What is the supremum and infimum of the set B = {x\in\mathbb{R} : sinx \geq 0}?

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Homework Help Overview

The discussion revolves around determining the supremum and infimum of the set B = {x ∈ ℝ : sin(x) ≥ 0}. Participants explore the implications of the periodic nature of the sine function and its effects on the bounds of the set.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the periodicity of the sine function and its impact on the supremum and infimum of the set. There are questions about whether the set is bounded above or below, and some participants suggest that the supremum could be positive infinity while the infimum could be negative infinity.

Discussion Status

The discussion is ongoing, with participants examining definitions and the implications of periodicity on the bounds of the set. Some guidance has been offered regarding the need for rigorous proof, and there is acknowledgment of the lack of a textbook and limited course materials, which may affect understanding.

Contextual Notes

Participants mention the extended reals, which include positive and negative infinity, as relevant to the discussion. There is also a note about the course context, which combines analysis and set theory for electrical engineers.

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let [tex]B = \{x\in\mathbb{R} : sinx \geq 0 \}[/tex]

find the supremum and infimum of this set.

Ok well, since it is periodic I guess the point would be to note that the set will repeat ever [tex]2\pi[/tex]

So then if we consider just between 0 and [tex]2\pi[/tex]

supremum = [tex]\pi[/tex]
infimum = 0

if we consider all [tex]\mathbb{R}[/tex]

here is where I'm confused. The supremum would just be the [tex]N\pi[/tex] when N is an odd integer. Should I just state the function is periodic it will repeat between 0 and [tex]2\pi[/tex]
 
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Proceed methodically from the definition. M is a supremum of B if it is the smallest superior bound. But is B even bounded superiorly?
 
quasar987 said:
Proceed methodically from the definition. M is a supremum of B if it is the smallest superior bound. But is B even bounded superiorly?


Ok so I think I see what your saying. The set will not be bounded above or below except by plus or minus infinity b/c the function is periodic. I can always find a larger number in the reals that satifies sin(x) greater than or equal to 0. Therefore the set would have a supremum or positive infinity and a infimum of negative infinity.

Oh, I made typo in the original problem [tex]x\in\mathbb{R}_e[/tex]
 
That is the idea, yeah. You'd have to write a few equations though for it to be considered a proof. You'd have to show rigorously that given any number in B, there is always another number in B that is superior(resp. inferior) to it.(What is [itex]\mathbb{R}_e[/itex]??)
 
Our professor stated that [tex]\mathbb{R}_e[/tex] is the extended reals which contains plus and minus infinity. This course is an analysis for electrical engineers we get a crash course in a little bit of set theory then a bunch about complex functions with linear algebra of complex functions. We don't have a textbook for this course and the professors only written some of the course notes so I'm kind of flying blind on what is going on here. Thanks for the help!
 

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