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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?
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...
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

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...
20180122 10:50:41 
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
20180122 11:18:30