$\epsilon$-criterion of infimum We have the statement:
$I$ is the infimum of set $M\iff$ $I$ is a lower bound and for each $\epsilon>0$ there exists an element $m\in M$ such that $I+\epsilon>m$.
I asked the tutor if we can replace the strict "$>$" by "$\geq$"?
He said no, because this would mean that $I$ would be always an element of $M$ which is not necessarily true.

Firstly, I think that my tutor is simply wrong and secondly I still think that it doesn't matter if we write $>$ or $\geq$. The reason is that I don't need this fact in the proof:
$I$ is the greatest lower bound so if there existed no $m\in M$ such that $I+\epsilon\geq m$, then $I<I+\epsilon$ would be a greater lower bound, which is a contradiction. Now we assume that $I$ is a lower bound and for each $\epsilon>0$ there exists an element $m\in M$ such that $I+\epsilon\geq m$. Let be $I'$ another lower bound with $I':=I+\frac{\epsilon}{2}$. This also leads to a contradiction because we find a $m\in M$ wich satisfies $I'>I+\frac{\epsilon}{4}\geq m$.
Am I wrong? I still don't see any reason why we must use $>$?
 A: You are correct and it doesn't matter.  It certainly wouldn't imply $I$ would always have to be an element of $M$ because $I + \varepsilon > m$ implies $I + \varepsilon \ge m$ so $I = \inf M$ would still imply $I$ is a lower bound for $M$ and for all $\varepsilon > 0$ there exists $m \in M$ with $I + \varepsilon \ge m$.  Your proof for the converse direction is correct as well.
A: $I$ is a lower bound and for each $\epsilon>0$ there exists an element $m\in M$ such that $I+\epsilon>m$. $\implies$ $I$ is a lower bound and for each $\epsilon>0$ there exists an element $m\in M$ such that $I+\epsilon\geq m$. Because "Greater or Equal" ($\geq$) is a less strict condition.

Blockquote

Now we want to prove that: $I$ is a lower bound and for each $\epsilon>0$ there exists an element $m\in M$ such that $I+\epsilon\geq m$. $\implies$ $I$ is a lower bound and for each $\epsilon>0$ there exists an element $m\in M$ such that $I+\epsilon> m$.  With $\geq$ we have to consider two possible cases (OR): 
1)$I+\epsilon> m$ then the implication it's trivial. 
2)$I+\epsilon= m$. For the Hypothesis we can always take  $\epsilon '>0$ such that $\epsilon>\epsilon'$ for which $I+\epsilon '\geq m'$. $I+\epsilon>I+\epsilon'\geq m'\implies \exists m'\in M $ such that $I+\epsilon>m'$
