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THIS IS PART OF MY HOMEWORK, all I need is some guidance:

I am trying to prove that 1 is the supremum of the function $\ f(x)= \frac{1}{(x+1)}$ for every $\ x\gt$ 0. I was able to show that 1 is an upper bound but I can't find a way to show that there's no smaller upper bound than 1 in order to show that it is a supremum.

AviAsks
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2 Answers2

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If $r<1$, $\frac1{x+1}=r\iff x=\frac1r-1$. So, since $\frac1r-1>0$…

José Carlos Santos
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  • Why r has to be equal to 1/x+1? If we assume r is a supremum then it has to be bigger than 1/x+1. Didn't quite get what you were trying to show here. – AviAsks Dec 08 '19 at 11:23
  • I did not claim that $r$ is the supremum. What I did was to prove that if $r<$, then $r$ is not an upper bound of your set. If $x\in\left(0,\frac1r-1\right)$, $f(x)>r$. – José Carlos Santos Dec 08 '19 at 11:25
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if p > 0 is the supremum p < 1, then p != 1/(x+1) (x is arbitrary real number)

so 1 > p > 1/(x+1) > 0

px + p > 1

px > 1 - p > 0

x > (1-p) / p

then for x =< (1-p) / p contradiction

but if supremum is 1, 1 >= 1/(x+1) --> x+1 >= 1 --> x >= 0 --> not contradiction

WhitePaper
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