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Let $A$ be the set of all order preserving bijections on the set of integers

Question. What is the cardinality of $A$?

Thoughts. For every integer $r$ $$f(n)=n+r$$ is an order preserving bijection. Hence $A$ is at least countably infinite. But I can't prove that it is not uncountable.

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    $\begingroup$ Can you construct a single order preserving bijection on the set of integers that IS NOT of the kind f(n)=n+r ? $\endgroup$ – Evargalo Jun 12 '17 at 15:36
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Hint: Let $f \colon \mathbb Z \to \mathbb Z$ be an order preserving bijection. Let $r = f(0)$. Show that $f(x) = x + r$ for all $x \in \mathbb Z$.

(You can - for example - show this by two inductions on $\mathbb Z^+$ and on $\mathbb Z^-$.)

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  • $\begingroup$ Btw. Once you've proved this, it pretty much automatically follows that there are only two ring isomorphisms $f \colon (\mathbb Z; +, \cdot, -, 0,1) \to (\mathbb Z; +, \cdot, -, 0,1)$, namely $f = \pm \operatorname{id}$. $\endgroup$ – Stefan Mesken Jun 12 '17 at 19:24

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