Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
up vote 1 down vote accepted

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.

share|cite|improve this answer

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.