Template:Short description In topology, the Lebesgue covering lemma is a useful tool in the study of compact metric spaces.
Given an open cover of a compact metric space, a Lebesgue's number of the cover is a number <math>\delta > 0</math> such that every subset of <math>X</math> having diameter less than <math>\delta</math> is contained in some member of the cover.
The existence of Lebesgue's numbers for compact metric spaces is given by the Lebesgue's covering lemma:
- If the metric space <math>(X, d)</math> is compact and an open cover of <math>X</math> is given, then the cover admits some Lebesgue's number <math>\delta > 0</math>.
The notion of Lebesgue's numbers itself is useful in other applications as well.
ProofEdit
Direct proofEdit
Let <math>\mathcal U</math> be an open cover of <math>X</math>. Since <math>X</math> is compact we can extract a finite subcover <math>\{A_1, \dots, A_n\} \subseteq \mathcal U</math>. If any one of the <math>A_i</math>'s equals <math>X</math> then any <math> \delta > 0 </math> will serve as a Lebesgue's number. Otherwise for each <math>i \in \{1, \dots, n\}</math>, let <math>C_i := X \smallsetminus A_i</math>, note that <math>C_i</math> is not empty, and define a function <math>f : X \rightarrow \mathbb R</math> by
- <math>f(x) := \frac{1}{n} \sum_{i=1}^n d(x,C_i). </math>
Since <math>f</math> is continuous on a compact set, it attains a minimum <math>\delta</math>. The key observation is that, since every <math>x</math> is contained in some <math>A_i</math>, the extreme value theorem shows <math>\delta > 0</math>. Now we can verify that this <math>\delta</math> is the desired Lebesgue's number. If <math>Y</math> is a subset of <math>X</math> of diameter less than <math>\delta</math>, choose <math>x_0</math> as any point in <math>Y</math>, then by definition of diameter, <math>Y\subseteq B_\delta(x_0)</math>, where <math>B_\delta(x_0)</math> denotes the ball of radius <math>\delta</math> centered at <math>x_0</math>. Since <math>f(x_0)\geq \delta</math> there must exist at least one <math>i</math> such that <math>d(x_0,C_i)\geq \delta</math>. But this means that <math>B_\delta(x_0)\subseteq A_i</math> and so, in particular, <math>Y\subseteq A_i</math>.
Proof by contradictionEdit
Suppose for contradiction that <math>X</math> is sequentially compact, <math>\{ U_{\alpha} \mid \alpha \in J \}</math> is an open cover of <math>X</math>, and the Lebesgue number <math>\delta</math> does not exist. That is: for all <math>\delta > 0</math>, there exists <math>A \subset X</math> with <math>\operatorname{diam} (A) < \delta</math> such that there does not exist <math>\beta \in J</math> with <math>A \subset U_{\beta}</math>.
This enables us to perform the following construction:
Note that <math>A_{n} \neq \emptyset</math> for all <math> n \in \mathbb{Z}^{+}</math>, since <math>A_{n} \not\subset U_{\beta}</math>. It is therefore possible by the axiom of choice to construct a sequence <math>(x_{n})</math> in which <math>x_{i} \in A_{i}</math> for each <math>i</math>. Since <math>X</math> is sequentially compact, there exists a subsequence <math>\{x_{n_{k}}\}</math> (with <math>k \in \mathbb{Z}_{> 0}</math>) that converges to <math>x_{0}</math>.
Because <math>\{ U_{\alpha} \}</math> is an open cover, there exists some <math>\alpha_{0} \in J</math> such that <math>x_{0} \in U_{\alpha_{0}}</math>. As <math>U_{\alpha_{0}}</math> is open, there exists <math>r > 0</math> with <math>B_{r}(x_{0}) \subset U_{\alpha_{0}}</math>. Now we invoke the convergence of the subsequence <math> \{ x_{n_{k}} \} </math>: there exists <math> L \in \mathbb{Z}^{+}</math> such that <math> L \le k</math> implies <math>x_{n_{k}} \in B_{r/2} (x_{0})</math>.
Furthermore, there exists <math>M \in \mathbb{Z}_{> 0}</math> such that <math> \delta_{M}= \tfrac{1}{M} < \tfrac{r}{2} </math>. Hence for all <math>z \in \mathbb{Z}_{> 0}</math>, we have <math>M \le z</math> implies <math>\operatorname{diam} (A_{M}) < \tfrac{r}{2}</math>.
Finally, define <math>q \in \mathbb{Z}_{> 0}</math> such that <math>n_{q} \geq M</math> and <math>q \geq L</math>. For all <math>x' \in A_{n_{q}}</math>, notice that:
- <math> d(x_{n_{q}},x') \leq \operatorname{diam} (A_{n_{q}})<\frac{r}{2}</math>, because <math>n_{q} \geq M</math>.
- <math>d(x_{n_{q}},x_{0})<\frac{r}{2}</math>, because <math>q \geq L</math> entails <math>x_{n_{q}} \in B_{r/2}\left(x_{0}\right)</math>.
Hence <math>d(x_{0},x')<r</math> by the triangle inequality, which implies that <math>A_{n_{q}} \subset U_{\alpha_{0}}</math>. This yields the desired contradiction.