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Exercise 1.9 Suppose $z = a + bi, w = c + di$. Define $z<w$ if $a<c$, and also if $a = c$ but $b<d$. Prove that this turns the set of all complex numbers into an ordered set. (This type of order relationis called a dictionary order, or lexicographic order, for obvious reasons.) Does this ordered set have the least upper bound property?

The solution of this question is the procedure this set has an ordered property such as "$x<y$ or $x=y$ or $x>y$", "$x<y$ and $y<z$ then $x<z$". enter image description here

But I think that this question already gives us that this set is ordered, so we don't need prove it. In addition, all of ordered sets have a least-upper-bound property, so we also need not prove they have a least-upper-bound property. I think this book asks us a wrong question. Is it right? ps. Why this solution doesn't prove there is a least upper bound?

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Hint: consider the subset $\{(0,y)\,:\,y\in\Bbb R\}$.

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  • $\begingroup$ Thank you. I realize that not all ordered sets have least upper bound property, but I don't know why this solution proves only whether this set is ordered or not. $\endgroup$ – 주혜민 Mar 29 at 9:48
  • $\begingroup$ @주혜민, because is only a proof of "$<$ is a relation of order". See you some related with the least-upper-bound property in i.stack.imgur.com/H8wBc.png? $\endgroup$ – Martín-Blas Pérez Pinilla Mar 29 at 9:52
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    $\begingroup$ @AkashPatalwanshi, $C$ is bounded in the lexicographic order. $\endgroup$ – Martín-Blas Pérez Pinilla Aug 21 at 10:39
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    $\begingroup$ @AkashPatalwanshi, see the correction. Any $(x,\cdots)$ with $x > 0$ is an upper bound of $C$. $\endgroup$ – Martín-Blas Pérez Pinilla Aug 21 at 14:20
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    $\begingroup$ Thank you so much sir for your last comment. Got it :-) $\endgroup$ – Akash Patalwanshi Aug 22 at 15:15

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