intervals in real number set in the calculus, we define the open and closed intervals by for all $a,b \in\ R$ then $$I=[a,b]= \{x \in R\ |a\le x\le b \}$$ is called closed interval of real numbers set , and $$I=(a,b)=\{x\in R\ |a\lt x\lt b\}$$ 
if $$I=[a,a]=\{a\}$$ is a singleton set,  and $$I=(a,a)=\emptyset$$ i asking, is that true? and how can prove it?
 A: That is one of these typical questions where everyone knows that the statement is true, but it is hard to write a formal proof. Or at least it is hard to see when a proof is really a proof. 
How about this: 
\begin{align*}
I=[a,a]&=\{x\in R\ |a≤ x≤ a\} \\&= \{x\in R\ | x≤ a\}∩\{x\in R\ |a≤ x\}  =\{a\}
\end{align*}
\begin{align*}
I=(a,a)&=\{x\in R\ |a\lt x\lt a\} \\&= \{x\in R\ | x\lt a\}∩\{x\in R\ |a\lt x\}  =∅
\end{align*}
Or you could use the standard method of showing set equality:
"$⊂$": Let $x∈I=[a,a]$, then it is $a≤x≤a$. Therefore it is $x=a∈\{a\}$.
 ⇒ $I⊂\{a\}$.
"$\supset$":  Let $x∈\{a\}$, then $x=a$. Then it holds $a≤x≤a$ and therefore $x∈I$.
 ⇒ $I\supset\{a\}$.
Thus  $I=\{a\}$. 
Same for the open set.
A: Following the definitions you mention:


*

*$[a,a]$ contains all the $x$-values satisfying $a \color{blue}{\le} x \color{blue}{\le} a$, so...

*$(a,a)$ contains all the $x$-values satisfying $a \color{red}{<} x \color{red}{<} a$, so...



I'm not sure what exactly you're looking for. Suppose $(a,a) \ne \emptyset$ which means you would have at least one $x \in (a,a)$. Because of the total order, you have $x \le a$ or $x \ge a$ but these directly contradict the (defining) conditions $x>a$ and $x<a$ respectively; so in fact $(a,a) = \emptyset$.
A: It is :
a)if $I=[\alpha,\alpha]$ let $x\in I$ then $\alpha\leq x \leq \alpha$ so $x= \alpha$. Hence $I=\{\alpha\}$. :P
b)if $I=(\alpha,\alpha)$ if $I\neq \emptyset$ then take $x\in I$ and you get $\alpha<x <\alpha$ which cant be true!
