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

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?

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".

• Assume $x,y$ are in the intersection with $x<y$. Then use the definition of $\lim_{n}(b_n-a_n)=0$ with $\epsilon=y-x$. Then, there is $N$ such that for all $n>N$ we have $b_n-a_n<\epsilon=y-x$. But $y,x\in I_n$, which implies that $y-x\leq b_n-a_n<y-x$. Contradiction. Dec 4, 2019 at 14:23

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 $$x=a (=b).$$

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

• If your argument were correct it will also prove that an intersection of nested open sets is non-empty. You cannot deduce that the infinite intersection is non-empty from the non-emptyness of the finite intersections alone. Since $I_n=(0,1/n)$ satisfies everything up until the last sentence, but their intersection is empty, then your argument cannot be correct. Dec 4, 2019 at 14:31
• @conditionalMethod In this case I think I am using Cantor's Intersection theorem? That intersection of compact nonempty nested intervals is nonempty Dec 4, 2019 at 14:42
• Yes, that the intervals are compact is crucial to proving their intersection is non-empty. Dec 4, 2019 at 14:43