In my textbook we have the theorem that
For each $n \in N$, assume we are given a closed interval $$I_n = [a_n, b_n] = \{x \in R : a_n ≤ x ≤ b_n\}.$$ Assume also that each $I_{n}$ contains $I_{n+1}$. Then, the resulting nested sequence of closed intervals$$I_1 \supseteq I_2\supseteq I_3\supseteq I_4\supseteq \cdots $$has a nonempty intersection.
What I am confused on: Can you explain what $I_n$ is? Am I right in thinking that $I_n$ is a subinterval of the set of natural numbers?
Then by showing their intersections, you are saying that the subinterval of $n=2$ contains all of interval $n=1$, hence the nested sequence of closed intervals will never be empty?
Thank you.
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$\begingroup$$I_n$ is a sequence of closed and bounded intervals.
$n \in \mathbb{N}$, hence $n$ is a natural number, it is used to index the sequence of intervals.
Now, let look at the definition of $I_n$, $I_n = [a_n, b_n] = \{x \in \color{blue}{\mathbb{R}}: a_n \le x \le b_n\}$. From the set builder notation, we can see that $I_n$ consists of real numbers.
In this case, the interval shrinks as you increases your index.
For example, we can have
$$I_1 = [-1, 1], I_2=[-\frac12, \frac12], I_3 = \left[-\frac13, \frac13 \right]\ldots, I_n = \left[ -\frac1n , \frac1n\right], \ldots$$ and so on.
For my particular example, you can see that if you intersect these sequence of intervals, you get $\{0\}$. It is not empty.
Is it possible that someone can construct such a sequence of shrinking intervals and the intersection is empty? The theorem tells us that it is impossible.
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