Metrik Uzaylarda Komşuluk

Question

$\mathbb{N}$ doğal sayılar kümesi olmak üzere

$$d(x,y):=\Big{|}\frac{1}{x}-\frac{1}{y}\Big{|}$$ kuralı ile verilen $$d:\mathbb{N}^2\to\mathbb{R}$$ metrik fonksiyonu için $$n\in\mathbb{N}\Rightarrow B\left(n,\frac{1}{n(n+1)}\right)=\{n\}$$ olduğunu gösteriniz.

Answer 1

$$B\left(n,\frac{1}{n(n+1)}\right)$$

$$=$$

$$\left\{x\Big{|}d(x,n)<\frac{1}{n(n+1)},x\in\mathbb{N}\right\}$$

$$=$$

$$\left\{x\Big{|}\big{|}\frac{1}{x}-\frac{1}{n}\big{|}<\frac{1}{n(n+1)},x\in\mathbb{N}\right\}$$

$$=$$

$$\left\{x\Big{|}-\frac{1}{n(n+1)}<\frac{1}{x}-\frac{1}{n}<\frac{1}{n(n+1)},x\in\mathbb{N}\right\}$$

$$=$$

$$\left\{x\Big{|}\frac{1}{n}-\frac{1}{n(n+1)}<\frac{1}{x}<\frac{1}{n}+\frac{1}{n(n+1)},x\in\mathbb{N}\right\}$$

$$=$$

$$\left\{x\Big{|}\frac{1}{n+1}<\frac{1}{x}<\frac{n+2}{n(n+1)},x\in\mathbb{N}\right\}$$

$$=$$

$$\left\{x\Big{|}\frac{1}{n+1}<\frac{1}{x},x\in\mathbb{N}\right\}\cap \left\{x\Big{|}\frac{1}{x}<\frac{n+2}{n(n+1)},x\in\mathbb{N}\right\}$$

$$=$$

$$\left\{x\Big{|}x<n+1,x\in\mathbb{N}\right\}\cap \left\{x\Big{|}x>\frac{n(n+1)}{n+2},x\in\mathbb{N}\right\}$$

$$=$$

$$\{1,2,3,\ldots, n\}\cap \{n,n+1,n+2,\ldots\}$$

$$=$$

$$\{n\}.$$