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

  1. Jun 22, 2011 #1
    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.
     
    Last edited: Jun 22, 2011
  2. jcsd
  3. Jun 22, 2011 #2

    micromass

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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.
     
  4. Jun 22, 2011 #3
    Thank you, micromass! :smile:
     
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