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Infimum and Supremum, when they Do not exist in finite sets

  1. Jan 17, 2017 #1

    RJLiberator

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    1. The problem statement, all variables and given/known data
    Give an example of each, or state that the request is impossible:

    1) A finite set that contains its infimum, but not its supremum.
    2) A bounded subset of ℚ that contains its supremum, but not its infimum.

    2. Relevant equations


    3. The attempt at a solution
    I either understand this perfectly, or am missing something with the definition of sup/inf not existing.

    for 1) I have: Set B = { x ∈ ℚ : 1 < x < sqrt(2)} so inf(B) = 1 and sup(B) = DNE.
    for 2) I have Set C = ( x ∈ ℚ : sqrt(2) < x < 2 } so inf(C) = DNE and sup(C) = 2.

    At first, I thought by the axiom of completeness that question 1 would be impossible, but I seem to have found a set rather easily that is sufficient since I defined x to exist only in Q.
     
  2. jcsd
  3. Jan 17, 2017 #2

    FactChecker

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    You should double check every requirement in the statements and be sure they are met.
     
  4. Jan 17, 2017 #3

    fresh_42

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    But a) it doesn't contain its infimum, unless you define ##1 \leq x## and b) why is it finite and c) why isn't ##\sup B = \sqrt{2}\,##?
    Same here: take ##x \leq 2## instead. And it has an infimum.
     
  5. Jan 17, 2017 #4

    RJLiberator

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    Ok, clearly I have some confirmations wrong.

    Let's analyze the first one.

    1) A finite set that contains its infinum, but not its supremum.

    My original example was: Set B = { x ∈ ℚ : 1 < x < sqrt(2)} so inf(B) = 1 and sup(B) = DNE.


    a) Ah... 'contain' being the key word there. Lack of experience with inf/sup led me to overlook that detail.

    Now, If I understand the wording here correctly. A finite set that 'contains' its infinum but NOT its supremum. So, all it is asking is for the set to contain it, not that it does not exist.

    If we have sets such as: Set B: { 1, 2, 3, 4, 5, 6} clearly we have a sup and inf.
    Now, if we take Set C: {1/x : x exists over the Natural numbers }
    Then we have 1 as a suprema and it does NOT contain its inf. But this is not finite.


    So, if we do Set D: {x : x exist over the Natural numbers } then we dont have a suprema, but we do have an inf of 1. But this is not finite.

    I'm not quite sure that a finite set allows this to occur.
     
  6. Jan 17, 2017 #5

    fresh_42

    Staff: Mentor

    If you have a finite set as subset of a partially ordered set, then you can number them. Now do all pairs ##(a,b)## satisfy one of the relations ##a \preccurlyeq b\, , \,b \preccurlyeq a## or ##a=b\;##? Or can there be a pair of non-comparable elements? So what would infimum and supremum be then?

    The "existence" of suprema and infima depends on what you allow as valid. E.g. does ##\sup D = \infty## mean it doesn't exist? It's easier if you restrict the set itself. Your example ##B = \{x\in \mathbb{Q}\,\vert \,1\leq x < \sqrt{2} \} \subseteq \mathbb{Q}## is a set, where the supremum doesn't exist, since we didn't allow ##\sup B = \sqrt{2} \notin \mathbb{Q}##. This can be easily adapted as an example for 2). The idea with intervals wasn't bad. You can choose them to be closed, open, or closed at one end and open at the other.

    Things become a little bit more interesting in an example as ##S := \{(x,y) \in \mathbb{R}^2\,\vert \, x \in [0,1] \wedge y \in (2,3) \}## which can be ordered as follows: ##(x,y) \preccurlyeq (x',y') \Longleftrightarrow x < x' \vee (x=x' \wedge y \leq y')\; ##. This ordering is called a lexicographical order as it is the way lexica order their keywords:

    Basically it is:
    infimum = approach from the left (small elements) as close as you can
    supremum = approach from the right (bigger elements) as close as you can.

    ____inf --->____(a____<____b)____<--- sup____
     
  7. Jan 18, 2017 #6

    RJLiberator

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    OK. So what I am seeing is:

    Let's start by taking my example and working it out for a bounded subset of Q that contains its supremum, but not its infimum.

    If we have ##B = \{x\in \mathbb{Q}\,\vert \, \sqrt{2}\< x \leq2 \} \subseteq \mathbb{Q}## then we see a set that contains its supremum (1) but does not contain its infimum. This set is bounded by sqrt(2) and 1.

    So, I feel strongly in saying that that example is correct for part 2.

    Now part 1, seems to be a bit more tricky. It must be done with using rationals. I'd like to first verify that my understanding for part 2 is correct, then I will re-attempt part 1.
     
  8. Jan 18, 2017 #7

    fresh_42

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    ##2## instead of ##1##.
    Minus the typo, yes.
    Are you sure there is an example? Try to answer the questions about orders I posed. The entire exercise is about the difference between maximum (resp. minimum) and supremum (resp. infimum).
     
  9. Jan 19, 2017 #8

    RJLiberator

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    When we talk about non-comparable elements are we talking about a set like (1, 2, egg, white) where "1" and "egg" are not comparable?

    I would assume, that in any example set I give, it would only contain comparable elements. Thus, we'd have a minimum and maximum in any finite set.

    If the set has to be finite and contain its infimum, but not its supremum, I can't think of a set that does this.
     
  10. Jan 19, 2017 #9

    FactChecker

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    Then maybe you can think of a reason that a finite set can not have one of those properties.
     
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