|
==Topological Statement==
'''Theorem.''' ''Let'' <math>''S</math>'' ''be a [[Topological Space|topological space]]. A decreasing nested sequence of non-empty compact,'' ''closed subsets of <math>S</math> has a non-empty intersection. In other words, supposing'' <math>(C_nC_k)</math> ''is a sequence of non-empty compact, closed subsets of'' <math>''S</math>'' ''satisfying''
:<math>C_0 \supset C_1 \supset \cdots C_kC_n \supset C_{kn+1} \cdots, </math>
''it follows that''
:<math>\left(\bigcap_{nk} C_nC_k\right) \neq \emptyset. </math>
'''Note:''': We may leave out the closedness condition in situations where every compact subset of <math>''S</math>'' is closed, for example when <math>''S</math>'' is [[Hausdorff space|Hausdorff]].
=== Proof ===
Assume, by way of contradiction, that <math>\bigcap C_nC_k=\emptyset</math>. For each <math>n</math>''k'', let <math>U_nU_k=C_0\setminus C_nC_k</math>. Since <math>\bigcup U_nU_k=C_0\setminus\left(\bigcap C_nC_k\right)</math> and <math>\bigcap C_nC_k=\emptyset</math>, we have <math>\bigcup U_nU_k=C_0</math>. Note that, since the <math>C_nC_k</math> are closed relative to <math>''S</math>'' and therefore, also closed relative to <math>C_0</math>, the <math>U_nU_k</math>, their set complements in <math>C_0</math>, are open relative to <math>C_0</math>.
Since <math>C_0\subset S</math> is compact and <math>(U_nU_k)</math> is an open cover (on <math>C_0</math>) of <math>C_0</math>, we can extract a finite cover <math>\{U_{n_1k_1}, U_{n_2k_2}, \ldots, U_{n_mk_m}\}</math>. Let <math>M=\max_{1\leq i\leq m} {n_ik_i}</math>. Then <math>\bigcup U_{n_ik_i}=U_M</math> because <math>U_1\subset U_2\subset\cdots\subset U_kU_n\subset U_{kn+1}\cdots</math>, by the nesting hypothesis for the collection <math> (C_nC_k).</math> Consequently, <math>C_0=\bigcup U_{n_ik_i} = U_M</math>. But then <math>C_M=C_0\setminus U_M=\emptyset</math>, a contradiction.
==Statement for Real Numbers==
A simple corollary of the theorem is that the [[Cantor set]] is nonempty, since it is defined as the intersection of a decreasing nested sequence of sets, each of which is defined as the union of a finite number of closed intervals; hence each of these sets is non-empty, closed, and bounded. In fact, the Cantor set contains uncountably many points.
'''Theorem.''' ''Let'' <math>(C_k)</math> ''be a family of non-empty, closed, and bounded subsets of'' <math>\mathbf{R}</math> ''satisfying''
:<math>C_0 \supset C_1 \supset \cdots C_n \supset C_{n+1} \cdots. </math>
''Then,''
:<math>\left(\bigcap_{k} C_k\right) \neq \emptyset. </math>
=== Proof ===
Each nonempty, closed, and bounded non-empty subset ''C''<submath>''k''C_k\subset\mathbf{R}</submath> of '''R''' admits a minimal element ''x''<submath>''k''x_k</submath>. Since for each ''k'', we have
:<math>x_{k+1} \in C_{k+1} \subseteq C_k</math>,
it follows that
:<math>x_k \le x_{k+1}</math>,
so (''x''<submath>''k''(x_k)</submath>) is an increasing sequence contained in the bounded set ''C''<submath>0C_0</submath>. The [[Monotonemonotone convergence theorem]] for bounded sequences of real numbers now guarantees the existence of a limit point
:<math>x=\lim_{k\to \infty} x_k.</math>
For fixed ''k'', we have that ''x''<sub>''j''</sub>∈''C''<sub>''k''</sub> for all ''j''≥''k'' and since ''C''<sub>''k''</sub> was closed, it follows that ''x''∈''C''<sub>''k''</sub>.
For fixed ''k'', <math>x_j\in C_k</math> for all <math>j\geq k</math> and since <math>C_k</math> was closed and ''x'' is a [[limit point]], it follows that <math>x\in C_k</math>. Our choice of ''k'' was arbitrary, hence ''x'' belongs to the intersection of all ''C''<submath>''k''\bigcap_k C_k</submath>'' and the proof is complete.
== Variant in complete metric spaces ==
| 