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I am trying to figure out what additional restriction I need to enforce to Cantor's intersection theorem for complete metric spaces to mimic nested interval theorem.


Cantor intersection "theorem": Given any decreasing sequence of non-empty closed nested subsets $\{C_k\}_{k=0}^{\infty}$ with $\color{red}{\text{some additional condition}}$, we have $\displaystyle \bigcap_{k=0}^{\infty} C_k = C$, where $C$ is closed & non-empty.


where $\color{red}{\text{some additional condition}}$ should give me non-emptiness.

Of course, $C$ is always closed. If I enforce, $\text{diameter}(C_n) \downarrow m$, as a counterexample, if we take $\mathbb{R}$ with $d(x,y) = \min \{\vert x- y\vert,1\}$ and $C_k = \left[k,\infty\right)$, we obtain that $C$ is empty, even though $\text{diameter}(C_n) = 1$.

Hence, I want a condition that replaces $\text{diameter}(C_n) \downarrow m$ that ensures that $C$ is non-empty.

However, when $m=0$, we see that $C$ has to be non-empty by completeness.

If we take $C_k$'s to be compact, then the theorem is true in any topological space.

Given that I am on a complete metric space, I want a weaker condition than compactness and it should be of the form closed + "something". If I take "something" = totally bounded, then closed + totally bounded will again give me compactness (since we are on a complete metric space).

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    $$\lim_{k\to\infty}\inf\{\varepsilon:C_k\text{ is the union of finitely many sets of diameter at most }\varepsilon\}=0$$2017-02-27
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    @bof Essentially "almost" close to totally bounded ?2017-02-27

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You need the concept of "completeness". If you are familiar with non-Archimedean analysis, you may see that there is also the concept "spherically completeness" which involves exactly with intersection of closed balls.