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?
Best Answer
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