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So, I have proven already that every open set in $\mathbb{R}$ can be written as a countable union of disjoint open intervals; i.e., that $\mathcal{O} = \bigcup_{i=1}^\infty (a_{x_i}, b_{x_i})$.

Now, consider the closed interval $\left [ a + \frac{1}{n}, b - \frac{1}{n}\right ] \subset (a, b)$.

Since the midpoint of $(a,b)$ is $\frac{b-a}{2}$, we need $\displaystyle n > \frac{2}{b-a}$.

Otherwise, $\displaystyle a + \frac{b-a}{2} = a + \frac{1}{ \frac{2}{b-a}} = b - \frac{1}{ \frac{2}{b-a}} = b - \frac{b-a}{2}$, and our interval will be degenerate.

Note that $\forall (a,b) \in \mathcal{O}$,

$\displaystyle \bigcup_{n = \frac{b-a}{2}+1}^{\infty} \left[ a + \frac{1}{n}, b - \frac{1}{n} \right] = (a,b)$.

So, we have $\displaystyle \mathcal{O} = \bigcup_{i=1}^\infty \left ( \bigcup_{n = \frac{b-a}{2} + 1}^\infty \left[ a_{x_i}+\frac{1}{n}, b_{x_i}-\frac{1}{n} \right] \right)$.

But, I don't like this. It looks like I'm still just taking a union of open sets. Please help me get this to look like the union of closed sets.

(For unbounded intervals, I know what to do - it should follow from this relatively easily).

Michael Hardy
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  • That's a countable union already. What exactly do you want? –  Sep 21 '15 at 04:45
  • No, you are taking a union of closed sets. You can rewrite your union as a single union of only closed sets. – 5xum Sep 21 '15 at 04:45
  • I guess because to me it looks like first, I've got the inner union, which gives me a bunch of open sets, and then the outer union is being taken over those resulting open sets. –  Sep 21 '15 at 04:46
  • Each $(i,n)$ in your final expression for $\mathcal{O}$ is an element of $\mathbb{N} \times \mathbb{N}$, which is countable. You could write $$\mathcal{O} = \bigcup_{(i,,n) \in S} \left[a_{x_i} + \frac{1}{n}, b_{x_i} - \frac{1}{n}\right]$$ where $S$ is a subset of $\mathbb{N} \times \mathbb{N}$. –  Sep 21 '15 at 05:02
  • The general idea is right, some details are not quite right. For example $\frac{b-a}{2}$ is generally not an integer. You have a bunch of $(a_i,b_i)$. To take care of this for a particular $i$, you want to use the union of the $[a_i+1/n,b_i-1/n]$, where $n$ ranges over all positive integers greater than say $\frac{3}{b_i-a_i}$. – André Nicolas Sep 21 '15 at 05:11
  • @André Nicolas, Why greater than $\frac{3}{b_{i}-a_{i}}$? Why not greater than $\frac{2}{b_{i}-a_{i}}$? The 3 seems kind of arbitrary to me. –  Sep 21 '15 at 12:32
  • It is arbitrary, I should have chosen $10$. The reason I don't like $2$ is that depending on details of wording it might not quite work (though it does in this case). For basically the same reason, in $\epsilon$-$N$ proofs I don't want to expend any energy finding a "small" $N$. – André Nicolas Sep 21 '15 at 13:20

2 Answers2

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This property actually holds in any metric space:

In a metric space, each closed set is a countable intersection of open sets and each open set is a countable union of closed sets.

Proof. Let $F$ be a closed set of the metric space $(E, d)$. Set, for each $n > 0$, $$ U_n = \bigcup_{x \in F}\ \bigl\{y \in E \mid d(x,y)< {1 \over n} \bigr\} $$ Then each $U_n$ is open and it is not difficult to see that $F = \bigcap_{n > 0} U_n$. This proves the first part of the statement. The second part follows by taking complements.

Tarc
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J.-E. Pin
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If you really want to see $"\bigcup_1^∞"$, it can be done (but it ain't pretty). Let $\iota:\Bbb N → \Bbb N\times \Bbb N$ be a bijection, with $\iota = (\iota_1,\iota_2)$. Then $$ \mathcal{O} = \bigcup_{\substack{k=1\\ \frac{1}{\iota_2(k)} \leq \frac{b(k)-a(k)}{3}} }^∞ \left[a(k) + \frac{1}{\iota_2(k)} \quad , \quad b(k) - \frac{1}{\iota_2(k)} \right] $$ where $a(k) = a_{x_{\iota_1(k)}}$ and $b(k) = b_{x_{\iota_1(k)}}$.

Calvin Khor
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  • I don't really care about starting the index at 1. But, thank you for letting me know how it could be done if I wanted it. –  Sep 21 '15 at 12:30