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Is there any order that can make complex numbers complete ordered set with least upper bound property?

I came up with $x+yi > m+ni$ if $x+y>m+n$ and if sum is equal then the complex number with greater real part is greater.

will this work? if not, are there any examples?

jnyan
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    See Total ordering on complex numbers for a negative answer and, if that doesn't answer your question, then please define what you mean by a complete ordered set. – dxiv Mar 12 '17 at 08:13
  • no. It doesnt answer. I mean ordered set, with no field axioms defined on them. – jnyan Mar 12 '17 at 08:19
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    If you mean a totally ordered set, just define a+bi<c+di if and only if a<c or (a=c and b<d), the lexicographical order. – Massimo Mar 12 '17 at 08:36
  • I have edited the question. Sorry. – jnyan Mar 12 '17 at 08:39
  • Yes, of course. Since the set $\mathbb C$ of all complex numbers has the same cardinality as the closed interval $[0,1],$ you can define an order on the complex numbers which makes $(\mathbb C,\lt)$ order-isomorphic to $([0,1],\lt).$ This is of no interest, because the ordering you get has nothing to do with the algebra. – bof Aug 22 '17 at 03:49

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Your proposal does not work. For instance, the set $\{a-ai:a\in\mathbb{R}\}$ is bounded above but has no least upper bound. Its upper bounds are all numbers $x+iy$ with $x+y>0$, and there is no least such number (since, for instance, there is no least value that $x+y$ can take).

For a simple correct example, just pick any bijection $f:\mathbb{C}\to\mathbb{R}$, and define an ordering $\preceq$ on $\mathbb{C}$ by $a\preceq b$ iff $f(a)\leq f(b)$, where $\leq$ is the usual order on $\mathbb{R}$. Since the usual order on $\mathbb{R}$ is complete, so is $\preceq$.

Eric Wofsey
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