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In Apostol, page $51$, he defines what he calls the component interval. I can't find any reference to it on the web. I have some problems with the definition:

Let $S \subseteq \mathbb{R}$. An open interval $I$ of $S$ is a component interval if there does not exist an open interval $J$ of $S$ such that $I \subset J$.

Intuitively, I get that $I$ is the largest possible open interval that is contained in $S$. I think that the set of all rationals between the end points of $S$, $\{\alpha \in (A,B)\ |\ \alpha \in \mathbb{Q}\}$, is a component interval. Is that true? If $D$ is dense in $S$, is $D$ in general a component interval of $S$?

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1 Answer 1

You’ve slightly misstated the definition: $I$ is a component interval of $S$ if there is no open interval $J\subseteq S$ that properly contains $I$. Moreover, in this context $S$ is a bounded open set in $\Bbb R$.

A component interval of $S$ is, first of all, an open interval in $\Bbb R$. It therefore can’t look like $(0,1)\cap\Bbb Q$, for instance, because that isn’t an interval in $\Bbb R$: it’s missing all of the irrationals between $0$ and $1$.

Consider the set $S=\{x\in\Bbb R:-1<x<1\text{ or }0<x<2\text{ or }4<x<5\}$; a little thought will show that $S=(-1,2)\cup(4,5)$. Here the intervals $(-1,2)$ and $(4,5)$ are the component intervals of $S$: each is an open interval, their union is all of $S$, and if you expand either of them even slightly to get a larger open interval, you pick up points that are not in $S$. $(-1,2)$ and $(4,5)$ are as big as they can be and still be open intervals contained in $S$. This set $S$ therefore has exactly two component intervals.

Now consider the set $$S=(0,1)\setminus\left\{\frac1n:n\in\Bbb Z^+\right\}\;.$$ $S$ is a bounded open subset of $\Bbb R$, and if you draw a sketch, you should be able to convince yourself pretty easily that

$$\begin{align*} S&=\left(\frac12,1\right)\cup\left(\frac13,\frac12\right)\cup\left(\frac14,\frac13\right)\cup\left(\frac15,\frac14\right)\cup\dots\\ &=\bigcup_{n\in\Bbb Z^+}\left(\frac1{n+1},\frac1n\right)\;. \end{align*}$$

Each of the intervals $\left(\frac1{n+1},\frac1n\right)$ is an open interval contained in $S$ that cannot be expanded to a larger open interval contained in $S$: if you try to expand $\left(\frac14,\frac13\right)$, for instance, you’ll get an interval that contains either $\frac14$ or $\frac13$ (or both) and therefore won’t be a subset of $S$ at all. By definition, therefore, each of these intervals $\left(\frac1{n+1},\frac1n\right)$ is a component interval of this set $S$. And since their union is all of $S$, they are all of the component intervals of $S$. This set $S$ has countably infinitely many component intervals, one for each $n\in\Bbb Z^+$.

What Apostol is going to prove there is that every bounded open subset of $\Bbb R$ can be decomposed in this way into component intervals, and that there are always at most countably infinitely many of these component intervals.

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Thanks ,Brian .Actually ,It's trivial ,I was abit hasty.So,the number of component intervals measure the number of disjoint connected subsets of an interval (Those subsets that are both close and open). –  nabil Sep 2 '12 at 0:09
    
@nabil: Yes, that’s right, provided that you mean both closed and open in the subspace topology on $S$. –  Brian M. Scott Sep 2 '12 at 6:26

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