Each open interval is a countable union of closed sets.

Question

How can we prove this $$(a, b)= \cup_{n=1}^\infty [a + 1/n, b - 1/n],$$ ?

First: it is clear that $\left [ a + \frac{1}{n}, b - \frac{1}{n}\right ] \subseteq (a, b)$ for every n, then $\cup_{n=1}^\infty [a + 1/n, b - 1/n] \subseteq (a, b) $. what about the other inclusion?

What have you tried? The statement can be (partially) rewritten as: for any $x$ such that $a<x$ there is $n$ such that $a+1/n<x$. — freakish, Dec 10 '19 at 11:39
one direction is clear for me, this one $\subset$ (sorry I meant the other inclusion but I do not know how to write it in latex) @freakish what about the other direction? — Intuition, Dec 10 '19 at 12:23
@Secretly if $x$ is such that $a<x<b$ then there is $n$ and there is $m$ such that $a+1/n\leq x$ and $x\leq b-1/m$. Let $k=max(n,m)$. Then $x\in [a+1/k, b-1/k]$. — freakish, Dec 10 '19 at 13:04

score 3 · Accepted Answer · answered Dec 10 '19 at 13:50

Let $x\in]a,b[$. Then $a<x<b$. Let $\epsilon_1 = x-a$ and $\epsilon_2 = b-x$. Choose $n^*\in\mathbb{N}$ such that $\frac{1}{n^*} < \min\{\epsilon_1,\epsilon_2\}$. Then $x\in [a+\frac{1}{n^*},b-\frac{1}{n^*}]\subset \bigcup\limits_{n=1}^\infty [a+\frac{1}{n},b-\frac{1}{n}]$.

Each open interval is a countable union of closed sets.

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