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In Royden's Real Analysis, he quoted the following:

Every nonempty open set is the disjoint union of a countable collection of open intervals.

Proof: Let $O$ be a nonempty open subset of $\mathbb{R}.$ Let $x$ belong to $O.$ There is a $y>x$ for which $(x,y) \subseteq O$ and a $z < x$ for which $(z,x) \subseteq O.$ Define the extended real number $a_x$ and $b_x$ by $$a_x = \inf \{ z|(z,x) \subseteq O \} and \space b_x = \sup\{ y|(x,y)\subseteq O \}.$$ Then $I_x = (a_x,b_x)$ is an open interval that contains $x.$ We claim that $$I_x \subseteq O \space but \space a_x \not\in O,b_x \not\in O. (2)$$ Indeed, let $w$ belong to $I_x,$ say $x < w < b_x.$ By the definition of $b_x,$ there is a number $y>w$ such that $(x,y) \subseteq O$ and so $w \in O.$ Moreover, $b_x \not\in P,$ for if $b_x \in O,$ then for some $r>0$ we have $(b_x - r,b_x + r) \subseteq O.$ Thus, $(x,b_x+r) \subseteq O$, contradicting the definition of $b_x.$ Note that $O = \cup_{x \in O}I_x.$ We infer from $(2)$ that $\{I_x\}_{x \in O}$ is disjoint.

Question: If $b_x = \infty,$ then $b_x + r = \infty,$ how does this contradicts with the definition of $b_x?$ Also, how to prove that $\{I_x\}_{x \in O}$ is disjoint?

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    $\begingroup$ If $b_x = \infty$, then $b_x \not\in O$ because $O \subseteq R$. $\endgroup$ – user49640 Jun 20 '17 at 8:54
  • $\begingroup$ Shouldn't the author include this in another sentence? $\endgroup$ – Idonknow Jun 20 '17 at 8:54
  • $\begingroup$ That's a weird demonstration. Obviously $\{I_x\}_{x\in O}$ is NOT disjoint, since $I_x=I_{(x+b_x)/2}$. (edit: oh, forget about that, I just saw user363464's answer) Also the case $b_x=\infty$, as you point, is not dealt with properly. $\endgroup$ – Evargalo Jun 20 '17 at 8:55
  • $\begingroup$ He says "for if $b_x \in O$, then" blah-blah-blah. He doesn't explicitly point out that this implies $b_x \in R$. I suppose he expects the reader to understand this. $\endgroup$ – user49640 Jun 20 '17 at 8:56
  • $\begingroup$ Meant is: for every $x,y\in\mathbb R$ the sets $I_x,I_y$ coincide or are disjoint. Then the sets form a partition of set $O$. $\endgroup$ – drhab Jun 20 '17 at 8:56
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If $b_{x}=\infty$ then (2) holds immediately.

To show that $\{I_{x}\}$ is disjoint, suppose $I_{x}\cap I_{y}\neq\emptyset$, and play with the sup and inf definitions to see that $a_{x}=a_{y}$ and $b_{x}=b_{y}$.

edit: Let $z\in I_{x}\cap I_{y}$. By definition of $I_{x}$, $(a_{x},z)\subset O$, so $a_{z}\leq a_{x}$ by definition of infimum. This inequality cannot be strict, since $I_{z}\subset O$ and $a_{x}\not\in O$. Thus $a_{x}=a_{z}$ and likewise $a_{z}=a_{y}$.

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  • $\begingroup$ "suppose $I_x=I_y$," I think you mean "suppose $I_x \cap I_y \neq \oslash$," $\endgroup$ – Evargalo Jun 20 '17 at 8:58
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    $\begingroup$ yes i mean that, thanks @Evargalo $\endgroup$ – user363464 Jun 20 '17 at 9:00
  • $\begingroup$ @user363464: Let $z \in I_x \cap I_y.$ To show $a_x = a_y,$ since $a_x \not\in O,$ by definition of infimum, we have $(a_x, z) \not\in O.$ As $a_y$ is an infimum, we have $a_x \leq a_y.$ $\endgroup$ – Idonknow Jun 20 '17 at 10:00
  • $\begingroup$ How to show that $I_x \cap I_y = \emptyset?$ Can give hint? $\endgroup$ – Idonknow Jun 20 '17 at 10:52
  • $\begingroup$ @Idonknow i've added some more details to my answer $\endgroup$ – user363464 Jun 20 '17 at 11:01
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Alternative route.

Prescribe a relation on $O$ by stating that $x\sim y$ if $(x,y)\subseteq O$ and $(y,x)\subseteq O$.

Then prove that $\sim$ is an equivalence relation on $O$.

Then prove that the equivalence classes (which form a partition of $O$) are open intervals.

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  • $\begingroup$ Or use that the connected components of $O$ are open in $\mathbb{R}$ as the reals are locally connected .The components are thus disjoint open intervals QED. $\endgroup$ – Henno Brandsma Jun 20 '17 at 12:13

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