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If I have a lexicographic ordering on $\mathbb{C}$, and I define a subset, $A = \{z \in \mathbb{C} : z = a + bi, a, b \in \mathbb{R}, a < 0\}$.

I have an upper bound, say $\alpha = 0 + di$. My question is does only the real part, $\Re(\alpha) = 0$ define the upper bound? Or does the $\Im(\alpha) = d$ have nothing to do with bounds in general?

Since it seems to me if I have the lexicographic ordering on $\mathbb{C}$ such as for any two $m, n \in \mathbb{C}$, where $m = a + bi$ and $n = c + di$ and I define the ordering as $m < n$ if $a < c$ or if $a = c$ and $b < d$.

The last bit, $b < d$ gives me the impression that $\Im(\alpha)$ would play a role in the upper bound. The reason I am asking is because in a proof I read, they prove this order has no least upper bound as there are infinitely many complex numbers with their real parts equal to $\Re(\alpha)$ but different imaginary parts. So, I guess if only the real parts of complex numbers define the bounds then it makes sense to me.

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2 Answers 2

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A least upper bound has to be a specific number with the LUB property. In this case there is no such number, since there are lots of upper bounds but none of them is the smallest.

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To expand on Pink Panda’s answer a bit, a complex number $a+bi$ is an upper bound for $A$ if and only if $a\ge 0$. Since any number of the form $bi$ is smaller in the lexicographic order than any number $x+yi$ with $x>0$, the only hope for a least upper bound would be some pure imaginary $bi$. But no matter what $b\in\Bbb R$ you try, $(b-1)i$ is a smaller upper bound for $A$. Thus, $A$ has no least upper bound in the lexicographic order, though it has infinitely many upper bounds.

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