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basic set theory

Semidevil

Diamond Member
I"m not understanding what the question is asking....

I"m gonna use the "n" = intersection sign....

find n(i=1, to infinite) of A(i). If A(i) = [0, 1/n].

so what is it asking?

from definition, this means {x : x belongs to [0, 1/n] for all n belong to N}.

but what is it asking? what am I suppose to do?
 
Please rewrite your question. You use n twice where one is intersection and the other is.... something. What the heck does 1/intersection mean?
 
Originally posted by: TuxDave
Please rewrite your question. You use n twice where one is intersection and the other is.... something. What the heck does 1/intersection mean?

When it comes to sets, summations, etc, just make an equation in Word and PDF it or just offer the DOC file as a download or a JPEG.
 
do you mean to say A(i) = [0,1/i]?

in that case, the answer is {0}. the set {1/n, n natural number} approaches 0 as n->infinity. Thus in this case, because the bounds are inclusive, 0 will be included on the "left" as the points on the "right" squeeze downard onto it...so in the end they will all share 0.
 
i am tempted to say the null set because n-> infinity 1/n -> 0 but never actually reaches it? it could just be 0 though

its one of those "its 0 because its close enough vs. its not 0 because its not exact" things
 
Originally posted by: eLiu
do you mean to say A(i) = [0,1/i]?

in that case, the answer is {0}. the set {1/n, n natural number} approaches 0 as n->infinity. Thus in this case, because the bounds are inclusive, 0 will be included on the "left" as the points on the "right" squeeze downard onto it...so in the end they will all share 0.

I agree... if A(i) = [0,1/i], then I would say they all share 0.
 
yes, it is 1/i. sorry.


what i'm more interested to know is, what is it asking...I'm not understanding what it is asking, in words...and how do you find out it is 0?
 
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