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Most sources that define an upper bound as:

If $A$ is a set of numbers and $b$ is a number, then $b$ is an upper bound if and only if $x \le b $ for all $x \in A$

I have three questions:

  1. Since it doesn't specify the set $b$ is in, is it that any number $b$ which satisfies $x \le b $ can be called an upper bound? So for example, the set $(0,1)$ has infinitely many upper bounds like $1, 2, \sqrt2 $??

  2. So if we further define the supremum as the least element of all the upper bounds of $(0,1)$ then the supremum would have to be $1$ even though $1 \notin (0,1)$?

  3. If the answer to the first two questions is yes, then can I say that $b$ could be a complex number greater than $1$ in my example? For example, can $4+3i$ be upper bound of $(0,1)$ since the definition implies $b$ can be any number?

Bunny
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  • $1$ is correct, $2$ is correct (try proving this), $3$ isn't (note that $\mathbb{C}$ cannot be ordered ) – Clarinetist Jan 19 '16 at 19:03
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    In general, the type of "number" depends. We might be talking about integers, or rationals, or reals. You can have upper bounds in any of these cases. You can define "upper bound" in practically any case where we have a meaningful definition for $\leq$. But we don't always have a least upper bound (supremum). – Thomas Andrews Jan 19 '16 at 19:04
  • @ThomasAndrews No you're right I'll correct that. – Bunny Jan 19 '16 at 19:05
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    If $b$ is a complex number, there is no definition for $a\leq b$ that is particularly useful. – Thomas Andrews Jan 19 '16 at 19:05
  • For example, if a set of integers has an upper bound, then there is an integer least upper bound. The same is not true for rational numbers - the set of rationals whose square is less than $2$ has no least upper bound in the rationals. This is why one reason the reals are special. – Thomas Andrews Jan 19 '16 at 19:09