Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $(X,\tau)$ a compact metric space and $\{ U_i : i \in I \}$ an open cover of $X$. Show that there is $r>0$ such that for all $a \in X$ there is an $i \in I$ such that $B_{r}(x) \subseteq U_{i}$.

My attempt:

By definition of compactness, $X$ is covered by some finite subset of $\{ U_{i} : i \in I \}$. Let $U_{1}, \ldots , U_{n}$ be such a finite subcover of $X$.

Choose any $x\in X$. Suppose $x$ lies in $U_1$. There is some number $r_{1}(x)$ such that for any $r < r_{1}(x)$, the ball $B_{r}(x)$ lies in $U_{1}$. For any $x \in X$, we may associate to $x$ the $n$ numbers $r_{1}(x), \ldots, r_{n}(x)$, noting that at least one of these is non-negative. Let $r(x)$ be the least non-negative member of $\{ r_{1}(x), \ldots , r_{n}(x) \}$.

Below, I prove that $r: X \to \mathbb{R}$ is a continuous function. As $r$ is continuous, $r(X)$ is a continuous subset of the real numbers. Therefore $r(X)$ is a finite union of finite closed intervals in $\mathbb{R}$. In particular, $r(X)$ contains a least element which we denote by $r_{0}$. Note that as $r(x)$ is greater than zero for all $x \in X$, we know that $r_{0} > 0$. For any $p<r_{0}$ and for any $x\in X$, we know that $B_{p}(x)$ lies in at least one $U_{i}$.

Is this correct? Can you give me another alternative solution?

share|cite|improve this question
This is very close to the result known as Lebesgue's covering lemma or Lebesgue's number lemma. There are a few questions on MSE related to this, e.g., Proof of the Lebesgue number lemma, Uses of Lebesgue's covering lemma or Explanations of Lebesgue number lemma. – Martin Sleziak Jun 26 '12 at 12:10

The argument does not work as it stands: your numbers $r_i(x)$ are not well-defined, so you’ve no guarantee that $r$ is continuous. Moreover, there’s nothing to keep you from choosing each $r_i$ to be $0$, since $B_0(x)=\varnothing\subseteq U_i$, and in that case $\{r_1(x),\dots,r_n(x)\}$ has no non-negative member. If you use this approach, you need to choose the $r_i(x)$ more systematically, and in a way that ensures that you choose a positive value if possible. One way is to let $r_i(x)$ measure the distance from $x$ to $X\setminus U_i$, as I now see that Arthur Fischer has just suggested, so I’ll say no more about that approach.

Here’s a completely different approach. Let $\mathscr{U}$ be an open cover of $X$. For each $x\in X$ there is an $\epsilon(x)$ such that $B_{\epsilon(x)}(x)\subseteq U$ for some $U\in\mathscr{U}$. Let $\mathscr{B}=\{B_{\epsilon(x)/2}(x):x\in X\}$; this is an open cover of $X$, so it has a finite subcover $\{B_{\epsilon(x_1)/2}(x_1),\dots,B_{\epsilon(x_n)/2}(x_n)\}$. Let $\epsilon=\frac12\min\{\epsilon(x_1),\dots,\epsilon(x_n)\}$.

Let $x\in X$ be arbitrary; $x\in B_{\epsilon(x_k)/2}(x_k)$ for some $k\in\{1,\dots,n\}$. Suppose that $y\in B_\epsilon(x)$, then $d(y,x_k)\le d(y,x)+d(x,x_k)<\epsilon+\epsilon(x_k)/2\le 2\epsilon(x_k)/2=\epsilon(x_k)$, so $y\in B_{\epsilon(x_k)}$, and therefore $B_\epsilon(x)\subseteq B_{\epsilon(x_k)}$. Thus, for each $x\in X$ the set $B_\epsilon(x)$ is a subset of some member of $\mathscr{U}$.

share|cite|improve this answer

You are pretty close. The main problem seems to be the ad hoc nature of your $r_i (x)$, which would preclude ever being able to prove the continuity of $r$. Instead, once you have your finite subcover $U_1 , \ldots , U_n$, for each $i \leq n$ define a function $f_i : X \to \mathbb{R}$ by $$f_i (x) = d ( x , X \setminus U_i ) = \inf \{ d(x,y) : y \in X \setminus U_i \}.$$ Show that these functions are continuous, and $f_i (x) > 0$ iff $x \in U_i$.

Next consider the function $f : X \to \mathbb{R}$ defined by $$f(x) = \max \{ f_1 (x) , \ldots , f_n (x) \}.$$ The compactness of $X$ will tell you something important about this continuous function, and this something important should lead you to finding $r$.

share|cite|improve this answer

By choosing a finite subcover $(U_i)_{1\leq i\leq N}$ right at the start one looses a lot of manoeuvrability. Here is another approach: Define the function $\rho: X\to{\mathbb R}_{>0}$ by $$\rho(x)\ =\ \sup\{\delta>0\ |\ \exists i\in I:\ U_\delta(x)\subset U_i\}\qquad(x\in X)\ .$$ So any ball with center $x$ and radius $\rho'<\rho(x)$ is contained in at least one $U_i$. By means of the triangle inequality one easily shows that the function $\rho$ is $1$-Lipschitz, whence continuous on $X$. As $\rho(x)>0$ for all $x$ there is a $\rho_{\min}>0$ with $\rho(x)\geq \rho_{\min}$ for all $x$. Now put $r:=\rho_{\min}/2$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.