If $A,B⊆X⊆ \mathbb R$, $A\cap B=∅$ $A,B$ open then $\sup A<\inf B$ or $\sup B< \inf A$

Question

Let $X\subseteq \mathbb R$ be a metric space. I'm trying to prove that if a subset $I$ is non connected, it implies that $\exists a,b \in I$, $a<b$, $\exists x, a<x<b$ such that $x\notin I$.

I am trying it, using the following definition of non connectedness:

$I$ is not connected iff $\exists A,B\subseteq X$ such that

• $A\cap B = \emptyset$
• $A\cup B = I$
• $A,B$ are open in $X$

So, starting with $I$ is not connected, there exists $A,B$ verifying the three above-mentioned points.

Pick $a\in A$ and $b\in B$ and suppose WLOG that $a<b$ (you can exchange them otherwise).

Now, I only need to pick a suitable $x$ that verifies $x \notin A\cup B$.

But for that I think that I'll need to prove the lemma:

If $A,B⊆X⊆\mathbb R$, $A\cap B=∅$ and $A,B$ open then $\sup A<\inf B$

I know that after proving the lemma I will be able to pick $x= \frac{\sup A + \inf B}{2}$.

I feel like this lemma is above my math skills, if anyone can give me hints in order to prove it I would be grateful.

Thanks in advance.

Answer 1

Let $s=\sup(I)$ and $i = \inf(I)$ where it might be that $s=\infty$ and $i=-\infty$ so these two objects are always defined.

If $I$ contains all the elements strictly between $s$ and $i$, then $I$ must be one of the following $4$: $$ (1) \text{ }\text{ }\text{ } I = (i,s)$$

$$ (2) \text{ }\text{ }\text{ } I = [i,s)$$

$$ (3) \text{ }\text{ }\text{ } I = (i,s]$$

$$ (4) \text{ }\text{ }\text{ } I = [i,s]$$

I will leave this as an exercise. Since all of these are connected, there must exists some $i<x<s$ so that $x\not\in I$.

By definition of infimum and supremum, there must exists some $a\in I$ with $i<a<x$ and some $b\in I$ with $x<b<s$. Hence, $a<x<b$ satisfies the requirements.