Question about proving singleton set has outer measure zero

Question

Proof:

$$\{x\}\subset \left(x-\frac{1}{n},x+\frac{1}{n}\right)$$ for all $n \in \mathbb{N}$

$$\{x\}\subset \cap_{n=1}^\infty \left(x-\frac{1}{n},x+\frac{1}{n}\right)$$ where $$\cap_{n=1}^\infty \left(x-\frac{1}{n},x+\frac{1}{n}\right)$$ is a $G_\delta$ set

and hence $$\lambda^*(\{x\}) \le \inf_{n \in \mathbb{N}}l\left(x-\frac{1}{n},x+\frac{1}{n}\right)=\inf_{n \in \mathbb{N}} \frac{2}{n}=0.$$

I don't understand why

$$\{x\}\subset \left(x-\frac{1}{n},x+\frac{1}{n}\right)$$ for all $n \in \mathbb{N}$

How do I know this is true?

Answer 1

This just says $$x-\frac1n<x<x+\frac1n$$ which is equivalent to $$-\frac1n<0<\frac1n$$ for all $n\in\mathbb N$. You believe that’s true, right?

Answer 2

Because, for each natural $n$, $x-\frac1n<x<x+\frac1n$. Therefore, $x\in\left(x-\frac1n,x+\frac1n\right)$, which is equivalent to asserting that $\{x\}\subset\left(x-\frac1n,x+\frac1n\right)$.