Prove that: $\bigcap_{t\in T} \space I_t \neq \varnothing $ for non-disjoint family of intervals from $\mathbb{R}$.

Question

Let $ (I_t)_{t \in T} $ be an indexed family of intervals of $\mathbb{R}$ that are closed, bounded and mutually non-disjoints.

1. Prove that: $$\bigcap_{t\in T} \space I_t \neq \varnothing $$

2. What happens if we replaced the open intervals with disks of $\mathbb{R}^2.$

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1. The indexed family being mutually non-disjoint means for every couples $ i,j \in T$

We have: $I_i \cap I_j \neq \emptyset \implies \bigcap_{t \in T} I_t \neq \varnothing$.

Which I am not if it is this simple.

2. By the definition of a disk: $ D(o,r) = \{ x \in \mathbb{R}^2 | \space d(o,x) \leq r \}$. If we replace the closed bounded intervals by closed bounded disks, the intersection will be the desk with the smallest radius. The smaller the radius gets, the more the disk gets closer to the center $o$.

$$ \bigcap_{t \in T} D_t = \lim_{r \to 0} D_t(o,r) = o$$. Which I am not sure of. Is it correct?

Answer 1

For 1. the solution is not this simple. Here is a sketch for a solution.

First, fix $t_0\in T$ and define $I = I_{t_0}$.

Suppose for a contradiction that $\bigcap_{t\in T} I_t$ is empty. Then $$ \bigcup_{t\in T} I_t^\complement = \left(\bigcap_{t\in T} I_t\right)^\complement = \mathbb{R} $$ In particular, $\{I^\complement_t\}_{t\in T}$ is a open cover for the compact set $I$. Then there exists a finite sub-cover of $I$. That is, there exists a finite set $S\subseteq T$ such that $$ \bigcup_{t\in S} I_t^\complement \supseteq I $$ It then follows that $$ I\setminus\bigcup_{t\in S}I_t^\complement =I\cap\left(\bigcup_{t\in S}I_t^\complement\right)^\complement = \bigcap_{t\in S} I\cap I_t = \emptyset $$ Using the fact that $S$ is finite, you can deduce that this is a contradiction (show that the finite intersection of sets $I_t$ cannot be empty).

For part 2, you can proceed similarly to show that if $D_t$ are mutually non-disjoint closed bounded disks then $$ \bigcap_{t\in T}D_t \neq\emptyset $$

EDIT: To show that the finite intersection of mutually non-disjoint closed, bounded intervals is non-empty, we proceed by induction. The base case is clear. Now, suppose $$[a_1, b_1], \dots, [a_n, b_n], [a_{n+1}, b_{n+1}] \in \{I_t\}_{t\in T}$$ By the induction hypothesis, $$ \bigcap_{i=1}^n [a_i, b_i] $$ is non-empty. Furthermore, notice that $$ \bigcap_{i=1}^n [a_i, b_i] = [a,b] $$ where $a = \max a_i$ and $b = \min b_i$. Now, since $$ [a_{n+1}, b_{n+1}]\cap [a_i, b_i] \neq \emptyset $$ for all $i=1,\dots, n$, we must have $a_{n+1} \leq b_i$ and $b_{n+1} \geq a_i$ for all $i = 1, \dots, n$. Ergo, $$ \bigcap_{i=1}^{n+1} [a_i, b_i] = [a,b]\cap[a_{n+1}, b_{n+1}] = [\max\{a, a_{n+1}\}, \max\{b, b_{n+1}\}] \neq \emptyset $$

Answer 2

I think we can argue as follows: your intervals are of the form $[a_t,b_t]$ and you know that $[a_t,b_t]\cap [a_s,b_s]\neq \emptyset$ for any pair $(s,t).\ $ This means that

$\tag1 \max (a_t,a_s)\le \min (b_t,b_s)$.

Now, take any finite sequence $((a_i,b_i))_i$ of intervals. If any one interval lies in another, we take the union of the two. Therefore, by $(1)$, we can order them so that $a_1\le \cdots \le a_n$ and $b_1\le \cdots \le b_n.$ By hypothesis, $I_1\cap I_n\neq \emptyset$ and this means that $\cap_iI_i\neq \emptyset$ so that $(I_t)_t$ has the finite intersection property.

Now take any $I_1$ from $(I_t)_t$ and consider $(I_1\cap I_t)_t$. By the first paragraph, these sets have the finite intersection property, and their union is $I_1$, a compact set in $\mathbb R$, which means that $\bigcap_t(I_1\cap I_t)_t=\bigcap_t I_t$ is nonempty.