The nested interval property states:
For each $n \in \mathbf{N}$, assume we are given a closed interval $I_n = [a_n, b_n] = \{x \in \mathbf{R} : a_n \leq x \leq b_n\}$. Assume also that each $I_n$ contains $I_{n+1}$. Then, the resulting sequence of closed intervals: $I_1 \supseteq I_2 \supseteq I_3 \supseteq I_4 \supseteq \dots$ has a nonempty intersection. (Abbot, Understanding Analysis)
I understand the proof of this statement. However, I was wondering if the proof is limited to $n \in \mathbf{N}$? Can the indices for the intervals instead be real, that is $n \in \mathbf{R}$?
The reason I ask this is because I've seen a proof that the real numbers are uncountable that relies on the nested interval property (see Using nested intervals to prove that $\mathbb{R}$ is not countable). While I understand this proof, I do not see how the countability of the reals plays a role in the contradiction that is reached (i.e. I do not see how assuming the reals are uncountable would allow us to avoid the contradiction — unless the nested interval property does not work with real indices).
Best Answer
I think you mean $\supseteq$ and not $\subseteq$, since otherwise $\displaystyle \bigcap_{n \in \mathbb{N}} I_n = I_1$.
Assuming you mean $\supseteq$, if you had $\mathbb{R}$ instead of $\mathbb{N}$, then the hypothesis would presumably be that $I_r \supseteq I_s$ for all $r<s$. In this case, you'd have $$\bigcap_{x \in \mathbb{R}} I_x ~=~ \bigcap_{n \in \mathbb{N}} I_n$$ since $I_x \supseteq I_{\mathrm{max}\{1, \lceil x \rceil\}}$ for each $x \in \mathbb{R}$.
So this property for $\mathbb{N}$-indexed intervals implies the same property for $\mathbb{R}$-indexed intervals.