$$[a,b]\subseteq\bigcap_{n\in\mathbb{N}}\left (a-\frac{b-a}{n},b+\frac{b-a}{n}\right)$$
ve
$$[a,b]\supseteq\bigcap_{n\in\mathbb{N}}\left (a-\frac{b-a}{n},b+\frac{b-a}{n}\right)$$ olduğunu göstermek yeterli ve gereklidir.
$$(a,b\in\mathbb{R})(a<b)$$
$$\Rightarrow$$
$$(\forall n\in\mathbb{N})\left(a-\frac{b-a}{n}< a\right)\left(b< b+\frac{b-a}{n} \right)$$
$$\Rightarrow$$
$$(\forall n\in\mathbb{N})\left([a,b]\subseteq \left (a-\frac{b-a}{n},b+\frac{b-a}{n}\right)\right)$$
$$\Rightarrow$$
$$[a,b]\subseteq\bigcap_{n\in\mathbb{N}}\left (a-\frac{b-a}{n},b+\frac{b-a}{n}\right)\ldots (1)$$
$$-----------------------------------$$
$$x\notin[a,b]$$
$$\Rightarrow$$
$$x<a\vee x>b$$
$$\Rightarrow$$
$$0<a-x\vee 0<x-b$$
$$\overset{\text{Arşimet Özelliği}}{\Rightarrow}$$
$$(\exists n_1\in \mathbb{N})(b-a\leq n_1(a-x))(\exists n_2\in \mathbb{N})(b-a\leq n_2(x-b))$$
$$\Rightarrow$$
$$(\exists n_1\in \mathbb{N})\left(x\leq a-\frac{b-a}{n_1}\right)(\exists n_2\in \mathbb{N})\left(x\geq b+\frac{b-a}{n_2}\right) $$
$$\Rightarrow$$
$$(n_0:=\max\{n_1,n_2\}\in \mathbb{N})\left(x\leq a-\frac{b-a}{n_0}\right)\left(x\geq b+\frac{b-a}{n_0}\right) $$
$$\Rightarrow$$
$$(n_0:=\max\{n_1,n_2\}\in \mathbb{N})\left (x\in \left(-\infty , a-\frac{b-a}{n_0}\right]\cup \left[b+\frac{b-a}{n_0},\infty\right)\right)$$
$$\Rightarrow$$
$$(n_0:=\max\{n_1,n_2\}\in \mathbb{N})\left (x\notin \left (a-\frac{b-a}{n_0},b+\frac{b-a}{n_0}\right)\right)$$
$$\Rightarrow$$
$$x\notin \bigcap_{n\in\mathbb{N}}\left (a-\frac{b-a}{n},b+\frac{b-a}{n}\right).$$
O halde $$[a,b]\supseteq\bigcap_{n\in\mathbb{N}}\left (a-\frac{b-a}{n},b+\frac{b-a}{n}\right)\ldots (2)$$
$$-----------------------------------$$
$$(1),(2)\Rightarrow [a,b]=\bigcap_{n\in\mathbb{N}}\left (a-\frac{b-a}{n},b+\frac{b-a}{n}\right).$$