Let $\alpha,\beta, \gamma$ be ordinals. I'm trying to prove that:

$$\alpha < \beta \implies \gamma+\alpha < \gamma + \beta$$

I managed to prove that

$$\alpha < \beta \implies \gamma + \alpha \leq \gamma + \beta$$

by showing there is a strictly order preserving function $\gamma+\alpha \to \gamma + \beta$

but I struggle to show equality is impossible.

So suppose to the contrary that:

$$\gamma+\alpha= \gamma + \beta$$

This implies that

$$(\gamma \times 0) \cup (\alpha \times 1) \cong (\gamma \times 0) \cup (\beta \times 1)$$

But I don't think we really can get a contradiction from this.

I read that ordinal addition is left addition, so we should somehow be able to prove that $\alpha = \beta$ which will be a good contradiction.


1 Answer 1


Let $f: (\gamma \times \{0\}) \sqcup (\alpha \times \{1\}) \to (\gamma \times \{0\}) \sqcup (\beta \times \{1\})$ be an order isomorphism. We show by induction on $\delta < \gamma$ that $f(\delta,0) = (\delta,0)$.

Base case: $f$ is an order isomorphism, so it must send the least element of $(\gamma \times \{0\}) \sqcup (\alpha \times \{1\})$ to the least element of $(\gamma \times \{0\}) \sqcup (\beta \times \{1\})$. The least element in either case is $(0,0)$.

Induction step: Suppose $f$ is the identity below $(\delta,0)$. Because $f$ is an order isomorphism, it must map $(\delta,0)$ to the least element of $(\gamma \times \{0\}) \sqcup (\beta \times \{1\})$ not in $f[\delta \times \{0\}]$; but $f[\delta \times \{0\}] = \delta \times \{0\}$ by hypothesis, so $f(\delta,0) = (\delta,0)$.

Thus, by induction $f$ is the identity on $\gamma \times \{0\}$. It follows that $f|_{\alpha \times\{1\}}$ must be an isomorphism of $\alpha \times \{1\}$ onto $\beta \times \{1\}$ for $f$ to be an isomorphism (injectivity implies elements of $\alpha \times \{1\}$ must map into $\beta \times \{1\}$, surjectivity implies all element of $\beta \times \{1\}$ are mapped to in this way, and $f$ being order-preserving implies that $f|_{\alpha \times \{1\}}$ is as well). Contradiction.

  • $\begingroup$ Looks correct on first sight! Thank you! I will come back tomorrow and see if everything is alright! $\endgroup$
    – user745578
    Apr 10, 2020 at 21:39

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