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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?

user642796
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Daniela del Carmen
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3 Answers3

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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 $\mathscr{U}$.

Brian M. Scott
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  • Perhaps one tiny detail is to notice that $U\subseteq \cup_{j=1}^m B_{r(x_j)/2}(x_j)$ for $m\leq n$. – user2820579 Dec 15 '20 at 18:36
  • @user2820579: For what $U$? If you mean some arbitrary $U\in\mathscr{U}$, that’s not in general true. In particular, it’s very unlikely to be true for any $U$ when $m=1$. – Brian M. Scott Dec 15 '20 at 18:59
  • If your cover are those $B_{\epsilon(x_1)/2}(x_1)$, how cannot be that any other element of some cover, say $\mathcal{U}$, does not fufill the property that I wrote if $X=\cup_{j=1}^n B_{\epsilon(x_j)/2}(x_j)$? – user2820579 Dec 15 '20 at 19:34
  • @user2820579: Every subset of $X$ is contained in that union for some $m\le n$, but quite possibly only for $m=n$. And in any case this is not relevant to the proof. – Brian M. Scott Dec 15 '20 at 19:48
  • Yes, that is probably the case in general. Actually the intention of my comment was to explicitly imply $B_\epsilon (x) \subseteq U$. In the proof you gave, it seemed to me that this tiny connection was needed to be said explicitly. – user2820579 Dec 15 '20 at 20:04
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    @user2820579: But that doesn’t actually follow from your comment. It does, however, follow from the definition of $B_{\epsilon(x_k)}$ that $B_\epsilon(x)$ is a subset of some member of $\mathscr{U}$. – Brian M. Scott Dec 15 '20 at 20:09
  • I still cannot see how you can ensure $B_\epsilon(x)\subseteq U$ for some $U\in\mathcal{U}$ if your prove only gets $B_\epsilon(x) \subseteq B_{\epsilon(x_k)}(x_k)$.

    Is it that your starting cover $\mathcal{U}$ simply doesn't matter and at the end one ends up with another cover that most likely $\mathcal{U}\neq \mathcal{V}$?

     – user2820579 Dec 15 '20 at 20:34
  • @user2820579: Because by definition each of the sets $B_{\epsilon(x)}(x)$ is contained in some member of $\mathscr{U}$. That’s how the numbers $\epsilon(x)$ were chosen. – Brian M. Scott Dec 15 '20 at 21:01
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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 $r$.

user642796
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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 $r:=\rho_{\min}/2$.

Christian Blatter
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