Is $\inf A$ and $\sup A$ belong $\bar A$?

Let $$A$$ be a anonempty and bounded subset of $$\mathbb{R}$$.

Now take $$A= (0,1)$$ in the discrete topology of $$\mathbb{R}$$.

My question is that :

Is $$\inf A$$ and $$\sup A$$ belong $$\bar A$$ ?

I thinks Yes

Is its true ?

No, it is not true. When we are working with the discrete topology, the closure of any set is that set itself. But $$\sup(0,1)=1\notin(0,1)$$ and $$\inf(0,1)=0\notin(0,1)$$.
Another angle to this questions is that $$\inf$$ and $$\sup$$ are concepts that require an order on the underlying set to be defined. A topology, however, has no structure that allows it to induce an order on the set it is defined on.
It just so happens that the usual order on $$\mathbb R$$ and the usual topology on $$\mathbb R$$ are 'closely linked'. That can be seen by the fact that the open intervals $$(a,b)$$ form a basis for the usual topology on $$\mathbb R$$, and any $$c\in (a,b)$$ has the property
$$a \le c \le b.$$
Such a connection does not exist between the usual order on $$\mathbb R$$ and the discrete topology on $$\mathbb R$$, so a priori there shouldn't be an expectation that $$\inf$$ and $$\sup$$ of a set have anything to do with the topological closure of that set under the discrete topology.