[a,b]⊆⋂n∈N(a−b−an,b+b−an)
ve
[a,b]⊇⋂n∈N(a−b−an,b+b−an) olduğunu göstermek yeterli ve gereklidir.
(a,b∈R)(a<b)
⇒
(∀n∈N)(a−b−an<a)(b<b+b−an)
⇒
(∀n∈N)([a,b]⊆(a−b−an,b+b−an))
⇒
[a,b]⊆⋂n∈N(a−b−an,b+b−an)…(1)
−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
x∉[a,b]
⇒
x<a∨x>b
⇒
0<a−x∨0<x−b
Arşimet Özelliği⇒
(∃n1∈N)(b−a≤n1(a−x))(∃n2∈N)(b−a≤n2(x−b))
⇒
(∃n1∈N)(x≤a−b−an1)(∃n2∈N)(x≥b+b−an2)
⇒
(n0:=max
\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).