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Let $\leq$ be the usual order in $\mathbb{Q}$ and $\mathbb{Q_-}=\{q\in \mathbb{Q}: q <0\}, \mathbb{Q_+}=\{q\in \mathbb{Q}: q >0\}$. Do the following hold?

(a) $\langle\mathbb{Q_-}, \leq\rangle \simeq \langle\mathbb{Q_+}, \leq\rangle$

(b) $\langle\mathbb{Q_-} \cup \{0\}, \leq\rangle \simeq \langle\mathbb{Q_+}\cup \{0\}, \leq\rangle$

where $\simeq$ denotes an isomorphism of ordered sets.

In order for two sets to be isomorphic, there has to be a bijection between them (1) and this bijection has to preserve order (2). If we define $f(x) = -{1 \over x}$, it maps from one set to another and preserves $\leq$. So, (a) would be true.

For (b), redefining $$ f(x) = \begin{cases} -{1 \over x}, & x \neq 0 \\ 0, & x=0 \end{cases} $$ doesn't help since $0$ breaks the order. It is my suspicion that (b) doesn't hold as we run into problems trying to send $0$ between the sets. I would appreciate any comments.

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    For (a) you can also use the nice fact that any two countable dense linear orders without endpoints are isomorphic. – 2017-02-01
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    @MatemáticosChibchas Yes, corrected. – 2017-02-01

1 Answers 1

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(a) is solved nicely, although it could do with an explicit proof that $f$ preserves the order.

For (b), the existence of minimal and maximal elements is preserved under order isomorphisms. One of them has a maximal element but no minimal, whlie the other has a minimal element but no maximal. Therefore they are not isomorphic.