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Call $S=\{x\in \mathbb{R} \mid x>-1\}$
I am trying to show that the $\inf\{S\}=-1$:

So here I am assuming that proceeding by contradiction is most efficient:
(Assuming that $-$1 is a lower bound) Assume $\exists\space l$ such that $l > -1$ and is a lower bound to $S$, we want to show that $l$ cannot be a lower bound, a contradiction. Is this the best way to start?

Take some arbitrary $\epsilon > 0$ and define $l=-1 + \epsilon > -1$

Which definition is more practical for showing such results? see this link:Definitions of supremum

I am quite confused as to what would be a procedure to solve this type of problem, I currently find it significantly harder than working through theory.

Any hints, ideas or full proof for the sake of having an example would be appreciated. Would the same ideas apply to finding the infimum of a sequence ${1 \over n}$?

EDIT______________________________________________________________ Lets assume that $l$ is the infimum, and $l > -1$. and want to show that such an $l$ cannot be a lower bound, setting a contradiction.
We define some $x$ such that $x = {l-1 \over 2}$, indeed $l > x > -1$ given x is the arithmetic mean of those two points.
Assuming $x \leq -1$, then ${l-1 \over 2}< -1$, then $l \leq -1$ showing that $l>-1$ is not an infimum.

Is this correct?

Community
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rannoudanames
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    Your idea is correct. Here you don't have to write $l=-1+\epsilon$. Just note that if $l>-1$ then $x:=\frac{l-1}{2}$ (the middle point of $[-1,l]$) is in $S$ and is less than $l$. – NeedForHelp Jan 24 '17 at 22:26
  • would $x$ not be ${l+1 \over 2}$ and would i try to show that x is a lower bound and have some contradiction forming? Would it be a similar reasoning to http://math.stackexchange.com/questions/386758/proving-rigorously-the-supremum-of-a-set ? – rannoudanames Jan 24 '17 at 22:38
  • If you have an interval $[a,b]$ with $a<b$ then $r:=\frac{b-a}{2}$ is the "radius" of the interval. The middle point is $m:=\frac{a+b}{2}$ (the arithmetic mean of $a$ and $b$). We have $m-r=a$ and $m+r=b$ as one would expect. From $x<l$ you can infer that $l$ is not a lower bound for $S$. But your hypothesis was that $l>-1$ was a lower bound for $S$. That's your contradiction. – NeedForHelp Jan 24 '17 at 22:42
  • hmmm, i see... ill give it shot update for further insight – rannoudanames Jan 24 '17 at 22:46
  • And yes the argument I gave is similar to the one in the link in your comment. The guy there used the point $x:=\frac{2-M}{2}+M$ which is nothing more than $\frac{2+M}{2}$, the middle point of $[M,2]$. The guy showed that that middle point is less than $2$ and greater than $M$. But that's geometrically very clear and I did not prove that. I guess in an ideal world you do want to prove that (the geometric interpretation is a heuristic, it wouldn't really be considered a proof in real analysis, strictly speaking). Note that the choice of the middle point is arbitrary. – NeedForHelp Jan 24 '17 at 22:50
  • @NeedForHelp is the answer under EDIT correct? – rannoudanames Jan 24 '17 at 23:38
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    Don't assume $l$ is the infimum. Assume $l$ is a lower bound. I don't understand the last sentence in your proof. You already set $x:=\frac{l-1}{2}$ and you noticed that $x>-1$. Why do you assume that $x\leq-1$ then? All you have to write is that $l>x>-1$ implies 1. $x>-1$ so $x\in S$ and 2. $l>x$ and by 1 $x\in S$ so $x$ is an element in $S$ which is less than $l$ so $l$ is certainly not a lower bound for $S$. So any lower bound of $S$ is $\leq-1$. But $-1$ is a lower bound, so it must be the greatest. So $-1$ is the greatest lower bound, aka infimum. End of story. – NeedForHelp Jan 24 '17 at 23:48

2 Answers2

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Since for all $s \in S$ we have $s \ge -1$, it is clear that $\inf S \ge -1$.

Since $-1+{1 \over n} \in S$ for all $n >0$, it is clear that $\inf S \le -1+{1 \over n}$.

Combining yields the desired result.

copper.hat
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If $x \in S$ then $x>-1$. So $-1$ is a lower bound of $S$.

If $l$ is any number greater than $-1$, then $l$ is not a lower bound, because the average between $l$ and $-1$ is in $S$ and is less than $l$.

So $-1$ is the infimum

Emilio
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