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

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In summary, the conversation discusses the need to prove the existence of sup(X) for a nonempty, bounded subset X of the continuum C. The attempted solution involves proving that A, the set of elements that are not upper bounds of X, is open, and that its complement is closed. The proof relies on the assumption that C is connected, and the conclusion is that sup(X) exists. The individual asking for help also asks if their proof is correct and if they have used the axiom of connectedness.
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math771
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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[itex]_{0}[/itex], is not an upper bound of X, then there exists an element of X, called x[itex]_{0}[/itex], such that a[itex]_{0}[/itex] < x[itex]_{0}[/itex]. Moreover, no element of C less that x[itex]_{0}[/itex] is an upper bound of X (so that all elements of C less than x[itex]_{0}[/itex] belong to A), and there exists at least one point of C, called y[itex]_{0}[/itex], that is less than a[itex]_{0}[/itex] (because it is admitted as an axiom that C has no first point) and therefore less than x[itex]_{0}[/itex]. Thus, y[itex]_{0}[/itex] < a[itex]_{0}[/itex] < x[itex]_{0}[/itex], and we may construct a neighborhood around a[itex]_{0}[/itex] 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={[itex]\alpha[/itex]|[itex]\alpha[/itex] 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 [itex]\alpha[/itex][itex]_{0}[/itex] such that for every x[itex]\in[/itex]X, [itex]\alpha[/itex][itex]_{0}[/itex] > x. Moreover, because sup(X) does not exist, there exists an [itex]\alpha[/itex][itex]_{1}[/itex] such that x < [itex]\alpha[/itex][itex]_{1}[/itex] < [itex]\alpha[/itex][itex]_{0}[/itex]. By the argument utilized above, there exists a neighborhood around [itex]\alpha[/itex][itex]_{0}[/itex] 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?
Thanks in advance.
 
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Hi math771! :smile:

You must use that C is connected, in fact the theorem you state is equivalent to saying that C is connected (well, in [itex]\mathbb{R}[/itex] 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
Thank you, micromass! :smile:
 

FAQ: Proving that sup(X) exists for X nonempty and bounded

1. What does it mean for a set to be nonempty and bounded?

A set is nonempty if it contains at least one element. A set is bounded if there is a finite number that serves as an upper or lower limit for all elements in the set.

2. How do you prove the existence of sup(X) for a nonempty and bounded set?

To prove the existence of sup(X), you must show that there exists an element in the set that is greater than or equal to all other elements in the set. This element is known as the supremum or least upper bound.

3. Can you give an example of a nonempty and bounded set and its supremum?

Yes, consider the set X = {1, 2, 3, 4, 5}. This set is nonempty and bounded, and its supremum is 5, as there is no element in the set that is greater than 5.

4. Is the supremum of a set always a member of the set?

Not necessarily. While the supremum must be greater than or equal to all elements in the set, it does not have to be a member of the set. For example, in the set X = {1, 2, 3}, the supremum is 3, even though 3 is not a member of the set.

5. Why is proving the existence of sup(X) important in mathematics?

The existence of sup(X) is important because it allows us to define the concept of limits and continuity in calculus. It also helps us establish the completeness property of real numbers, which states that every nonempty and bounded set of real numbers has a least upper bound.

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