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I want to prove that $(\mathbb{N},\le)$ and $(\mathbb{Z}, \le)$ are not order isomorphic. So what I want to show is that the following is not true:

$$x \le_\mathbb{N} y \iff f(x) \le_\mathbb{Z} f(y)\qquad\forall x,y \in \mathbb{N}$$

But I think it is true, isn't it? Or have I done a mistake in the definition of the orde isomorphism?

I could also want to show that $\mathbb{N}$ and $\mathbb{Z}$ are not bijections. Is this what I have to do rather than show that it contradicts the definition of orde isomorphisms? Do I have to show that $\mathbb{N}$ does not have a surjection $\phi$ to $\mathbb{Z}$?

Asaf Karagila
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FlyingDutchman
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2 Answers2

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$\mathbb{N}$ has a minimal element; $\mathbb{Z}$ doesn't. An order isomorphism preserves minimality of an individual element. Can you show this?

Marc
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You can't prove that there is no bijection between $\Bbb N$ and $\Bbb Z$, because there is one.

On the other hand, you can certainly prove that their natural orders are not isomorphic. Simply ask yourself, if $f$ is an order preserving injection from $\Bbb N$ to $\Bbb Z$, and $k=f(0)$, what could $f^{-1}(k-1)$ be?

Asaf Karagila
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  • Thank you very much! I think I understand it now. Because there is no $x \in \mathbb{N} | x = f^{-1}(k-1)$ the two sets can not be isomorph? – FlyingDutchman Oct 23 '14 at 07:52
  • Yes, if they were isomorphic, there would be some $x\in\Bbb N$ such that $f(x)=k-1$, but $x>0$ and $k-1<k$. – Asaf Karagila Oct 23 '14 at 07:54