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 X {\displaystyle X}, a Lebesgue's number of the cover is a number δ > 0 {\displaystyle \delta >0} such that every subset of X {\displaystyle X} having diameter less than δ {\displaystyle \delta } 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 ( X , d ) {\displaystyle (X,d)} is compact and an open cover of X {\displaystyle X} is given, then the cover admits some Lebesgue's number δ > 0 {\displaystyle \delta >0}.

The notion of Lebesgue's numbers itself is useful in other applications as well.

Proof

Direct proof

Let U {\displaystyle {\mathcal {U}}} be an open cover of X {\displaystyle X}. Since X {\displaystyle X} is compact we can extract a finite subcover { A 1 , … , A n } ⊆ U {\displaystyle \{A_{1},\dots ,A_{n}\}\subseteq {\mathcal {U}}}. If any one of the A i {\displaystyle A_{i}}'s equals X {\displaystyle X} then any δ > 0 {\displaystyle \delta >0} will serve as a Lebesgue's number. Otherwise for each i ∈ { 1 , … , n } {\displaystyle i\in \{1,\dots ,n\}}, let C i := X ∖ A i {\displaystyle C_{i}:=X\setminus A_{i}}, note that C i {\displaystyle C_{i}} is not empty, and define a function f : X → R {\displaystyle f:X\rightarrow \mathbb {R} } by

f ( x ) := 1 n ∑ i = 1 n d ( x , C i ) . {\displaystyle f(x):={\frac {1}{n}}\sum _{i=1}^{n}d(x,C_{i}).}

Since f {\displaystyle f} is continuous on a compact set, it attains a minimum δ {\displaystyle \delta }. The key observation is that, since every x {\displaystyle x} is contained in some A i {\displaystyle A_{i}}, the extreme value theorem shows δ > 0 {\displaystyle \delta >0}. Now we can verify that this δ {\displaystyle \delta } is the desired Lebesgue's number. If Y {\displaystyle Y} is a subset of X {\displaystyle X} of diameter less than δ {\displaystyle \delta }, choose x 0 {\displaystyle x_{0}} as any point in Y {\displaystyle Y}, then by definition of diameter, Y ⊆ B δ ( x 0 ) {\displaystyle Y\subseteq B_{\delta }(x_{0})}, where B δ ( x 0 ) {\displaystyle B_{\delta }(x_{0})} denotes the ball of radius δ {\displaystyle \delta } centered at x 0 {\displaystyle x_{0}}. Since f ( x 0 ) ≥ δ {\displaystyle f(x_{0})\geq \delta } there must exist at least one i {\displaystyle i} such that d ( x 0 , C i ) ≥ δ {\displaystyle d(x_{0},C_{i})\geq \delta }. But this means that B δ ( x 0 ) ⊆ A i {\displaystyle B_{\delta }(x_{0})\subseteq A_{i}} and so, in particular, Y ⊆ A i {\displaystyle Y\subseteq A_{i}}.

Proof by contradiction

Suppose for contradiction that X {\displaystyle X} is sequentially compact, { U α ∣ α ∈ J } {\displaystyle \{U_{\alpha }\mid \alpha \in J\}} is an open cover of X {\displaystyle X}, and the Lebesgue number δ {\displaystyle \delta } does not exist. That is: for all δ > 0 {\displaystyle \delta >0}, there exists A ⊂ X {\displaystyle A\subset X} with diam ⁡ ( A ) < δ {\displaystyle \operatorname {diam} (A)<\delta } such that there does not exist β ∈ J {\displaystyle \beta \in J} with A ⊂ U β {\displaystyle A\subset U_{\beta }}.

This enables us to perform the following construction:

Note that A n ≠ ∅ {\displaystyle A_{n}\neq \emptyset } for all n ∈ Z + {\displaystyle n\in \mathbb {Z} ^{+}}, since A n ⊄ U β {\displaystyle A_{n}\not \subset U_{\beta }}. It is therefore possible by the axiom of choice to construct a sequence ( x n ) {\displaystyle (x_{n})} in which x i ∈ A i {\displaystyle x_{i}\in A_{i}} for each i {\displaystyle i}. Since X {\displaystyle X} is sequentially compact, there exists a subsequence { x n k } {\displaystyle \{x_{n_{k}}\}} (with k ∈ Z > 0 {\displaystyle k\in \mathbb {Z} _{>0}}) that converges to x 0 {\displaystyle x_{0}}.

Because { U α } {\displaystyle \{U_{\alpha }\}} is an open cover, there exists some α 0 ∈ J {\displaystyle \alpha _{0}\in J} such that x 0 ∈ U α 0 {\displaystyle x_{0}\in U_{\alpha _{0}}}. As U α 0 {\displaystyle U_{\alpha _{0}}} is open, there exists r > 0 {\displaystyle r>0} with B r ( x 0 ) ⊂ U α 0 {\displaystyle B_{r}(x_{0})\subset U_{\alpha _{0}}}. Now we invoke the convergence of the subsequence { x n k } {\displaystyle \{x_{n_{k}}\}}: there exists L ∈ Z + {\displaystyle L\in \mathbb {Z} ^{+}} such that L ≤ k {\displaystyle L\leq k} implies x n k ∈ B r / 2 ( x 0 ) {\displaystyle x_{n_{k}}\in B_{r/2}(x_{0})}.

Furthermore, there exists M ∈ Z > 0 {\displaystyle M\in \mathbb {Z} _{>0}} such that δ M = 1 M < r 2 {\displaystyle \delta _{M}={\tfrac {1}{M}}<{\tfrac {r}{2}}}. Hence for all z ∈ Z > 0 {\displaystyle z\in \mathbb {Z} _{>0}}, we have M ≤ z {\displaystyle M\leq z} implies diam ⁡ ( A M ) < r 2 {\displaystyle \operatorname {diam} (A_{M})<{\tfrac {r}{2}}}.

Finally, define q ∈ Z > 0 {\displaystyle q\in \mathbb {Z} _{>0}} such that n q ≥ M {\displaystyle n_{q}\geq M} and q ≥ L {\displaystyle q\geq L}. For all x ′ ∈ A n q {\displaystyle x'\in A_{n_{q}}}, notice that:

  • d ( x n q , x ′ ) ≤ diam ⁡ ( A n q ) < r 2 {\displaystyle d(x_{n_{q}},x')\leq \operatorname {diam} (A_{n_{q}})<{\frac {r}{2}}}, because n q ≥ M {\displaystyle n_{q}\geq M}.
  • d ( x n q , x 0 ) < r 2 {\displaystyle d(x_{n_{q}},x_{0})<{\frac {r}{2}}}, because q ≥ L {\displaystyle q\geq L} entails x n q ∈ B r / 2 ( x 0 ) {\displaystyle x_{n_{q}}\in B_{r/2}\left(x_{0}\right)}.

Hence d ( x 0 , x ′ ) < r {\displaystyle d(x_{0},x')<r} by the triangle inequality, which implies that A n q ⊂ U α 0 {\displaystyle A_{n_{q}}\subset U_{\alpha _{0}}}. This yields the desired contradiction.

  • Munkres, James R. (1974), , Prentice-Hall, p. , ISBN 978-0-13-925495-6