If a set $A$ is a bounded, nonempty subset of $\mathbb R$, prove that $\sup(A)\in\partial A$ If a set $A$ is a bounded, nonempty subset of $\mathbb R$, prove that $\sup(A)\in\partial A$. 
I know how to prove that $\sup(A)\in Cl(A)$, so i'm thinking if I prove that $\sup(A)\in Cl(X/A)$ then I'll be finished because the boundary of $A$ is defined to be $\partial A=Cl(A) \cap Cl(X/A)$. But I don't know how to prove this.
Also, if I show that $\sup(A)\notin int(A)$ then I'll be finished, since $\partial A$= $Cl(A)/int(A)$ but I don't know how to prove this either.  
Is there an easier way? 
 A: Fix $\varepsilon>0$ and define $\ell:=\sup(A)$. Clearly $(\ell,\ell+\varepsilon) \cap A=\emptyset$ so
$$
(\ell-\varepsilon,\ell+\varepsilon) \cap A^c \neq \emptyset.
$$
Moreover, since $\ell$ is the sup of $A$ then $(\ell-\varepsilon,\ell]\cap A \neq \emptyset$ so
$$
(\ell-\varepsilon,\ell+\varepsilon) \cap A \neq \emptyset.
$$
Hence, since $\varepsilon$ is arbitrary, $\ell \in \partial A$.
A: $\DeclareMathOperator{\Int}{Int}$
Let's prove $\sup A \notin \Int A$.
Assume $\sup A \in \Int A$. Since $\Int A$ is an open set, there exists $r > 0$ such that $\langle \sup A - r, \sup A + r\rangle \subseteq \Int A \subseteq A$. But $\sup A + \frac{r}{2} \in A$ and it is greater than $\sup A$, which is a contradiction.
A: $x \in \partial A$ if forall $\epsilon>0$ we have $B(x,\epsilon) \cap A \neq \emptyset$ and $B(x,\epsilon) \cap A^c \neq \emptyset$
Now in the real line an open ball with center $x$ and radious $\epsilon$ is an interval $(x- \epsilon,x+\epsilon)$
$A$ is bounded above so exists $s=\sup A$
We have to prove that $\forall \epsilon>0, (s-\epsilon,s+\epsilon) \cap A \neq \emptyset$
Now assume that $\exists \epsilon_0>0$ such that $(s-\epsilon_0,s+\epsilon_0) \cap A=\emptyset$
Fror the property of supremum exists $a \in A$ such that $$s- \epsilon_0<s -\frac{\epsilon_0}{2}<a \leq s$$
So $a \in (s-\epsilon_0,s+\epsilon_0)$ which is a contradiction.
Now assume that $\exists \epsilon_0>0$ such that $(s-\epsilon_0,s+\epsilon_0) \cap A^c=\emptyset$
Thus $(s-\epsilon_0,s+\epsilon_0) \subseteq A$ so $s+\frac{\epsilon_0}{2} \in A$
which is a contradiction because $s=\sup A$
We are done.
This not the quickest proof or the easiest and it is elemtary and uses only the definition of the boundary and the propert of supremum of a set in $\mathbb{R}$ which is a very useful property for someone to remember.
