Show that $I=(−\infty,\sup I]$ as $I$ is bounded above but not bounded below

Question

Let $I$ be a non-empty interval. Suppose $I$ is not bounded below, I is bounded above, and $\sup I ∈ I$. Show that $I=(−\infty,c]$, where $c=\sup I$.

My attempt:($\Longrightarrow$) Since $I$ is not bounded below but bounded above, to see that $I=(−\infty,c]$, let $x \in (-\infty,c]$, then there exists two real numbers $x_1$ and $x_2$ such that $x_1<x<x_2\leq c:= \sup I$. But since $I$ is an interval, so we have $(-\infty,c] \subset I$.

($\Longleftarrow$)It is trivially true by showing $x \in (-\infty,c]$ if $x\in I$. Since the interval $I$ is bounded above but not bounded below, denote $c:=\sup I$ such that $x \leq c := \sup I$. Then we have $ I \subset (-\infty,c]$.

Combining these we can obtain $I=(-\infty, c]$ where $c=\sup I$.

I feel weird for my proof for the statement above because it is trivial to me. Is that any case that I have not been discussed in my proof or my proof was getting in the wrong direction?

Answer 1

In the first line of your answer there is a typo. In the second line you should say three exists $x_1<x$ such that $x_1 \in I$. You can take $x_2=c$.

Answer 2

Kavi's comments are valid ;and after "But" you might add-and and interval is convex (contains all points between any two ) .However you are right to think that the result is trivial basically for the as you said .