# Proving that sup(X) exists for X nonempty and bounded

1. Jun 22, 2011

### math771

To be more precise, I must prove that sup(X) exists if X is a nonempty, bounded subset of the continuum C. I cannot find any problems with my attempted solution. However, I think that I have to use the fact (admitted as an axiom of the continuum) that C is connected, something my proof does not do.
My attempted solution:
Let A={a|a is not an upper bound of X}. We will prove that A is open (not just for a bounded X, but any nonempty X that is a subset of C).
If an element of C, called a$_{0}$, is not an upper bound of X, then there exists an element of X, called x$_{0}$, such that a$_{0}$ < x$_{0}$. Moreover, no element of C less that x$_{0}$ is an upper bound of X (so that all elements of C less than x$_{0}$ belong to A), and there exists at least one point of C, called y$_{0}$, that is less than a$_{0}$ (because it is admitted as an axiom that C has no first point) and therefore less than x$_{0}$. Thus, y$_{0}$ < a$_{0}$ < x$_{0}$, and we may construct a neighborhood around a$_{0}$ that is a subset of A. This proves that every point of A is an interior point of A and therefore that A is open.
The complement of A, C/A={$\alpha$|$\alpha$ is an upper bound of X}, is therefore closed.
We will now show that if sup(X) does not exist, C/A is open.
Because X is bounded, there exists an $\alpha$$_{0}$ such that for every x$\in$X, $\alpha$$_{0}$ > x. Moreover, because sup(X) does not exist, there exists an $\alpha$$_{1}$ such that x < $\alpha$$_{1}$ < $\alpha$$_{0}$. By the argument utilized above, there exists a neighborhood around $\alpha$$_{0}$ that is a subset of C/A. Thus every point of C/A is an interior point of C/A, and C/A is open.
C/A is nonempty proper subset of C (because A is nonempty). Thus, C/A cannot be both open and closed. (This statement obviously requires a proof, which I will not present here. I will note, however, that my proof of this statement relies on the axiom that C is connected (something I hadn't realized until now). If all of the above work is correct, this may resolve my underlying concern that the proof I have just presented should but does not utilize the axiom that C is connected.)
sup(X) exists.
To reiterate my principal questions: must I use the axiom that C is connected? Are there any problems with my work?

Last edited: Jun 22, 2011
2. Jun 22, 2011

### micromass

Staff Emeritus
Hi math771!

You must use that C is connected, in fact the theorem you state is equivalent to saying that C is connected (well, in $\mathbb{R}$ at least).

Your proof is correct and you did use the fact that C is connected (as you already noticed) by saying that there are no nontrivial (i.e. not empty and not C) sets in C that are both open and closed. This statement is equivalent to connectedness.

3. Jun 22, 2011

### math771

Thank you, micromass!