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I am trying to find an open cover for an interval $I = [0,2]$ such that the open sets that cover this interval have a $20\%$ overlap.

For example: I have the interval $I=[0,2]$. I want the length of the open sets that create the cover to have length $l = 1$ with a $p = \frac{2}{3}$ overlap. So the sets are: $[0,1],[0.33,1.66],[1,2]$. However when I try to do $p=\frac{2}{10}$ I get

$[0,1],[0.8,1.8],[1.6,2.6]$.

I am failing to see why any $p$ will not work because I was able to get intervals for values of $p = 1/2$ and $p=8/10$.

Any help or insight would be appreciated!

Also what I would like to do is be able to apply this to the interval $I=[-1,1]$ and break it into $5$ intervals with 20% overlap or generally any interval $I$ broken into $n$ intervals with $x\%$ overlap.

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  • $\begingroup$ the usual convention for open intervals is $$]a,b[$$ $\endgroup$ – janmarqz Jan 10 '15 at 2:39
  • $\begingroup$ Or $(a,b)$... ${}{}{}$ $\endgroup$ – copper.hat Jan 10 '15 at 3:45
  • $\begingroup$ What do you mean by overlap? Do you mean that any two elements of the cover overlap by that amount? $\endgroup$ – copper.hat Jan 10 '15 at 3:47
  • $\begingroup$ I want the ends of the sub intervals to over lap by that much so in the example the first interval is (0,1) and then the next starts at (1-2/3,1-2/3+1) $\endgroup$ – RDizzl3 Jan 10 '15 at 17:11

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