Let $(M,d)$ a metric space and $\mathcal{U}=\{U_i\}_{i\in I}$ an open cover. A number $\lambda>0$ is called a Lebesgue number for $\mathcal{U}$ if for every $x\in M$ the ball $B_\lambda(x)\subseteq U_{i_x}$ for any $i_x\in I$. If all open covers have a Lebesgue number, we say that $(M,d)$ has the property of Lebesgue number.

Remark: Given an open cover $\mathcal{U}=\{U_i\}_{i\in I}$ for every $x\in M$ exist $U_{i_x}$ such that $x\in U_{i_x}$ and, because is open, exist $r_x>0$ such that the ball $B_{r_x}(x)\subseteq U_{i_x}$. The importance of having a Lebesgue number is that this radius works for all $x\in M$ at the same time.

Results: It can be shown that:

  • (Lebesgue number's lemma) If $(M,d)$ is compact, then it's has the property of Lebesgue number.

Furthermore, a characterization of the spaces that have the property of the Lebesgue number can be given:

  • (Theorem -UC spaces or Atsuji spaces-) Let $(M,d)$ a metric space. These statements are equivalent:
  1. $(M,d)$ has the property of the Lebesgue number.
  2. For every metric space $(Y,d')$ if $f:M\to Y$ is continuous then $f$ is uniformly continuous.
  3. $(i)$ $M'$ (the set of accumulation points of $M$) is compact and $(ii)$ for every $\varepsilon>0$ exist $\delta>0$ such that if $x,y\in M-M'$ satisfy that $d(x,M')\geq\varepsilon$ and $d(y,M')\geq\varepsilon$, then $d(x,y)>\delta$.

With this, we can say that a Atsuji space can be write as a compact set union a discrete set.

Question: What are the applications of this?

For example, if $(M,d)$ is Atsuji and chainable, we can prove that $M$ is connected. In fact, if $M=A\cup B$ with $A,B$ disjoint, non empty and open sets the cover $\{A,B\}$ has a Lebesgue number $\lambda>0$. For $a\in A, b\in B$ we can select a $\lambda$-chain (ie, $x_0=a, x_1, \dots,x_{n-1},x_n=b$ such that $d(x_i,x_{i+1})<\lambda$).

Because $\lambda$ is a Lebesgue number and $a\in A$, then $B_\lambda(a)\subseteq A$ (and not in $B$). Note that $x_1\in B_\lambda(a)\subseteq A$ and therefor $B_\lambda(x_1)\subseteq A$. Thus following, we come to that $b=x_n\in B_\lambda(x_{n-1})\subseteq A$, a contradiction because $A\cap B=\emptyset$. The contradiction came from assuming that $M$ is not connected and therefore is connected.

More in general, this idea can be used for example to prove that every interval of $\mathbb{R}$ is connected. What are other uses of the Lebesgue number in the theory of metric spaces?

  • 1
    $\begingroup$ are you looking for applications of Lebesgue number like arguments to only metric space or in another fields? $\endgroup$
    – Sebathon
    Commented Oct 29, 2021 at 22:06
  • $\begingroup$ @Sebathon in general maybe $\endgroup$ Commented Oct 29, 2021 at 22:50

2 Answers 2


Here I present some examples from topology. If I have time, I will add another ones.

Some examples of Munkres' Topology:

In Theorem 28.2.2, we use it to prove that in a metrizable space, sequentially compactness implies compactness. Se also $\S 50$ abut topological dimension. He also use it in the following result.

First let me recall that given a continuous map $f:X \to B$ and a covering map $p:E \to B$, a lifting of $f$ is a map $\tilde{f}:X \to E$ s.t. $p \circ \tilde{f}=f$.

Theorem. Let $p:E \to B$ be a covering map with $p(e_{0})=b_{0}$. Then any path $f:[0,1] \to B$ beginning at $b_{0}$ has a unique lifting to a path $\tilde{f}$ in $E$ beginning at $e_{0}$.

Proof. Cover $B$ by open sets $U$ each of which is evenly covered by $p$ (i.e., $p^{-1}(U)$ is the disjoint union of open sets, each one homeomorphic by $p$ to $U$). By Lebesgue's lemma, we can find a partition of $[0,1]$, say $s_{0}, \ldots,s_{n}$, such that for each $i$ the set $f([s_{i},s_{i+1}])$ lies in such an open set $U$.

Define $\tilde{f}(0)=e_{0}$. Now, supposing $\tilde{f}(s)$ is defined for $0 \leq s \leq s_{i}$, define $\tilde{f}$ on $[s_{i},s_{i+1}]$ as follows: by construction, the set $f([s_{i},s_{i+1}])$ lies in some $U$ evenly cover by $p$. Let $\{V_{\alpha}\}$ be the disjoint open sets whose union is $p^{-1}(U)$. Say that $\tilde{f}(s_{i})$ is in some of this sets, call it $V_{0}$. For $s \in [s_{i},s_{i+1}]$ define $$ \tilde{f}(s)=(p|_{V_{0}})^{-1}(f(s)). $$ Since $p|_{V_{0}}:V_{0} \to U$ is a homeomorphism, then $\tilde{f}$ is continuous on $[s_{i},s_{i+1}]$. Following in this way you can define $f$ in all the unit interval.

The continuity follows by the pasting lemma. The uniqueness is because from the base point, you can translate the uniqueness to the whole interval using that if $g([s_{i},s_{i+1}])$ is in some $V_{\alpha}$ and $p|_{V_{\alpha}} \to U$ is an homeomorphism. See Munkre's book for more details.

In a similar way, using Lebesgue lemma on $[0,1] \times [0,1]$, it can be proved the following:

Theorem. Let $p:E \to B$ be a covering map with $p(e_{0})=b_{0}$. Let $F:[0,1]^{2} \to B$ be continuous, with $F(0,0)=b_{0}$. Then there is a unique lifting $\tilde{F}:[0,1]^{2} \to E$ such that $\tilde{F}(0,0)=e_{0}$.

Finally, but not less interesting, Munkres also use it to prove a "baby" version of Seifert–-Van Kampen theorem, see Theorem 59.1.

From Topology and Geometry by Glen E. Bredon

Theorem. $S^{n}$ is simply connected for $n \geq 2$.

Proof. Cover $S^{n}$ with open hemispheres. Let $f:I \to S^{n}$ be a loop and consider the cover of $I$ by the inverse images of the hemispheres by $f$. Since $I$ is compact, by Lebesgue's number lemma there exists $n \in \mathbb{N}$ such that each interval of length $1/n$ is mapped by $f$ to an open hemisphere. Let $g(s)$ be a parametrization of the line segment in $\mathbb{R}^{n+1}$ from $f(i/n)$ to $f((i+1)/n)$. Note that since we are in an open hemisphere, $g$ is never zero. Let $$ F_{i}(s,t)=\frac{ tg(s)+(1-t)f(s) }{\| tg(s)+(1-t)f(s) \|}. $$ gluing these homotopies together, you obtain a homotopy from $f$ to a loop made up of a finite number of great circles arcs. Let $p \in S^{n}$ be a point that this loop does not touch. All this construction implies that $[f]$ is in the image of the homomorphism $\pi_{1}(S^{n} \setminus \{p\}) \to \pi_{1}(S^{n})$. Since $S^{n} \setminus \{p\}$ is contractible (recall the stereographic projection), then $\pi_{1}(S^{n} \setminus \{p\})$ is trivial. Hence $[f]$ is trivial and we are done.

From Geometric Theory of Dynamical Systems by Jacob Palis Jr and Welington de Melo:

Theorem. $\mathrm{Diff}(M)=\{f \in C^{r}(M,M):f \text{ is a diffeomorphism}\}$ is open (in $C^{r}(M,M)$). Here $M$ is a compact manifold.

Proof. Let $f \in \mathrm{Diff}(M)$. W.L.O.G. $M \subset \mathbb{R}^{s}$ (by Whitney's Theorem). Given $p \in M$, by IFT there exist neighborhoods $V_{p} \ni p$ on $M$ and $\mathscr{V}_{p}\ni f$ in $C^{r}(M,M)$ such that if $g \in \mathscr{V}_{p}$, then $g|_{V_{p}}$ is a diffeomorphism. By compactness, there exists a finite subcover $V_{p_{1}},\ldots,V_{p_{j}}$ of $M$. Define $\mathscr{V}= \cap_{i=1}^{j} \mathscr{V}_{p_{i}}$. Let $\lambda>0$ be the Lebesgue number of that cover. Then, by construction, if $0<d(p,q)<\lambda$ then $g(p) \neq g(q)$ for $g \in \mathscr{V}$. On the other hand, $$\inf\{d(f(p),f(q)): p,q \in M \text{ and }d(p,q) \geq \lambda\}$$ is positive. By shrinking $\mathscr{V}$, we can assume that if $g \in \mathscr{V}$ then $g$ is injective. Therefore $g$ is a bijective local diffeomorphism, we conclude that $g$ is a diffeomorphism.

Here are some applications from do Carmo's Differential Geometry of Curves & Surfaces

The author uses the Lebesgue's number lemma in a part of the proof of the following result:

Proposition. Let $S \subset \mathbb{R}^{3}$ be a regular compact orientable surface. Then there exists a number $\varepsilon>0$ such that whenever $p,q \in S$ the segments of the normal lines of length $2\varepsilon$, centered in $p$ and $q$, are disjoint.

Then, he uses this result (among other lemmas) to prove:

Theorem. Let $S \subset \mathbb{R}^{3}$ be a regular compact orientable surface. Then there exists a differentiable function $g:V \to \mathbb{R}$, defined in a open set $V \subset \mathbb{R}^{3}$ which has zero as a regular value and is such that $S=g^{-1}(0)$.

In Mitrea's Distributions, Partial Differential Equations, and Harmonic analysis, is used in one step to prove the following

Lemma 14.56. Let $\alpha>0$. Suppose $\Omega$ is either an upper-graph Lipschitz domain in $\mathbb{R}^{n}$, or a bounded Lipschitz domain in $\mathbb{R}^{n}$. Then there exists some constant $C=C(\Omega,\alpha) \in (0,\infty)$such that for each $x \in \partial \Omega$ and each $r>0$ we have $$\int_{\partial \Omega} \frac{\chi_{\{|x-y|<r\}}}{|x-y|^{n-1-\alpha}}d\sigma(y) \leq C r^{\alpha}.$$ In particular, corresponding to the special case when $\alpha=n-1$, this becomes $$\sigma(\partial \Omega \cap B(x,r))\leq Cr^{n-1}, \quad \forall x \in \Omega, \forall r \in (0,\infty).$$

If you are interested, this implies

Theorem 12.44. Let $\Omega$ be either an upper-graph Lipschitz domian in $\mathbb{R}^{n}$, or a bounded Lipschitz domain in $\mathbb{R}^{n}$. Then the following are true

  1. If $\varphi \in \mathrm{Lip}_{\mathrm{comp}}(\partial \Omega)$ then $\varphi u \in H^{1/2}(\partial \Omega)$ for every $u \in H^{1/2}(\partial \Omega)$ and $$ \| \varphi u \|_{H^{1/2}(\partial \Omega)} \leq C(\varphi,\Omega) \| u \|_{H^{1/2}(\partial \Omega)}, $$ for some constant $C(\varphi,\Omega) \in (0,\infty)$ independent of $u$.
  2. The set $\mathrm{Lip}_{\mathrm{comp}}(\partial \Omega)$ is dense in $H^{1/2}(\partial \Omega)$.

Finally, Lebesgue's number arguments are used in Ergodic Theory when you are dealing with pressure. See Walter's An Introduction to Ergodic Theory and Viana and Oliveira's Fundamentos da Teoria Ergódica (the version in portuguese from the Sociedade Brasileira de Matemática). This was communicated to me by the user Ángela Flores.


I've seen this used a lot in dynamical systems to dominate open covers. In the most general setting possible lets say that I have some function $E$ that sends open covers to some number. One example can be the topological entropy of a continuous transformation $T:X \to X$ on some topological space.

We will also say that $E$ has the following condition:

If $A$ and $B$ are open covers and if for any $A_i \in A$ we have a $B_j \in B$ such that $A_i \subset B_j$ then we want $E(A) \leq E(B)$. This gives us some sort of monotonicity.

Now let $r$ be the Lebesgue number for the open cover $B$. If A is a collection of open balls with radius less that $r$ then we have $E(A) \leq E(B)$. Notice that we have found a bound for $E(A)$ in terms of $B$ for any open cover that is sufficiently small. This idea is how you prove the kolmogorov-sinai entropy theorem.


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