A open subset of $\Bbb R$

Question

Given the definitions in Open Subsets of open sets

I need to prove that $\{x \in \Bbb R : |x|>2\}$ is open in $(\Bbb R , d_E)$

This seems to be true, however I don't know how to prove it without referencing close sets.

Answer 1

Assuming you want to prove that $S:=\mathbb R \setminus [-2,2]$ is open.
Given $x\in S$ set $\delta := \min(|x-2|, |x+2|)$ the "Distance of $x$ to the complement". Since $x\neq 2$ and $x\neq -2$, $\delta > 0$ so now $$(x-\frac\delta2, x+\frac\delta2) = B_{\frac\delta2}(x) \subset S$$ Thus $x$ is an interior point. Since $x\in S$ was arbitrary, $S = S^\circ$ and thus $S$ is open.

Answer 2

For this kind of questions is useful theorem(you might take it as definition of continuous function)

Let $f:X \rightarrow Y$ be continuous function. Than for any open $U\subset Y$ is $f^{-1}(U)$ open.

Now take continuous function $f(x) = |x|$. open interval $(2,\infty]$ is open therefore $f^{-1}( (2,\infty])$ is open.