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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.

Community
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Mathman
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  • Pick a point $x$ in your set, i.e. $|x| > 2$. Then show that there's some $\epsilon$-ball around $x$ which fully lies within your set. Hint: If $2 < x$, then $x - 2 > 0$... – fgp Mar 24 '14 at 11:37

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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.

AlexR
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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.

tom
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    This may be begging the question somewhat. After all, if they know that $(2, \infty)$ is open, then they probably also know that $(-\infty, -2)$ is open, and so since the union of open sets is open... – Simon Rose Mar 24 '14 at 12:00
  • Ups silly me, I should read and think first :D I just saw question of type "is ${ x : f(x)<2 }$ open?". So I typed standard answer – tom Mar 24 '14 at 12:03