Content deleted Content added
Alsosaid1987 (talk | contribs) mostly formatting |
Alsosaid1987 (talk | contribs) m formatting -- no need to put proof in own subsection |
||
Line 11:
'''Note:''' We may leave out the closedness condition in situations where every compact subset of ''S'' is closed, for example when ''S'' is [[Hausdorff space|Hausdorff]].
''Proof.'' Assume, by way of contradiction, that <math>\bigcap C_k=\emptyset</math>. For each ''k'', let <math>U_k=C_0\setminus C_k</math>. Since <math>\bigcup U_k=C_0\setminus\left(\bigcap C_k\right)</math> and <math>\bigcap C_k=\emptyset</math>, we have <math>\bigcup U_k=C_0</math>. Note that, since the <math>C_k</math> are closed relative to ''S'' and therefore, also closed relative to <math>C_0</math>, the <math>U_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_k)</math> is an open cover (on <math>C_0</math>) of <math>C_0</math>, we can extract a finite cover <math>\{U_{k_1}, U_{k_2}, \ldots, U_{k_m}\}</math>. Let <math>M=\max_{1\leq i\leq m} {k_i}</math>. Then <math>\bigcup U_{k_i}=U_M</math> because <math>U_1\subset U_2\subset\cdots\subset U_n\subset U_{n+1}\cdots</math>, by the nesting hypothesis for the collection <math> (C_k).</math> Consequently, <math>C_0=\bigcup U_{k_i} = U_M</math>. But then <math>C_M=C_0\setminus U_M=\emptyset</math>, a contradiction. [[Q.E.D.|∎]]
Line 40:
:<math>\left(\bigcap_{k} C_k\right) \neq \emptyset. </math>
:
''Proof.'' Each nonempty, closed, and bounded subset <math>C_k\subset\mathbf{R}</math> admits a minimal element <math>x_k</math>. Since for each ''k'', we have ▼
▲Each nonempty, closed, and bounded subset <math>C_k\subset\mathbf{R}</math> admits a minimal element <math>x_k</math>. Since for each ''k'', we have
:<math>x_{k+1} \in C_{k+1} \subseteq C_k</math>,
Line 72 ⟶ 71:
''for some x in X.''
''Proof (sketch).'' A proof goes as follows. Since the diameters tend to zero, the diameter of the intersection of the ''<math>C_k</math>'' is zero, so it is either empty or consists of a single point. So it is sufficient to show that it is not empty. Pick an element <math>x_k\in C_k</math> for each ''k''. Since the diameter of ''<math>C_k</math>'' tends to zero and the ''<math>C_k</math>'' are nested, the ''<math>x_k</math>'' form a Cauchy sequence. Since the metric space is complete this Cauchy sequence converges to some point ''x''. Since each ''<math>C_k</math>'' is closed, and ''x'' is a limit of a sequence in ''<math>C_k</math>'', ''x'' must lie in ''<math>C_k</math>''. This is true for every ''k'', and therefore the intersection of the ''<math>C_k</math>'' must contain ''x''. ∎▼
▲A proof goes as follows. Since the diameters tend to zero, the diameter of the intersection of the ''<math>C_k</math>'' is zero, so it is either empty or consists of a single point. So it is sufficient to show that it is not empty. Pick an element <math>x_k\in C_k</math> for each ''k''. Since the diameter of ''<math>C_k</math>'' tends to zero and the ''<math>C_k</math>'' are nested, the ''<math>x_k</math>'' form a Cauchy sequence. Since the metric space is complete this Cauchy sequence converges to some point ''x''. Since each ''<math>C_k</math>'' is closed, and ''x'' is a limit of a sequence in ''<math>C_k</math>'', ''x'' must lie in ''<math>C_k</math>''. This is true for every ''k'', and therefore the intersection of the ''<math>C_k</math>'' must contain ''x''. ∎
A converse to this theorem is also true: if ''X'' is a metric space with the property that the intersection of any nested family of non-empty closed subsets whose diameters tend to zero is non-empty, then ''X'' is a complete metric space. (To prove this, let ''<math>(x_k)</math>'' be a Cauchy sequence in ''X'', and let ''<math>C_k</math>'' be the closure of the tail of this sequence.)
| |||