The question is (let e stand for epsilon): \"Let A = (1, 3) U { - 4} , so lubA =

Question

The question is (let e stand for epsilon):

"Let A = (1, 3) U { - 4} , so lubA = 3 and glbA = - 4. For each e > 0, find x1 in A such that x1 > lubA - e. Also, for each e > 0, find x2 in A such that x2 < glbA + e. "

My teacher emphasized that x1 and x2 must both be in A, and the answers can either be specific numbers or formulas depending on e.

I am pretty confused by this problem, and I'd appreciate anyone's help and explanations. I will quickly give the points to the best answer. Thank you!

Explanation / Answer

A = {-4} U (1, 3)

lubA = 3 and glbA = - 4

i think you know that   a < (a+b)/2 < b    when ever a,b

x1>lubA-e

x1>3-e

so,x1 =[(3-e)+3]/2 whenever 0<e<2

see that x1=[(3-e)+3]/2 belongs to A

because 1< [(3-e)+3]/2   xince e<2

and [(3-e)+3]/2<3   since (a < (a+b)/2 < b    when ever a,b)

x1=2 whenever e>=2

x1 =2 belongs to A

so,x1 =[(3-e)+3]/2 whenever 0<e<2

x1=2     whenever e>=2

x2 < glbA + e

x2 < -4+ e

-4 belongs to A and -4 <-4+e when ever e>0

so,x2=-4   for e>0

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