The nested intervals theorem says the following.
If a sequence of intervals $\langle I_n\rangle$ is decreasing, then $\bigcap_{n=1}^{\infty} I_n$ is not empty.
However, I'm trying to modify the theorem, say, if the sequence is strictly decreasing, then $\bigcap_{n=1}^{\infty} I_n$ should be a single point.
What I tried:
Let $I_n = [ a_n, b_n]$. Then $\langle a_n\rangle$ is increasing, while $\langle b_n\rangle$ is decreasing. Since $\langle a_n\rangle$ is bounded, it converges to a point, say $\alpha$. And, since $b_k$ is an upper bound $\forall k$, thus $\alpha \le b_k$.
Even if $\langle b_n\rangle$ is decreasing and bounded, I don't know how I can say $\lim_{n \to \infty} b_n = \alpha$.
Is there anyone to help me?
The statement is false. Take $a_n=-1-\frac1n$ and $b_n=1+\frac1n$. Then the sequence $(I_n)_{n\in\mathbb N}$ is strictly decreasing, but$$\bigcap_{n\in\mathbb N}[a_n,b_n]=[-1,1].$$