I'm doing an excercise about infimums and supremums and I've seen different examples of proving a =inf(S) $\iff \forall \epsilon >0\exists s\in S\colon a+\epsilon>s$ and to me they just seem to be making it up like:
Let S := { $x+1, x\in (0,1)$}
Here the infimum is clearly 1 , so let's prove the hypothesis inf(S) = 1. Which means $1+\epsilon >x+1 $ for some $x \in (0,1)$.
Then it usually goes as, solving for x such that if $0<x<\epsilon, \forall\epsilon >0$ , then you add 1 on both sides and reach the first inequality and it's "proven", but I don't see it, what about if we believe inf(S) = -1.
Doing the same, obviously -1 is a lower bound of S and then, $-1 + \epsilon > x+1$, solve for x $if, x<\epsilon -2 $ then , I know I can't say that -1 is the infimum but it doesn't make sense to me.