Ordinal Numbers: is it possible to have an isomorphic copy of an ordinal $\alpha$ inside an ordinal $\beta < \alpha$?

Question

Could there be a function $f:\alpha \to \beta$ order-preserving and injective such that $\alpha$ and $\beta$ are ordinal numbers with $\alpha > \beta$?

Answer 1

No, you cannot. For one thing, what should $f(0)$ be?

Now it may be tempting to let $dom(f)$ be the nonzero ordinals to try and fix this. However, this still doesn't work: think about where $\omega$ goes. We must have $f(\omega)=n$ for some finite $n$, but then the infinitely-many ordinals $<\omega$ have to be squished into the finitely-many ordinals $<n$.

Indeed, no version of this can happen: Fodor's lemma says that if $F$ is a function from ordinals to ordinals (or $\omega_1$ to $\omega_1$, or . . . ) which is regressive - $F(x)\le x$ - then $F$ is constant on a stationary set.

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EDIT: David Ullrich's comments below have led me to worry that I've interpreted the question incorrectly. I read it as: $f$ maps ordinals to ordinals, and $f$ maps each $\alpha$ to a smaller $\beta$. (That's the usual meaning of "$\mapsto$".)

If instead you just are asking about a map from $\alpha$ to $\beta$ - that is, domain $\alpha$, codomain $\beta$ (usually represented by "$f: \alpha\rightarrow\beta$"), then the answer is still no: suppose such an $f$ existed. Since $\beta<\alpha$, we have $\beta\in\alpha$; consider the sequence $$\gamma_0=\beta, \gamma_1=f(\beta), . . . , \gamma_{n+1}=f(\gamma_n).$$ It's easy to show that $\gamma_0>\gamma_1>\gamma_2> . . . $, but this contradicts the well-foundedness of ordinals.