Proof $\bigcap_{j \in \mathbb{N}} I_j $ consists of one point

Question

Let $I_n := [a_n,b_n], n \in \mathbb{N}$ a sequence of intervals in $\mathbb{R}$ with $I_{j+1} \subset I_j$ for all $j \in \mathbb{N}$ and $\lim_{n \to \infty} (b_n - a_n) = 0.$

How can one prove that

$$\bigcap_{j \in \mathbb{N}} I_j $$

consists of one point?

In another thread I've read the following:

I arrived at the above problem while trying to show the following (known as Bonferroni inequalities):

Let $A_1, \ldots, A_n$ be events of a probability space. For a subset $I$ of $\{1,\ldots, n\}$, write $A_I$ to denote $\bigcap_{j\in I}A_j$. Further, denote $\sum_{|I|=i}P(A_I)$ as $\sigma_i$. We agree by convention that $\sigma_0=1$.

But here it is only written with "We agree by convention".

Answer 1

We have $a_1 \le a_n \le b_n \le b_1$. Hence the sequences $(a_n)$ and $(b_n)$ are bounded.

Forthermore we have $a_n \le a_{n+1}$ and $b_{n+1} \le b_n.$ It follows that both sequences are momotone. Consequence: both sequences are convergent.

Let $a$ be the limit of $(a_n)$ and $b$ be the limit of $(b_n)$. From $\lim_{n \to \infty} (b_n - a_n) = 0$, we see that $a=b.$

Let $x \in \bigcap_{j \in \mathbb{N}} I_j$, then we have $a_n \le x \le b_n$ for all $n$. This gives (with $n \to \infty$) that $x=a (=b).$

Answer 2

Notice that if the intersection contained some point $c$, then we are forced to have $\lim_{n\to\infty} a_n = c = \lim_{n\to\infty} b_n$. It follows that the intersection cannot have more than one distinct point (try to prove this). In other words $\cap_{j\in\mathbb{N}} I_j$ contains at most one point

Now we have the task of proving the intersection is non-empty, which will complete our proof. To do this we notice that for all $j$, $I_j$ is non-empty. Further, we have that the intervals are nested, so by induction $$ I_1 \cap I_2 \neq \emptyset $$ and $$\mbox{For any finite } N>1 \mbox{, we have } \left( \bigcap_{1 \leq j < N} I_j \right) \cap I_N \neq \emptyset $$ In the limit as $N\to\infty$, together with the fact that each $I_j$ is non-empty, we get the desired result