# Approximate countable union of open intervals by a finite covering of open intervals.

Given an open interval $$I=(a, b)\subset \mathbb R$$ for some $$a, b\in \mathbb R$$ and a countable union of its open subintervals $$U=\bigcup\limits_{i=1}^\infty J_i$$ such that $$\sum\limits_{i=1}^\infty L(J_i)<\infty$$($$L(J_i)$$ is the length of the interval $$J_i$$), for any $$\epsilon>0$$ can we choose an finite union of open subintervals $$K=\bigcup_{i=1}^n K_i\subset I$$ and an integer $$N$$ such that $$\bigcup\limits_{i=N+1}^\infty J_i\subset K\text{ and } \left| L(K)-\sum_{i=N+1}^\infty L(J_i)\right|<\epsilon?$$

I came up with this question when I was attempting to prove a problem in real analysis, and this would finish the proof if it is true.

• Can you assume the $J_i$ are pairwise disjoint? (Otherwise $\sum L(J_i)$ could be infinite.) – David Mitra Apr 20 at 8:02
• I can assume that the sum $\sum L(J_i)<\infty$. – William Sun Apr 20 at 8:06
• Even if the $J_i$ were pairwise disjoint, I don't think this is possible. Take $L_1$, $L_2$, $\ldots$ to be disjoint open intervals in $(0,1/8)$ and $R_1$, $R_2$, $\ldots$ to be disjoint open intervals in $(7/8,1)$. Take $(J_i)=R_1,L_1,R_2,L_2,\ldots$. Then $|L(K)-\sum_{i={N+1}}^\infty L(J_i)|$ is always at least $1/2$. – David Mitra Apr 20 at 8:10
• We can assume that the set $K$ is a finite union of open intervals instead of a single one. Sorry it takes efforts to examine the least condition I need in proof and I just found out. – William Sun Apr 20 at 8:16
• Where is $U$ used again? Are the $J_i$ contained in $I$? – zhw. Apr 20 at 18:43

Let $$(q_i)$$ be an enumeration of the rationals in $$I$$. Any such enumeration has the property that the tails of $$(q_i)$$ are always dense in $$I$$. (Any point of $$I$$ is the limit of a sequence of distinct rationals - a subsequence of $$(q_i)$$. But a tail of $$(q_i)$$ will contain a tail of the subsequence.)
Let $$\epsilon > 0$$ and let $$J_i = (q_i - \epsilon 2^{-i-1}, q_i + \epsilon 2^{-i-1})\cap I$$ Then $$L(J_i) \le \epsilon 2^{-i}$$, so for any $$N > 0,$$ $$\sum_{i>N} L(J_i) \le \sum_{i>N} \epsilon 2^{-i} = \epsilon 2^{-N} < \epsilon$$
By choosing $$\epsilon$$ small enough, we can make $$\sum_{i} L(J_i)$$ as small as we please. In particular, make it smaller than $$\frac{b-a}2$$.
Because $$(q_i)_{i > N} \subseteq \bigcup_{i>N}J_i$$ is dense in $$I$$, any finite union of open intervals $$K_N \subseteq I$$ containing $$\bigcup_{i>N}J_i$$ can miss at most a finite number of points of $$I$$, so $$L(K_N) = b-a$$, a fixed value with respect to $$N$$. Therefore, for all $$N$$,$$\left| L(K_N)-\sum_{i>N} L(J_i)\right| > \frac {b-a}2$$