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?

  • $\begingroup$ @Riley That is true, however, in the case that a maximum exists, the maximum is also the minimal upper bound to the set because every element of the set is equal to or less than the maximum. If there weren't a maximum, then I wouldn't consider it as a method to get to the Supremum. $\endgroup$
    – Davy M
    Nov 15, 2017 at 17:21
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    $\begingroup$ Sorry, I deleted my comment because I thought you were talking about the supremum of $(0,5/3)$, not $\mathbb{Z}\cap (0,5/3)$. In the latter, a maximum exists. $\endgroup$
    – Riley
    Nov 15, 2017 at 17:24

1 Answer 1


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?

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    $\begingroup$ Just to mention the general statement: let $S := \mathbb{Z} \cap (a,b)$, then $sup S = max S = \lfloor b \rfloor$ if $\lfloor b \rfloor \in (a,b)$ and $b \notin \mathbb{Z}$; if $b \in \mathbb{Z}$ and $b-1 \in (a, b)$, then $sup S = max S = b-1$; otherwise $S = \emptyset$ and hence $sup S = -\infty$ by convention. $\endgroup$ Nov 15, 2017 at 17:38

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