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Cantor's Intersection theorem says if $F_{n+1}\subset F_n$ $\forall n\in\Bbb{N}$, then $\bigcap_{i=1}^{\infty}F_i\neq\emptyset$.

This is valid only in complete metric spaces, and the proof is based on a cauchy sequence formed by choosing one $x_k$ from each $F_k$. The number of terms in a convergent sequence is countable, and hence, it makes sense that the number of nested closed sets $F_i$ is also countable.

Is there a way to generalise this to an uncountable number of nested closed sets? Say $F_m\subset F_n\forall m,n\in\Bbb{R},n>m$? If this were possible, it woud prove $\Bbb{R}$ is not the union of an uncountable number of nowhere dense sets, which would be a contradiction.

Thanks in advance!

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    Your $F_n$ in the original statement should be closed, bounded, nonempty. So you will have to put that in your uncountable version, too. – GEdgar Aug 12 '13 at 03:23

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Let $I$ be an uncountable index set endowed with a well ordering $\succeq$ (this is possible by the axiom of choice). For any $i\in I$, let $F_i$ be a nonempty, bounded, and closed subset of a complete and totally bounded metric space $S$ (hence, $F_i$ is compact for any $i\in I$). Moreover, suppose that $F_i\subseteq F_j$ whenever $i,j\in I$ such that $i\succeq j$. I will show that $F\equiv\bigcap_{i\in I} F_i$ is nonempty.

Suppose that $F$ is empty. Let $V_i\equiv S\bigcap F_i^c$ for any $i\in I$. Then, \begin{align*} \bigcup_{i\in I}V_i=S\bigcap\left(\bigcup_{i\in I} F_i^c\right)=S\bigcap\left(\bigcap_{i\in I}F_i\right)^c=S\bigcap\varnothing^c=S. \end{align*} Therefore, $\{V_i\}_{i\in I}$ is an open cover of $S$ and hence also of any subset of $S$.

Let $k$ be the least element in $I$ according to the well order $\succeq$. Then, $F_i\subseteq F_k$ for any $i\in I$. Since $F_k$ is compact and $\{V_i\}_{i\in I}$ is an open cover of it, there is an open subcover $\{V_i\}_{i\in J}$, where $J\subset I$ is a nonempty finite set. Let $\ell$ be the greatest element in $J$. Then, $\bigcup_{i\in J}V_j=V_{\ell}$ by construction. Hence, $F_k\subseteq V_{\ell}$, which implies that $F_i\subseteq V_{\ell}$ for any $i\in I$. But this is a contradiction, since $F_{\ell}$ is nonempty (so that $x\in F_{\ell}$ should imply that $x\in V_{\ell}\subseteq F_{\ell}^c$).

Therefore, $F\neq\varnothing.$

triple_sec
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  • Actually, the assumption that $\succeq$ is a well ordering can be dispensed with. The proof can be modified for any given total order $\succeq$ that satisfies the following property: $I$ can be decomposed as the union of two disjoint, nonempty subsets $I_1$ and $I_2$ such that $I_2$ contains a least element and for any $i_1\in I_1$ and $i_2\in I_2$, $i_2\succeq i_1$. (continued below) – triple_sec Aug 12 '13 at 05:50
  • (continued from above) For example, if $I=(1,\infty)$, the proof can be slightly modified so that it goes through for the standard ordering $\geq$ with $I_1=(1,2)$ and $I_2=[2,\infty)$. – triple_sec Aug 12 '13 at 05:52
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Suppose the $F_k$ are closed and bounded, for each real $k\ge 1$, and that $F_m \subset F_n$ whenever $m > n.$ Then the subsequence $F_1,F_2,F_3,\cdots$ satisfies the hypothesis of Cantor's intersection theorem, so that the intersection of these $F_k$ is nonempty, so contains some specific point $z.$ But then the intersection of all the $F_m$, $m \in [1,\infty)$ is also nonempty, since for any $m$ we can choose a $k>m$ with $k$ an integer, so that $F_k \subset F_m$ and thus $z \in F_m.$

coffeemath
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  • But does this not prove that $\Bbb{R}$ is not the union of uncountable nowhere dense sets? Isn't this a contradiction? –  Aug 12 '13 at 04:04
  • My assumption is based on the fact that $\forall r\in\Bbb{R}$, ${r}$ is nowhere dense, and $\Bbb{R}=\bigcup{r}$. –  Aug 12 '13 at 04:13
  • @AyushKhaitan What makes you think that this result implies that $\mathbb{R}$ is not the union of an uncountable number of nowhere dense sets? – triple_sec Aug 12 '13 at 04:23
  • Let $A_i$ be one of the nowhere dense sets $\Bbb{R}$ is the union of. Then there is an open set $S_i$ which is disjoint with $A_i$. Let $F_i\subset S_i$ be a closed set. For $r>i$, there is an open set $S_r\subset F_i$ which is disjoint with $A_r$. Let $\lim_{n\infty}dia(F_n)=0$. Then we get an uncountable number of nested closed bounded sets which must be non-empty, as you say. The point $p\in\bigcap^\infty F_k$ is not present in $\bigcup^\infty A_i$. –  Aug 12 '13 at 04:30
  • The proof is similar to the one which proves Baire's Category Theorem. –  Aug 12 '13 at 04:34
  • How did you conclude that $F_i$ must contain a nonempty open set? What if all of the $F_i$'s are nowhere dense? I'm not sure I understand your argument. Can you present a more systematic proof? – triple_sec Aug 12 '13 at 04:36
  • Let's see the Baire category theorem. The first part says that if ${U_n}{n\in\mathbb{Z}+}$ is a countable sequence of open dense sets, then their intersection is dense. This need not be true if you consider uncountably many open dense sets, which I think is what causes your confusion. Indeed, let, for any $x\in\mathbb{R}$, $U_x\equiv\mathbb{R}\setminus{x}$. Then, $U_x$ is open and dense, but $\bigcap_{x\in\mathbb{R}} U_x$ is empty. (continued below) – triple_sec Aug 12 '13 at 04:49
  • (continued from above) Therefore, you cannot use the first part of the BCT to conclude that $\mathbb{R}$ cannot be an uncountable union of nowhere dense sets. I see no contradiction here. – triple_sec Aug 12 '13 at 04:50
  • Now I skimmed through the proof of the first part of the Baire category theorem. It uses ordinary induction. To prove your statement for uncountably many sets, I conjecture that you will need transfinite induction. I did not try to work out the details, but my intuition suggests that that's where the logic of your counterargument might go astray. – triple_sec Aug 12 '13 at 05:00
  • let us continue this discussion in chat – triple_sec Aug 12 '13 at 05:35
  • Thanks for the comments and insights @triple_sec! I'm not a big fan of the chat software here. Moreover I am yet to fully understand your argument. It would be great if I could try and assess your comments on my own in some time and get back to you in the comments section itself? –  Aug 12 '13 at 05:41
  • Actually, I didn't suggest the chat session, it seems to have been an auto-generated message. Sure, I'm eager to see exactly why this result still does not contradict the fact that $\mathbb{R}$ is an uncountable union of nowhere dense sets. – triple_sec Aug 12 '13 at 05:45
  • A family $ F$ of sets has the F.I.P. (Finite Intersection Property) iff $\cap G \neq \phi$ whenever $ G $ is a finite non-empty subset of $ F$. Theorem (Heine-Borel): A bounded closed subspace of the reals is compact. It is immediate from the definition of "compact" that if a non-empty family $ F$ of closed subsets of a compact space has the F.I.P., then $\cap F \neq \phi$. (Without the axiom of choice we would say "Tarski-finite" for "finite".) – DanielWainfleet Sep 15 '15 at 04:21