# Relatively open subsets of the unit interval can be expressed uniquely as unions of disjoint open intervals

From "An Introduction to Lebesgue Integration and Fourier Series" by Howard J. Wilcox and David L. Myers:

7.2 Theorem: Every non-empty open set $$G \subset \mathbb{R}$$ can be expressed uniquely as a finite or countably infinite union of pairwise disjoint open intervals.

The meaning of 'uniquely' is defined in Exercise 9.11: Prove uniqueness in Theorem 7.2; if $$G = \cup_{n} I_{n} = \cup_{k} J_{k}$$, where $$I_{n}$$, $$J_{k}$$ are open intervals and $$I_{n} \cap I_{m} = \varnothing$$, $$J_{k} \cap J_{l} = \varnothing$$ for all $$n$$, $$m$$, $$k$$, $$l$$, then show that for every $$n$$ there is a $$k$$ such that $$I_{n} = J_{k}$$, and for every $$k$$ there is an $$n$$ such that $$J_{k} = I_{n}$$.

The author then states:

Whenever we use open sets in the unit interval $$E$$, we shall use "open" to mean open in the relative topology of $$E$$ as a subspace of $$\mathbb{R}$$. That is, $$G$$ is open in $$E$$ if and only if $$G = E$$ intersected with some open subset of $$\mathbb{R}$$. Then it is clear that the theorem holds in the relative topology of $$E$$, in the sense that if $$G$$ is open in $$E$$, there is a unique representation $$G = \cup_{i} I_{i}$$, where the $$I_{i}$$ are disjoint intervals of the form $$(a_{i}, b_{i}) \cap E$$.

I am attempting to prove that this holds. I believe I have been able to demonstrate that every non-empty subset $$G$$ of $$[0,1]$$ that is open in the relative topology of $$[0,1]$$ can be expressed as a finite or countably infinite union of pairwise disjoint intervals of the form $$(a_{i}, b_{i}) \cap [0,1]$$:

Let $$G$$ be an arbitrary non-empty subset of $$[0,1]$$ that is open in the relative topology of $$[0,1]$$. Then there exists a non-empty set $$U$$, open in $$\mathbb{R}$$, such that $$G = U \cap [0,1]$$. By Theorem 7.2, $$U$$ can be expressed as a finite or countably infinite union of pairwise disjoint open intervals. That is, $$U = \cup_{i} (a_{i}, b_{i})$$. Then $$G = U \cap [0,1] = [\cup_{i} (a_{i}, b_{i})] \cap [0,1] = \cup_{i} [(a_{i}, b_{i}) \cap [0,1]]$$.

I am struggling to prove uniqueness. The problem seems to be that if $$G = U \cap [0,1]$$, then although $$U$$ can be expressed uniquely as a finite or countably infinite union of pairwise disjoint open intervals, it does not appear that this implies that $$U \cap [0,1]$$ can.

EDIT

Let $$G$$ be an arbitrary non-empty subset of $$[0,1]$$ that is open in the relative topology of $$[0,1]$$. Let $$I$$ and $$J$$ be finite or countably infinite sets of pairwise disjoint intervals open in the relative topology of $$[0,1]$$ such that $$G = \cup I$$ and $$G = \cup J$$.

Suppose $$0 \in G$$ and $$1 \notin G$$. Then $$0 \in \cup I$$ and $$0 \in \cup J$$. Since the intervals in $$I$$ are disjoint, then there is a single interval $$I_{0} \in I$$ that contains $$0$$. Since the intervals in $$J$$ are disjoint, then there is a single interval $$J_{0} \in J$$ that contains $$0$$.

$$I_{0} = [0, e)$$ for some $$e \in (0,1)$$. $$J_{0} = [0, f)$$ for some $$f \in (0,1)$$.

Let $$a < 0$$. Let $$I_{0}' = (a, 0] \cup [0, e)$$. Then $$I_{0}'$$ is an open interval. Let $$J_{0}' = (a, 0] \cup [0, f)$$. Then $$J_{0}'$$ is an open interval.

Let $$I'$$ be the set formed by replacing $$I_{0}$$ by $$I_{0}'$$ in $$I$$. Let $$J'$$ be the set formed by replacing $$J_{0}$$ by $$J_{0}'$$ in $$J$$.

Then $$\cup I' = \cup J'$$. Then since $$I'$$ and $$J'$$ are finite or countably infinite sets of pairwise disjoint open intervals, $$I' = J'$$ by Theorem 7.2. Then since $$I_{0}'$$ and $$J_{0}'$$ are the only sets in $$I'$$ and $$J'$$ that contain $$0$$, $$I_{0}' = J_{0}'$$. Then $$I_{0} = J_{0}$$. Then $$I = J$$.

• Check out this question and its answers. – Henno Brandsma Sep 27 '18 at 4:50
• The statement of Theorem 7.2 is valid only if we include $\Bbb R, \emptyset,$ and all $(r,\infty)$ and $(-\infty,r)$ among the "open intervals", i.e. it would be clearer if it said "convex sets" instead of "intervals" – DanielWainfleet Sep 28 '18 at 3:47

Both in $$\mathbb{R}$$ and in $$[0,1]$$ the decomposition is just the unique decomposition of an open set into its connected components, which are in both cases relatively open due to local connectedness.

• I apologize, it has been a while since I studied connectedness, so I don't quite understand this yet. – Patrick Sep 28 '18 at 16:21
• @Patrick Look up connected components. Also, local connectedness. – Henno Brandsma Sep 28 '18 at 21:21

Let us consider $$G \subset [0,1]$$ which is open in the relative topology of $$[0,1]$$. What does $$G$$ prevent from being open in $$\mathbb{R}$$? If $$G \subset (0,1)$$, then $$G$$ is open in $$\mathbb{R}$$, but if one of the boundary points $$0,1$$ is contained in $$G$$, then $$G$$ is not open in $$\mathbb{R}$$. We consider the case that $$0 \in G$$ and $$1 \notin G$$, the other cases are treated similarly. We know that there exists $$\varepsilon \in (0,1)$$ such that $$[0, \varepsilon) = (-\varepsilon, \varepsilon) \cap [0,1] \subset G$$. Then $$G' = G \cup (-\varepsilon,0]$$ is open in $$\mathbb{R}$$. Apply to theorem to express $$G'$$ uniquely as a finite or countably infinite union of pairwise disjoint open intervals $$I'_n$$. Then the $$I_n = I'_n \cap [0,1]$$ are the desired subintervals of $$[0,1]$$.

Now assume you have another family $$J_k$$ for $$G$$. Exactly one of the $$J_k$$, call it $$J_0$$, contains $$0$$. All other intervals are contained in $$(0,1)$$ (recall $$1 \notin G$$) and are therefore open in $$\mathbb{R}$$. Define $$J'_0 = J_0 \cup (-\varepsilon,0]$$ and $$J'_k = J_k$$ else. Then the theorem applies to show that the families $$I'_n$$ are $$J'_k$$ are identical up to a bijection between the indices. W.l.o.g. assume that $$0 \in I'_0$$. Then necessarily $$I'_0 = J'_0$$, hence $$I_0 = J_0$$, and we are done.

• I wasn't entirely certain that I understood the second part of your answer, so I tried to expand it as an edit to mine. Is my interpretation correct? – Patrick Sep 28 '18 at 15:21
• Yes, it is correct. – Paul Frost Sep 28 '18 at 15:53
• Thank you. Was the original existence part of my proof correct? – Patrick Sep 28 '18 at 16:04
• Yes, that was also correct. Note that Henno Brandsma's proof is much more elegant then reducing the $[0,1]$-case to $\mathbb{R}$. – Paul Frost Sep 28 '18 at 16:07
• It has been a while since I studied connectedness, so I don't quite understand his answer yet. – Patrick Sep 28 '18 at 16:13