this seems doesn't always work... (REAL ANALYSIS) | Mathematics | Page 1

rapid rampart Jan 9, 2025, 5:45 PM

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supremum value prove problem...

rare bloomBOT Jan 9, 2025, 5:45 PM

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rapid rampart Jan 9, 2025, 5:46 PM

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now given that $supA = \beta$ how to prove that any $x \in A$ is alwyas $x<=\beta$

sweet otterBOT Jan 9, 2025, 5:46 PM

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Waleed Al-Thqfi

restive python Jan 9, 2025, 5:48 PM

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that would just follow from definition, no?

narrow parrot Jan 9, 2025, 5:49 PM

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isn't this precisely the definition of supA=b

restive python Jan 9, 2025, 5:49 PM

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cant prove a definition

rapid rampart Jan 9, 2025, 5:51 PM

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there was a trick (that seems doesn't work always and i will explain why) assuming that there's $\alpha < \beta$ so $0<\beta - \alpha$
choose $\epsilon = \beta - \alpha$, and replace it in in the definition
$ x_o > \beta - \epsilon$ where $x_o \in A$ so $ x_o > \beta - (\beta - \alpha)$ and by this $x_o > \alpha$ where $x_o \in A$

sweet otterBOT Jan 9, 2025, 5:52 PM

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Waleed Al-Thqfi

narrow parrot Jan 9, 2025, 5:52 PM

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The supremum is defined as the smallest number $\beta$ that is greater than or equal to every element of $A$. If any $x \in A$ were greater than $\beta$, then $\beta$ would fail to be an upper bound, contradicting its definition.

sweet otterBOT Jan 9, 2025, 5:52 PM

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trigonometria

narrow parrot Jan 9, 2025, 5:53 PM

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I'm confused, is there any missing context to this question?

rapid rampart Jan 9, 2025, 5:53 PM

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wait

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narrow parrot Jan 9, 2025, 5:56 PM

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According to wikipedia

restive python Jan 9, 2025, 5:56 PM

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just because x0>alpha doesnt imply that x0>beta

rapid rampart Jan 9, 2025, 5:56 PM

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i want to give an example
$A = (0,1)$ 2 is greater than any point in A, right?

sweet otterBOT Jan 9, 2025, 5:56 PM

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Waleed Al-Thqfi

rapid rampart Jan 9, 2025, 5:57 PM

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it has the property (a)

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u know what

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i'm dumb

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.close

#this seems doesn't always work... (REAL ANALYSIS)