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"Every open subset of $\mathbb{R}$ can be written as a countable union of open intervals."

Why is the above statement true? Is it because the topology of $\mathbb{R}$ is defined to be the topology generated by intervals of the form $(a,b)?$

I want, next, to prove that $$ (r,\infty)=\bigcup_{n=1}^{\infty}\left(r + \dfrac{1}{2^n},\infty \right). $$ My attempt at a proof: To show that $\bigcup_{n=1}^{\infty}\left(r + \dfrac{1}{2^n},\infty \right)\subseteq (r,\infty)$, is trivial: $x>r+\frac{1}{2^n}>r$, for some $n\in \mathbb{Z}^+$.

To exhibit the other direction, can I use the fact that $n\leq 2^n$ (provable by induction) and then use the Archimedean property?

EEEB
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3 Answers3

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To show: $(r,\infty)\subset \bigcup_{n\in \mathbb{N}}(r + \frac{1}{2^n},\infty)$.

Proof: Let $x\in (r,\infty)$. This implies $x > r$, in other words $x-r > 0$. When we choose $$n_0 > \text{log}_2 (\frac{1}{x-r}),$$ we have $$x-r > \frac{1}{2^{n_0}}.$$ In other words, $x \in (r+\frac{1}{2^{n_0}}, \infty)$ and therefore $$x \in (r + \frac{1}{2^{n_0}},\infty) \subset \bigcup_{n\in \mathbb{N}}(r + \frac{1}{2^n}, \infty).$$ The claim now follows since $x$ was arbitrary.

Fritz Hefter
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The first statement is true because a neighborhood in $\mathbb{R}$ is an open interval and $\mathbb{R}$ is a second-countable space. Your idea for the proof is fine, let $x\in ]r,\infty[$. Then $x > r$ and so $\exists n\in\mathbb{N}$ such that $x-r\geq\frac{1}{n}\geq \frac{1}{2^n}$. Thus $x\in]r+\frac{1}{2^n}, \infty[$ and so $]r,\infty[\subset \bigcup_{n=1}^{\infty}\left]r + \dfrac{1}{2^n},\infty \right[$.

Community
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Jürgen Sukumaran
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Observe that if $x \in (r, \infty)$ then $x-r>0.$ So by the Archimedean property of real numbers $\exists$ $k \in \Bbb N$ such that $$k(x-r)>1 \implies 0<\frac {1} {k} < x-r.$$ Since $\frac {1} {2^n} < \frac {1} {n},$ for all $n \in \Bbb N$ we have $$0<\frac {1} {2^k} < x-r \implies r<r + \frac {1} {2^k} < x< \infty.$$ This shows that $$x \in \left ( r + \frac {1} {2^k} , \infty \right ) \subseteq \bigcup\limits_{n=1}^{\infty} \left ( r + \frac {1} {2^n} , \infty \right ).$$

Therefore $$(r,\infty) \subseteq \bigcup\limits_{n=1}^{\infty} \left ( r + \frac {1} {2^n} , \infty \right )$$

as required.

little o
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