Every point of a non-empty open set $S$ belongs to exactly one component interval of $S$

I don't understand this theorem:

Every point of a non-empty open set $S$ belongs to exactly one component interval of $S$.

Proof: 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

$$a(x)=inf\{ a:(a,x)\subseteq S\}$$ $$b(x)=sup\{ b:(x,b)\subseteq S\}$$

Here $a(x)$ might be $-\infty$ and $b(x)$ might be $+\infty$. Clearly, there is no open interval $J$ such that $I_x\subseteq J \subseteq S$. So $I_x$ is a component interval of $S$ containing $x$. If $J_x$ is another component interval of $S$ containing $x$, then the union $I_x\cup J_x$ is an open interval contained in $S$ and containing both $I_x$ and $J_x$. Hence, by the definition of component interval, it follows that $I_x\cup J_x=I_x$ and $I_x\cup J_x=J_x$ So $I_x=J_x$

How do I prove that this largest interval is $I_x=(a(x),b(x))$, where

$$a(x)=\inf\{a:(a,x)\subseteq S\}$$ $$b(x)=\sup\{b:(x,b)\subseteq S\}?$$

I don't understand how $a(x)$ might be $-\infty$ and $b(x)$ might be $+\infty$. Clearly, there is no open interval $J$ such that $I_x\subseteq J \subseteq S$.

• That should say: Clearly there is no open interval $J$ such that $I_x\subsetneqq J\subseteq S$. – Brian M. Scott Sep 3 '16 at 0:14

Let $x\in S$. $S$ is open, so there are real numbers $c$ and $d$ such that $x\in(c,d)\subseteq S$; this shows that the sets $\{a\in\Bbb R:(a,x)\subseteq S\}$ and $\{b\in\Bbb R:(x,b)\subseteq S\}$ are not empty: $c$ is in the first one, and $d$ is in the second. Thus, it’s meaningful to define

$$a(x)=\inf\{a\in\Bbb R:(a,x)\subseteq S\}$$

and

$$b(x)=\sup\{b\in\Bbb R:(x,b)\subseteq S\}\;.$$

(We do have to allow $b(x)$ to be $\infty$ if necessary in order to take care of cases like $S=\Bbb R$ and $x=0$, when $\{b\in\Bbb R:(x,b)\subseteq S\}$ is the entire set of positive real numbers. Similarly, we have to allow for the possibility that $a(x)=-\infty$.)

We can now let $I_x=\big(a(x),b(x)\big)$; this is an open interval (where we include open rays and $\Bbb R$ itself as open intervals). Moreover, $a(x)\le c<x<d\le b(x)$, so $x\in(c,d)\subseteq I_x$, meaning that $I_x$ is an open interval containing $x$. We still have to show that $I_x\subseteq S$, and that if $J$ is any strictly larger open interval (i.e., any open interval such that $I_x\subsetneqq J$), then $J\nsubseteq S$. This will show that $I_x$ is the largest open interval containing $x$ that is a subset of $S$.

To show that $I_x\subseteq S$, suppose that $y\in I_x$; we want to show that $y\in S$. This is certainly true if $y=x$, so suppose that $y\ne x$; then either $a(x)<y<x$, or $x<y<b(x)$. (Why?)

• Show that if $a(x)<y<x$, then there is an $a\in\Bbb R$ such that $a<y$ and $(a,x)\subseteq S$, and conclude that $y\in S$.
• Make a similar argument for the case $x<y<b(x)$.

Once you’ve done this, you’ll have shown that $I_x\subseteq S$.

Now suppose that $J$ is a strictly larger open interval.

• Show that either there is a $y\in J$ such that $y\le a(x)$, or there is a $y\in J$ such that $b(x)\le y$.
• Use the fact that $J$ is open to show that in the first case there is a $z\in J$ such that $z<a(x)$, and in the second case there is a $z\in J$ such that $b(x)<z$.
• By considering the interval $(z,x)$ in the first case and the interval $(x,z)$ in the second, show that $J\nsubseteq S$.