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Upper bound

  1. Feb 16, 2008 #1
    Let [itex] E [/itex] be a nonempty subset of an ordered set; suppose [itex] \alpha [/itex] is a lower bound of [itex] E [/itex] and [itex] \beta [/itex] is an upper bound of [itex] E [/itex]. Prove that [itex] \alpha \leq \beta [/itex].

    So do I just use the following definition: Suppse [itex] S [/itex] is an ordered set, and [itex] E \subset S [/itex]. If there exists a [itex] \beta \in S [/itex] such that [itex] x \leq \beta [/itex] for every [itex] x \in E [/itex], then [itex] \beta [/itex] is an upper bound for [itex] E [/itex], and similarly for lower bound?
     
  2. jcsd
  3. Feb 16, 2008 #2

    arildno

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    Indeed, so it is fairly trivial.

    However, you should exert yourself a bit more for your own sake:
    WHY is the condition of non-emptiness of E crucial?
     
  4. Feb 16, 2008 #3

    mathman

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    Easy proof: Let x be an element of E. a<=x<=b. Therefore a<=b.
     
  5. Feb 16, 2008 #4
    If [itex] E [/itex] was empty, then it would not have a upper or lower bound.
     
  6. Feb 16, 2008 #5

    arildno

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    Incorrect!
    Try again..
     
  7. Feb 16, 2008 #6
    If [tex] E [/tex] was empty, then there would be no order relation? Or the upper bound and lower bound would be the same?
     
  8. Feb 16, 2008 #7
    Didn't someone above prove the original problem via transitivity?
     
    Last edited: Feb 16, 2008
  9. Feb 16, 2008 #8
    no but he asked why E must be non empty.
     
  10. Feb 16, 2008 #9

    arildno

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    Nope.

    The trick is to rewrite the definitions of upper bounds and lower into EQUIVALENT expressions that makes it clear why, for an empty subset a lower bound can be greater than an upper bound, say:

    Now, the standard def. of the upper bound for some set E is that U is an upper bound for E if and only if for every element x in E, we have x<=U.

    But, this can be rephrased equaivalently as:

    A number U is the upper bound of E if and only if there exists NO elements x in E so that x>U

    Similarly can the lower bound L be rephrased.

    Now, let us say that E is an empty subset of a set of integers, for simplicity.

    Since there are no elements in E at all, there certainly don't exist any elements in E that are greater than 1, so 1 is certainly an upper bound for E.

    But, 2 is definitely a LOWER bound of E, by a similar argument.

    But 2>1, so a lower bound of E is greater than an upper bound of E. :smile:
     
  11. Feb 16, 2008 #10
    The result was proved in post #3 via transitivity. As for why E must be not empty why would we define a lower bound for an empty subset of the real numbers?
     
  12. Feb 16, 2008 #11

    arildno

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    Why do maths at all?

    Why develop precision in our thinking and understand why conditions stated were NECESSARY for the proposition to be true?
     
  13. Feb 16, 2008 #12
    While, I haven't decide water I agree with the first definition, can you prove to me it is logically equivalent to the second?
     
  14. Feb 16, 2008 #13
    There are clearly a multitude of reasons and I don't consider why not sufficient.
     
  15. Feb 16, 2008 #14

    arildno

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    In the last one, remember if there can't exist any elements greater than U, then for whatever existing elements in E, they must be equal or less than U. But thereby, the last one implies the first one.

    Now, the other way around:
    Suppose that whatever elements is in E IS less than or equal to U. Then there can't be any elements in E that are greater than U, meaning the first implies the second statement.

    They are equivalent statements, and contrapositives of each other.
     
  16. Feb 16, 2008 #15

    arildno

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    Incoherent and irrelevant remark.
     
  17. Feb 16, 2008 #16
    Or, equivalently, the transitivity proof that goes:

    "Easy proof: Let x be an element of E. a<=x<=b. Therefore a<=b."

    clearly doesn't work if there is no actual x in E. Maybe it isn't entirely obvious, but you really have to have an actual x, here. You aren't doing something like "IF x WERE an element of E, then...," but rather are actually getting an x from E to use. That requires the x to actually exist (i.e. for E to be nonempty).
     
  18. Feb 16, 2008 #17

    arildno

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    That is indeed why you get the proof working for non-empty sets.

    The prposition (i.e that lower bounds are less than or equal to greater bounds) is not meaningless, but just false, for the empty set.
     
  19. Feb 16, 2008 #18
    Now that I think about it, just to elaborate on what I'm saying, "Let x be an element of E," is not actually an assertion -- a proposition that is either true or false. When someone says this in a proof we have to interpret it (and usually it is no big deal). But, in this case, it should be interpreted as the assertion that "There exists x, an element of E," which would be clearly false if E was empty. So, strictly speaking, I think the error with the easy proof (if it was not already known that E was nonempty) would be that first step: "Let x be an element of E."

    Now, it is possible to reinterpret that statement to mean something more like "IF there existed x, an element of E, THEN...," but that would just bring us right back to E having to be nonempty for the result to hold. In other cases, you might interpret "Let x be an element of E," this way if it was a proof that, for instance, went like "Let x be an element of E, blah blah blah (contradiction), so E must be empty." In such an argument, the correct interpretation is the conditional one and x doesn't actually have to exist.

    At any rate, I think the whole problem originates from the fact that it is such a knee jerk response to think of "Let x be an element of E," as an always true assertion or something. In fact, it, strictly speaking, is not actually an assertion at all and means different things depending on the context in which it is used.
     
  20. Feb 16, 2008 #19
    Well, I think that your argument really drives home that it certainly cannot be said to be true about empty sets, in any case, whether it is false or just meaningless. However, I think I might agree with you, here. It is the same reason we tend to take "Let x be an element of E," as if it were a proposition. You have to ascribe all of the meaning you can to statements or maybe the best meaning you can. So, it is not meaningless to talk about the empty set being bounded not because it is natural to do so, but because there exists a way to meaningfully do so that is consistent with the same discussion about nonempty sets. So, if the set could be empty or someone is just talking about the empty set this way, you would have to, strictly speaking, imagine that they must mean exactly what you were saying (using the contrapositive of the normal definition of bounded).

    However, I do think that at this point, we are definitely well out of mathematics and into the foundations and/or entirely in philosophy at this point (as to whether it is false or just meaningless). (Nevertheless, it is an interesting discussion.)
     
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