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RolyPoly

  • 11 months ago

Suppose \(E\subset R\), \(z\in R\) and \(x\le z \forall x\in E\). The upper bound \(z_0\) of the set E is called the least upper bound of E if every other upper bound z of E is greater than or equal to \(z_0\). Why is \(z_0\) unique?

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  1. Spacelimbus
    • 11 months ago
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    Consider the following. Suppose \(z_0 \) and \(z_1\) are both relative suprema (least upper bounds) of the given set \(E\). Now since we said that \(z_1\) is the supremum of \(E\) this means that \(z_1 \geq z_0\) by its definition, but we can cycle this and say that we also defined \(z_0\) to be the least upper bound of the set \(E\). Therefore \(z_0 \geq z_1 \) must also be true. This can only be true for all cases if and only if \(z_1=z_0\) which shows uniqueness.

  2. Spacelimbus
    • 11 months ago
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    Also make sure that you understand that \(z_0 \notin E\) is possible, a least upper bound doesn't have to be in the set, it can be though and then we would call it the maximum, so the proof would be analogous. For \(z_0 \in E \implies \sup E=\max E=z_0\)

  3. RolyPoly
    • 11 months ago
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    Ah! The order axiom plays a role here. So, if \(E \subset R\) is bounded from above, E has a least upper bound. If the supremum is also an element in E, then the supremum is also the maximum of E, right?

  4. Spacelimbus
    • 11 months ago
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    Yes that is correct, to make the proof more elegant and rigorous you can/should include ordered/partial ordered sets and use antisymmetry and partial ordering. The Set \(E\) has to be bounded from above and be nonempty. Now most of the time the proof goes as follows by assuming - without the loss of generality - that the given supremum lays in \(E\) and therefore is the maximum of the given set \(E\) so you can apply the axion of order (as by ZFC). Then the rest is as always. A bit tedious to write it all down but you got it right :-)

  5. RolyPoly
    • 11 months ago
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    Wait wait wait. Can you explain the terms "ordered/partial ordered sets" and "antisymmetry" in your context, as well as the part "without the loss of generality"? Also, how do you know if it is really "without the loss of generality"?And, what is ZFC?

  6. Spacelimbus
    • 11 months ago
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    I will try, during your proof you will establish the following identity: \(\Large z_1 \leq z_0 \wedge z_0 \leq z_1 \implies z_0 = z_1 \) this is what we call antisymmetry and we call \(\leq\) a relation, in fact it is a partial ordering relation, that means \(\leq\) is reflexive, antisymmetric and transitive. I only wrote it all down to make sure that you use the right context during your proof. It would be a bit vague to just have a set \(E\) without any ordering relation on it. So you want to use the partial order relation \(\leq\) and its consequences. As to without the loss of generality. This is a general principle of Mathematics to assume an easier case and prove the statement in that manner. For the proof above it is not really necessary to assume wether or not the least upper bound lays in the given set \(E\) or not. However if it does we can treat it as a Maximum and most people find that easier to deal with, since the Maximum is only a special case of the Supremum we don't violate the generality of the proof. ZFC is Zermelo Fraenkel (Choice). Big contributors of our modern set theory and the ultimate defeat of the Russel Paradox. I thought you already referred to them by saying "The order axiom plays a role"

  7. RolyPoly
    • 11 months ago
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    Oh. I'm sorry. I should have written those relations on that little summary as well. As for the order axiom, I just read it from the book. It says that for every pair x and y in R, \(x\le x \ \forall x\in R \), and \((x\le y) \wedge (y \le x) \implies x=y\) Is \(x\le x \ \forall x\in R \) the ZFC you mentioned?

  8. Spacelimbus
    • 11 months ago
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    The \(( x \leq y) \wedge (y \leq x) \implies x=y\) is the antisymmetric property of the \(\leq\) relation. However in general all those definitions and statement result from the work and rigorous definitions of Zermelo and Fraenkel, so you're right :-) I hope I did not confuse you too much with that, in general you can keep it away, it was more of a side note. In general ZFC is only a list of axioms that most Mathematicians use nowadays in set theory (there are other axioms too which are not a part of ZFC). I only mentioned it to say that I am relying on their definitions and axioms when I am talking about things like ordered, or symmetric or antisymmetric, bound and supremum.

  9. RolyPoly
    • 11 months ago
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    Thanks for your help! I think I got that antisymmetric property. I remember I did it once in a test. I'm still confused by ZFC since I haven't seen it or heard of it before. Thanks for bringing it out though! :)

  10. Spacelimbus
    • 11 months ago
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    You are welcome, it's really no big deal if you haven't heard about Zermelo Fraenkel yet, you will in Analysis II or in Function Theory :-)

  11. RolyPoly
    • 11 months ago
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    What is Analysis II about?

  12. RolyPoly
    • 11 months ago
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    Oh, and what is that function theory?

  13. Spacelimbus
    • 11 months ago
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    The bounds are very open, as for different countries too, I am European, but Analysis II mainly deals with a mathematical approach to Multivariable Calculus and different approximations like Taylor and Bernstein Polynomials to approximate functions and deal with their limits. Function Theory is about different measurements in mapping, basically you enhance the regular Mapping/Functions you already know as in being injective,surjective,bijective with some additional properties like connectivity, convex under a mapping, analytic functions and so on. A lot of vocabulary, big topic for Mathematicians during the 3rd Semester usually.

  14. RolyPoly
    • 11 months ago
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    Ah, ok. Probably it's one of the "Analysis Trio". Thanks again for your help!

  15. Spacelimbus
    • 11 months ago
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    Welcome :) !

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