Order isomorphism between $(-\infty,0)$ to $(0, \infty)$

Question

Is there an explicit order isomorphic function between $(-\infty,0)$ to $(0, \infty)$? There should be one, because both are order isomorphic to $\mathbb{R}$, but I can't think of one.

Answer 1

Do you mean something like this?

Suppose $a, b \in (-\infty, 0)$ with $a < b$ in the usual sense. Define $f : (-\infty, 0) \to (0, \infty)$ by

$$f(x) = \tan\left(\arctan(x)+{\pi\over 2}\right)$$

and in fact$^{(*)}$ $$f(x) = -{1\over x}$$

Then

$$a < b \Longleftrightarrow f(a) < f(b)$$

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(*) Kudos to Henning for pointing this out, see below.