Union of open overlapping intervals is an open interval

I am taking my first real analysis course and I read about the fact that every open set of real numbers can be represented by a countable union of disjoint open intervals (although the proofs are hard for me to understand at this point in time). Now what about overlapping intervals? This should be easier to prove. Assume that for all $$n \in \mathbb{N}$$ we have an interval $$I_n = (x_n,y_n) \subseteq \mathbb{R}$$. Furthermore, assume that for all these intervals the intersection is not empty, i.e. $$I_n \cap I_k \not = \emptyset$$ for all $$n,k \in \mathbb{N}$$. Now take the union as $$T= \cup_{n \in \mathbb{N}}$$. Then $$T$$ should be an open interval. However, I do not know how to prove that $$T$$ is an interval. I would be grateful for any suggestions on how to start.

• Hint to start: Suppose we have a number $p$ such that there exists $a\in T$ with $a<p$, and there exists $b\in T$ with $b>p$. Then every interval $I_n$ is contained in either $(-\infty,p)$ or $(p,\infty)$. Must there be at least one $I_n$ interval contained in each? Can they intersect each other? What would we conclude if such a $p$ could not exist (consider the interval $(\inf T,\sup T)$)? Jun 17, 2022 at 12:58

Let us begin with the definition of an interval on real line: a subset $$I$$ of the real line is an interval if and only if for any two points $$a in $$I$$ and any $$a, one has $$c\in I$$.
Note that the union $$T$$ of a sequence of open intervals $$I_n$$ is automatically open. So what remains is to check that $$T$$ is still an interval. Let $$a,b$$ be any two point in $$T$$ and suppose that $$a. The two points $$a$$ and $$b$$ must lie in some given intervals, say $$a\in I_1$$ and $$b\in I_2$$. Let $$c$$ be any point in $$\mathbb{R}$$ such that $$a. But as $$I_1\cap I_2$$ is not empty, so one can take $$d\in I_1\cap I_2$$ which satisfies $$a (there are only finitely possibile situations for two open sets to intersect and you can compute it out explicitly by considering the sup and inf of the intervals). If $$c=d$$, then it is done as $$c\in I_1\cap I_2 \subset T$$; otherwise either $$a or $$d will imply that $$c\in I_1\cup I_2\subset T$$ as $$I_1$$ and $$I_2$$ are intervals. The proof is complete.