Supremum and infimum of a set

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  • #1
Valhalla
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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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  • #2
quasar987
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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?
 
  • #3
Valhalla
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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]
 
  • #4
quasar987
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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]??)
 
  • #5
Valhalla
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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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