# Specific part (not entire question) of the proof that O can be written as a union of disjoint open intervals

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?

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.

• 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! – bomb456 Oct 8 '20 at 11:19
• @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? – Henno Brandsma Oct 8 '20 at 17:10
• 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! – bomb456 Oct 8 '20 at 18:54

Hint: prove that $$(a_x+\frac1n,\, b_x-\frac1n)\subseteq O$$ for all $$x\in O$$ and big enough $$n\in\Bbb N$$.

• 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. – bomb456 Oct 7 '20 at 14:27
• Why not? Every single element of $(a_x,b_x)$ has positive distance from both end points, so is contained in one of $(a_x+\frac1n,b_x-\frac1n)$. – Berci Oct 7 '20 at 14:30
• 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? – bomb456 Oct 7 '20 at 14:32
• Exactly. In other words, their union is just $(a_x,b_x)$. – Berci Oct 7 '20 at 14:32
• 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. – bomb456 Oct 7 '20 at 14:34

Let $$x\in O.$$

Consider any $$v\in (x,b_x).$$ We have $$v Now $$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 So $$v\in O.$$

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

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

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