I've found in an article the following statement:

Consider a compact metric set covered by $n$ balls of radius less than $\epsilon/3$. Now consider the balls with the same center and doubled radius (implying that all $r$ are $<2\epsilon/3$). The Lebesgue number of this new cover is $\geq \epsilon/4$.

Note that this is subtler than it could appear: we are not saying that the Lebesgue number of any cover of radius $2\epsilon/3$ is more or equal than $\epsilon/ 4$: this is true only if the cover comes from an $\epsilon/3$- cover by rescaling; note that rescaling by $1/2$ a cover we don't necessarily obtain a cover of the space.

Now can someone help me proving the given statement?

Thank you in advance.


If a set $A$ has diameter $\leq\epsilon/4$, find a ball $B$ in your original cover that contains a point $x$ in $A$. Then every $y\in A$ is within $\epsilon/4$ of this $x$ and therefore lies in the doubled version of $B$.

  • $\begingroup$ You mean $\leq \epsilon$, right? And why does every $y$ lie there? Perhaps I'm missing something obvious. $\endgroup$ – W. Rether Jun 28 '17 at 17:35
  • $\begingroup$ You're right about the reversed inequality; I just corrected it. Now is $x_0$ is in $B$ and therefore within $\epsilon/3$ of the center $c$ of $B$, and if $y$ is within $\epsilon/4$ of $x_0$ (because they're both in $A$), then the distance between $y$ and $c$ is, by the triangle inequality, at most $$\frac\epsilon3+\frac\epsilon4=\frac{7\epsilon}{12}<\frac{2\epsilon}3,$$ which puts $y$ in the ball of radius $2\epsilon/3$ centered at $c$, i.e., the doubled version of $B$. $\endgroup$ – Andreas Blass Jun 28 '17 at 18:48

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