Veblen function larger than its input?

Question

I am looking to prove that:

$$ \forall \gamma < \Gamma_0 . \forall \alpha < \Gamma_0 . \varphi_\alpha(\gamma) \geq \gamma $$

Where $\varphi$ is the Veblen function and $\Gamma_0$ is the Feferman–Schütte ordinal. Which is to say that the Veblen functions is at least it's input.

This seems true to me and I can't come up with any counter examples. I can prove some simple cases. But I also can't seem to prove it in general.

Answer 1

As BrianO brought up, every strictly increasing function $f:\textrm{Ord}\rightarrow\textrm{Ord}$ has this property (*), so every Veblen function $\varphi_\alpha$ (including $\alpha>\Gamma_0$ as well!) is strictly increasing.

Proof of (*): Assume towards a contradiction that there exists some $\gamma$ with $f(\gamma)<\gamma$. Due to the fact $f$ is strictly increasing, along with $f(\gamma)<\gamma$ this gives us some properties:

• For all $\gamma'<\gamma$, $f(\gamma')<\gamma$. (Otherwise $f(\gamma')\ge f(\gamma)$ contradicts $f$'s strictly increasingness)
• $f$ is injective. (Otherwise we'd have $\delta<\varepsilon\le\gamma$ such that $f(\delta)=f(\varepsilon)$, contradicting $f$'s strictly increasingness)

Since we have a strictly increasing (i.e. for $\delta,\varepsilon<\gamma$, $\delta<\varepsilon\leftrightarrow f(\delta)<f(\varepsilon)$) injection that maps $\gamma$ to $f(\gamma)$, we have an order-preserving injection from $\gamma$ to $f(\gamma)<\gamma$, which is a contradiction.