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.


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