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 $.

  • 1
    $\begingroup$ That should say: Clearly there is no open interval $J$ such that $I_x\subsetneqq J\subseteq S$. $\endgroup$ – 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

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


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

(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 $J\nsubseteq S$.

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.