The problem I'm working on involves more than just the question I've put in the title. I'm proving that $1$ is the supremum of the set $S$ where $S$ is $\text {the set of all integers in the open interval}$ (0, 5/3).
This set obviously only contains one element, 1, and I could prove this using the fact that this interval only has one element, so it is the maximum, and if there's a maximum it is also the supremum, yadda yadda yadda. But it got me curious about how to prove the maximum of a set that is defined by an interval where we're talking about the set of integers on an interval where the defined with non-integer numbers.
So, in the case of the set $S$ where $S$ is $\text {the set of all integers on the open interval}$ (a,5/2) where $a$ is arbitrary but less than 2, how could I prove that 2 is the maximum of $S$?
My attempt would be to show first that there is no value $x \in S$ such that $ x \gt 2$ , then show that $2 \in S$, but is there a better way to do this?
The "greatest integer function" seems to be an ideal tool for this job. We define $\lfloor x\rfloor$ as the greatest integer less than or equal to $x$. Since $\frac52$ is not an integer, we have that $\left\lfloor\frac52\right\rfloor<\frac52$, and then we note that it equals $2$, which is greater than $a$, and thus in the interval.
Does that work?