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Ok so I was goin over the structure theorem proof for open sets in my book, and there seems to be one point where I am completely lost.

Theorem: Structure theorem for open set For a nonempty set $S$ in $\mathbb{R}$, the following are equivalent

$1)$ $S$ is an open set

$2)$ every element of $S$ is an interior point of $S$ i.e. $S^o = S$.

$3)$ S is a countable union of pairwise disjoint open intervals.

Proof of the theorem: First we prove that $1$ $\implies$ $2$, that is S open $\implies$ $S = \bigcup_{\alpha} (a_\alpha, b_\alpha)$. $\forall x \in S$, $x \in (a_\alpha, b_\alpha)$ for at least one $\alpha$. Let $r=\min(x-a_\alpha, b_\alpha-x)$, then $r>0$, and $(x-r,x+r) \subseteq (a_\alpha, b_\alpha) \subseteq S \implies x \in S^o$, thus $S=S^o$. Next we show that $2 \implies 3$. Suppose $S^o=S$, if $x \in S$, then $(x-r,x+r) \subseteq S$ for some $r.0$. Let $I_x$ be the largest interval containing $x$ such that such that $I_x \subseteq S$. Here $I_x$ is an interval from $m_x$ to $M_x$ where $m_x= \inf \{m:(m,x] \subseteq S\}$ and $M_x= \sup \{M:[x,M) \subseteq S\}$.

Now if $a = M_x$ or $m_x$ is a real number then $a\notin S$ (otherwise $(a-r',a+r')\subseteq S$ for some $r'>0$, then $I_x \bigcup (a-r',a+r')$ is a larger interval in $S$ containing $x$ which is a contradiction). For $x,y\in S$, if $I_x \bigcap I_y \ne \varnothing$, then $I_y=I_x$. For if it was not then $I_x \bigcup I_y$ is a larger interval in $S$ containing $x$ and $y$). So $S$ is the union of these pairwise disjoint open intervals.

In each of these open pairwise disjoint intervals, we can choose a rational number by the density of $\mathbb{Q}$. For different pairwise disjoint open intervals the rational numbers are different. Since $\mathbb{Q}$ is countable there are countably many open pairwise disjoint intervals.

Question:

$1)$ Why is that $M_x$ and $m_x$ cannot be a real number? Because any number is considered real, so does that mean $M_x$ and $m_x$ cannot take any value?, this really confuses me.

$2)$ Why do we choose a rational number $\mathbb{Q}$ for the pairwise disjoint open intervals?

Basically I dont get it at all after the part of the proof where it starts: Now if $a = M_x$ or $m_x$ is a real number then $a\notin S$ ....

Can anyone please help me clarify this confusion for me? I really need to understand this part so I can go further on with other theorems.

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1) Here $M_x , m_x$ cannot be real number means they can be $\infty$ or -$\infty$. For example, if $S=\mathbb R$ then $M_x=\infty$ and $m_x=-\infty$ for any $x\in S$.

2) Instead of $\mathbb Q$ you can take any countable dense subset of $\mathbb R$. But $\mathbb Q$ is one of the easiest example of such subset, so we generally consider $\mathbb Q$. As $\mathbb Q$ is dense in $\mathbb R$, every open interval contains at least one rational number. Now as $\mathbb Q$ is countable, open subintervals of $S$ are countable because we have surjective map from $\mathbb Q$ to open subintervals of $S$.

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  • $\begingroup$ Thank you for the reply, but I wonder why cant they be a real number?, so it means it can only be $\infty$? What does taking $ \mathbb{Q}$ actually really mean? Could you like give a numerical example? $\endgroup$ – Aurora Borealis Nov 24 '17 at 7:01
  • $\begingroup$ @AuroraBorealis I have edited my answer. Take a look at it. $\endgroup$ – Mayuresh L Nov 24 '17 at 11:01

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