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Let $(X,d)$ be a metric space. The diameter of a set $A\subset X$ is defined to be
$\operatorname{diam}(A)= \sup\{d(x,y):x,y\in A\}$.

Suppose $A_1, \dots, A_n$ is a finite collection of subsets of $X$ each with finite diameter.
Prove that $\bigcup\limits_{i=1}^n A_i$ has finite diameter.

My approach:
I am trying to show that $\operatorname{diam}(\bigcup\limits_{i=1}^n A_i)\leq \operatorname{diam}(A_1)+\operatorname{diam}(A_2)+\cdots+\operatorname{diam}(A_n)$. If we can show that since each $A_i$ has finite diameter then $\operatorname{diam}(\bigcup\limits_{i=1}^n A_i)$ must also have a finite diameter.

But I stumbled on how to show that inequality, should we use the triangle inequality somewhere? But not very sure how to write it.

Thanks very much!

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  • $\begingroup$ The inequality you are trying to show is wrong (except if $X$ has only one point ). For example, take $A_1=\{x\}$, $A_2=\{y\}$ for $x\ne y$. $\endgroup$ – Nitrogen Oct 29 '15 at 0:13
  • $\begingroup$ @Travis The diameter is defined even if $\bigcup A_i$ is not connected. $\endgroup$ – Nitrogen Oct 29 '15 at 0:14
  • $\begingroup$ Also, note that by induction it's sufficient to prove the claim for $n = 2$. $\endgroup$ – Travis Oct 29 '15 at 0:15
  • $\begingroup$ @Nitrogen You're right, I had confused the geodesic distance function with a general distance function. $\endgroup$ – Travis Oct 29 '15 at 0:16
  • $\begingroup$ @MichaelHardy : You are correct. I have deleted the comment. $\endgroup$ – steven gregory Oct 29 '15 at 1:34
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It is not true that $\operatorname{diam}\left(\bigcup\limits_{i=1}^n A_i\right)\leq \operatorname{diam}(A_1)+\operatorname{diam}(A_2)+\cdots+\operatorname{diam}(A_n)$. For example, suppose the diameter of each of ten sets is two inches. But one of those ten sets is in Constantinople and another is in Adelaide.

Pick a point $a_i$ in each of the sets $A_i$. Find the the two indices $k,\ell$ such that $d(a_k,a_\ell) = \max \{ d(a_i,a_j) : i,j \in \{ 1,\ldots,n \} \}$. Given points $x\in A_k$, $y\in A_\ell$, we have $$ d(x,y) \le d(x,a_k)+d(a_k,a_\ell) + d(a_\ell, y) \le \underbrace{\operatorname{diam}(A_k) + d(a_k,a_\ell) + \operatorname{diam}(A_\ell)}. $$ Then show that the quantity over the $\underbrace{\text{underbrace}}$ is an upper bound on the diameter of the union.

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Hint: Show that $A\subset X$ has finite diameter iff it is bounded. Then, it is easy to see that a finite union of bounded sets is bounded.

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  • $\begingroup$ A simple example that your hypothesis is incorrect can be found in $\mathbb R$. $diam([0,1] \cup [2,3]) = 3$ and $diam([0,1]) + diam([2,3]) = 2$ $\endgroup$ – steven gregory Oct 29 '15 at 1:40

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