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I reading about Representation Theorem for Open Sets on the Real Line from Apostol's Mathematical analysis. Consider the following definitions and Theorems from the text.

Definition. Let $S$ be an open set of $\mathbb{R}$. An open interval (which may be finite or infinite) is called a component interval of $S$ if $I\subseteq S$ and if there is no open interval $J\ne I$ such that $I\subseteq J\subseteq S$.

Theorem 3.10. Every point of a nonempty open set $S$ belongs to one and only one component interval of $S$.

Theorem 3.11. Every nonempty open set $S$ in $\mathbb{R}$ is the union of a countable collection of disjoint component intevals of $S$.

Note. This representation of $S$ is unique. In fact, if $S$ is a union of disjoint open intervals, then these intervals must be the component intervals of $S$. This is an immediate consequence of theorem 3.10.

My Questions

$1$. In the definition, it says "open intervals (which may be finite or infinite)...". Can open intervals be finite!? I assume he wants to say bounded or unbounded, right?

$2$. About the Note, I don't get that how the uniqueness follows immediately from theorem 3.10. Can someone shed some light on this?

$3$. Does the Note really raises a uniqueness question of such a representation? I mean if I were going to pose such a question I would say, if there were two countable collection of disjoint component intervals of $S$ such that their union is $S$ then the collections would be the same. I mean that the claim in the Note is just another theorem but not about the uniqueness of the representation mentioned in theorem 3.11.

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  • $\begingroup$ He is talking about the finite or infinite $length$ of an interval, not the number of points in it. He should have said bounded/unbounded interval rather than finite/infinite interval. $\endgroup$ – DanielWainfleet Nov 13 '17 at 7:51
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It's not obvious what Apostol has in mind, but it could be that, although as you say the Note does not seem to follow "immediately" from the statement of Theorem 3.10, it does follow quite quickly from the proof of the theorem, which begins thus:

Assume $x \in S$. Then $x$ is contained in some open interval $I$ with $I \subseteq S$. There are many such intervals but the "largest" of these will be the desired component interval. We leave it to the reader to verify that this largest interval is $I_x = (a(x), b(x))$, where $$ a(x) = \inf\{a: (a, x) \subseteq S\}, \quad b(x) = \sup\{b: (x, b) \subseteq S\}. $$ Here $a(x)$ might be $-\infty$ and $b(x)$ might be $+\infty$. [$\ldots$]

Suppose $\mathscr{S}$ is a collection of disjoint open intervals whose union is $S$, and $x \in (c, d) \in \mathscr{S}$. Then $(c, x) \subseteq S$ and $(x, d) \subseteq S$, whence $c \geqslant a(x)$ and $d \leqslant b(x)$, by the definitions of $a(x)$ and $b(x)$. If $(c, d) \ne I_x$, then either $c > a(x)$ or $d < b(x)$. If $c > a(x)$, then by the definition of $a(x)$, there exists $a < c$ such that $(a, x) \subseteq S$. But by definition $c < x$, therefore $c \in S$, therefore there exists $J$ such that $c \in J \in \mathscr{S}$. But then $J$ and $(c, d)$ are distinct yet not disjoint, which contradicts the definition of $\mathscr{S}$. Similarly, it cannot be true that $d < b(x)$. Therefore, $(c, d) = I_x$.

This proves not only that every interval in $\mathscr{S}$ is a component interval of $S$, but also the converse, because every component interval of $S$ is of the form $I_x$ for some $x \in S$, and for any such $x$ there exists $(c, d)$ such that $x \in (c, d) \in \mathscr{S}$, therefore $I_x = (c, d)$. Against what you say in your Question 3, this justifies Apostol's description of the Note as a proof of uniqueness - but you do have a point, because he doesn't actually state the converse part of the proposition.

The ease with which the argument may be pictured as a single geometrical image may account for Apostol's sense that it is "immediate", but I agree with you and Eric Wofsey that it is not strictly so.

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  • $\begingroup$ (+1), I really enjoyed your argument. :) But shouldn't we rule out the cases $c<a(x)$ or $d>b(x)$ too? I can see that we can also obtain a contradiction in these cases. $\endgroup$ – H. R. Nov 13 '17 at 7:41
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    $\begingroup$ You're right, I should have made this explicit. Rather than argue indirectly, I would prefer to point out that from the hypothesis $x \in (c, d) \in \mathscr{S}$, it follows that $(c, x) \subseteq S$ and $(x, d) \subseteq S$, whence $c \geqslant a(x)$ and $d \leqslant b(x)$, by the definitions of $a(x)$ and $b(x)$. $\endgroup$ – Calum Gilhooley Nov 13 '17 at 11:15
  • $\begingroup$ please add that part to your answer. :) $\endgroup$ – H. R. Nov 13 '17 at 13:05
  • $\begingroup$ Done. ......... $\endgroup$ – Calum Gilhooley Nov 13 '17 at 14:14
  • $\begingroup$ I notice we've been a bit unfair to Apostol, because he does write, "These intervals must be the component intervals of $S$" [emphasis added], which is a statement of uniqueness (of the collection that I've denoted by $\mathscr{S}$). $\endgroup$ – Calum Gilhooley Nov 13 '17 at 22:03
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(1) That's correct: "finite" here means "finite length" (i.e., bounded), not "finite cardinality".

(2) I wouldn't exactly say it's an "immediate consequence". Here's one way to prove it. Suppose $S=\bigcup U_i$ where the $U_i$ are disjoint open intervals. Given $x\in U_i$, by Theorem 3.10 there is a unique component interval $I$ of $S$ such that $x\in I$. Now note that $U_i\cup I$ is also an open interval (as the union of two open intervals that overlap at $x$) and $I\subseteq U_i\cup I\subseteq S$. By definition of component intervals, this means $I=U_i\cup I$, so $U_i\subseteq I$.

So, we have shown that each $U_i$ is contained in some component interval $I$. Since $S$ is the disjoint union of the $U_i$, each component interval $I$ of $S$ is therefore the disjoint union of all the $U_i$ which it contains. But $I$ is connected, so it cannot be written as a nontrivial disjoint union of open sets. Thus $I$ can only contain one $U_i$, so it is one of the $U_i$. That is, every component interval is one of the $U_i$, and therefore $\{U_i\}$ is the set of component intervals.

If you want to do it without connectedness, here's how you can show an open interval $I=(a,b)$ cannot be a union of more than one disjoint open intervals. Say $(c,d)$ is one of your disjoint open intervals. Then either $a\neq c$ or $b\neq d$ (otherwise we would have just one interval); without loss of generality suppose $b\neq d$. Then $d\in I$, so it is contained in one of our disjoint open intervals. But that is impossible, since any open interval containing $d$ would intersect $(c,d)$.

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  • $\begingroup$ (+1) Thanks Eric. :) I totally follow your answer till the second paragraph of 2. Why the sentence "Since $S$ is the disjoint union of ..." is true? I can imagine it intuitively but how to write it down? Also, as the text has not discussed connectedness, I would be happy if you change the reasoning that follows this sentence. :) $\endgroup$ – H. R. Nov 11 '17 at 23:42
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    $\begingroup$ Let $V$ be the union of the $U_i$ that are contained in $I$. Clearly $V\subseteq I$. On the other hand, if $x\in I$, then $x\in U_i$ for some $i$, and so as shown above, $U_i\subseteq I$. Thus $U_i\subseteq V$, and so $x\in V$. $\endgroup$ – Eric Wofsey Nov 11 '17 at 23:55

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