intervals in real number set

2018-01-22 09:54:09

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...

    2018-01-22 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

    2018-01-22 11:18:30