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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

Based on Paul Frost's answer:

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 $I = J$.

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  • $\begingroup$ Check out this question and its answers. $\endgroup$ Commented Sep 27, 2018 at 4:50
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    $\begingroup$ 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" $\endgroup$ Commented Sep 28, 2018 at 3:47

2 Answers 2

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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.

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  • $\begingroup$ I apologize, it has been a while since I studied connectedness, so I don't quite understand this yet. $\endgroup$
    – Patrick
    Commented Sep 28, 2018 at 16:21
  • $\begingroup$ @Patrick Look up connected components. Also, local connectedness. $\endgroup$ Commented Sep 28, 2018 at 21:21
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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.

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  • $\begingroup$ 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? $\endgroup$
    – Patrick
    Commented Sep 28, 2018 at 15:21
  • $\begingroup$ Yes, it is correct. $\endgroup$
    – Paul Frost
    Commented Sep 28, 2018 at 15:53
  • $\begingroup$ Thank you. Was the original existence part of my proof correct? $\endgroup$
    – Patrick
    Commented Sep 28, 2018 at 16:04
  • $\begingroup$ Yes, that was also correct. Note that Henno Brandsma's proof is much more elegant then reducing the $[0,1]$-case to $\mathbb{R}$. $\endgroup$
    – Paul Frost
    Commented Sep 28, 2018 at 16:07
  • $\begingroup$ It has been a while since I studied connectedness, so I don't quite understand his answer yet. $\endgroup$
    – Patrick
    Commented Sep 28, 2018 at 16:13

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