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I have completed a proof of the following statement:

Prove that any bounded open subset of R is the union of disjoint open intervals Let O $\subset$ R We want to write O as a disjoint union of open intervals

Since O is bounded, we can define $\forall x \in O$

\begin{align} a_x &= \inf\{ y \in R: (y,x) \subset O\}\\ b_x &= \sup\{z \in R: (x,z) \subset O\} \end{align} Let $$\ U = \bigcup_{x \in O} (a_x,b_x),$$

where $(a_x,b_x) \subset O$.

There is more to the proof, but this is the part I need help on. I have been told by my instructor the proof is correct. However, he says it is not evident that $(a_x,b_x) \subset O$. He wants me to prove this specific part, and I cannot figure out how to do it. I know I need to take an element in $(a_x,b_x)$, and show that it is in O, but I am not sure how to do this and how to connect it. I've been trying this for a week, and I've decided it is about time I ask for help. Can anyone help me here?

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So $a_x = \inf\{y \in \Bbb R\mid (y,x) \subseteq O\}$

and $b_x = \sup\{y \in \Bbb R\mid (x,y) \subseteq O\}$

which are well-defined (there are $a' < x < b'$ such that $(a',b') \subseteq O$ as $x$ is in the interior of $O$ so the sets are non-empty); and $O$ is bounded so any lower bound for $O$ is one for the set defining $a_x$ etc.

Why $(a_x, b_x) \subseteq O$? Let $z \in (a_x, b_x)$. If $z=x$ we are done, so assume (first case) that $z < x$. Because $a_x$ is the greatest lower bound for the set $A_x:=\{y \in \Bbb R\mid (y,x) \subseteq O\}$, we are sure that $z$ is not a lower bound for $A_x$, so there is some $z' \in A_x$ such that $z' < z$. But then $z' \in A_x$ implies $(z',x) \subseteq O$ and as $z \in (z',x)$, we know $z \in O$.

A similar argument can be held for $b_x$ and $B_x = \{y \in \Bbb R\mid (x,y) \subseteq O\}$ when $z > x$ instead, using that $b_x$ is the smallest upper bound for $B_x$ etc.

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  • $\begingroup$ Hey there! So I'm reading through this and I have a few questions. My professor is extremely picky, and he will mark us off for things that are correct but aren't in the way he wants. So, when you define A_x, I can already see he is going to say, "How do you know a_x is the GLB for A_x? This is unclear." Can you explain how we know? On the next line, you say we are sure that z is not a lower bound for A_x. I have a feeling he is going to ask why. Why are we sure z is not a lower bound for a_x? Can you explain those two things? Thanks! $\endgroup$ – bomb456 Oct 8 '20 at 11:19
  • $\begingroup$ @DominicBlanco You defined it yourself: $a_x = \inf\{y \in \Bbb R\mid (y,x) \subseteq O\}$. $\inf$ means greatest lower bound; So $A_x$ is the set it is the glb of. I just give it a name for convenience (so I can refer to it later, easier than "that set $a_x$ is the glb of"). Because $x \in (a_x, b_x)$ we know that $z > a_x$ by definition. It follows that $z$ is not a lower bound of $A_x$ because that would contradict $a_x$ being the greatest (largest) lower bound for $A_x$. That's all really. I only apply the definitions, and even the most petty professor wouldn't mind that, right? $\endgroup$ – Henno Brandsma Oct 8 '20 at 17:10
  • $\begingroup$ I hope he will be ok with it! He would be the one to question it in all honesty, but I feel like it is fully justified. We'll just have to see! $\endgroup$ – bomb456 Oct 8 '20 at 18:54
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Hint: prove that $(a_x+\frac1n,\, b_x-\frac1n)\subseteq O$ for all $x\in O$ and big enough $n\in\Bbb N$.

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  • $\begingroup$ I feel like that is trivial. If a_x is the infimum and b_x is the supremum, then of course those elements are in there. What makes the main case tricky is that a_x and b_x themselves are not in O. But (a_x,b_x) is. This didn't really help me, unfortunately. $\endgroup$ – bomb456 Oct 7 '20 at 14:27
  • $\begingroup$ Why not? Every single element of $(a_x,b_x)$ has positive distance from both end points, so is contained in one of $(a_x+\frac1n,b_x-\frac1n)$. $\endgroup$ – Berci Oct 7 '20 at 14:30
  • $\begingroup$ Hmm. So I'm thinking since (a_x + 1/n,b_x - 1/n) is in O (trivially), then as n becomes large, we approach (a_x,b_x). Is that the argument you're making? $\endgroup$ – bomb456 Oct 7 '20 at 14:32
  • $\begingroup$ Exactly. In other words, their union is just $(a_x,b_x)$. $\endgroup$ – Berci Oct 7 '20 at 14:32
  • $\begingroup$ Hmm. Interesting. I believe it, and it seems to work, I just worry because of how my professor is. He told me the way to do it is take an element (say c) in (a_x,b_x) and show it is also in O. He will give you no credit if you don't do the problem in his way. I worry he may not like this. But in all reality, this seems to work. It makes sense when I thought about it more. $\endgroup$ – bomb456 Oct 7 '20 at 14:34
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Let $x\in O.$

Consider any $v\in (x,b_x).$ We have $v<b_x.$ Now $$ v\not \in O\implies \forall z\in \Bbb R\,(\,(x,z)\subset O\implies z\le v)\implies b_x\le v$$ contrary to $v<b_x.$ So $v\in O.$

So $(x,b_x)\subset O.$

Similarly we obtain $(a_x,x)\subset O.$

So $(a_x,b_x)=(a_x,x)\cup \{x\}\cup (x,b_x)\subseteq O.$

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