# All closed intervals but in between there are few open intervals

Le us consider a a nest of closed intervals. Suppose that I convert few of the closed intervals in the nest into open intervals (except the first one). Now in such a case is the cantor's completeness principle still going to hold?

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The "0% accept rate" is somewhat troubling... –  Ｊ. Ｍ. Nov 18 '11 at 13:38
what do you mean by few? If you only convert finitely many, then Cantor completeness still hold trivially... –  Willie Wong Nov 18 '11 at 13:38
Also, converting "a few" of the closed intervals to open ones may make your nest sequence of intervals no longer nested. Consider $I_n = [0,1/n]$. Now convert $I_3$ to be open: $I_3 \mapsto (0,1/3)$. Now $I_4$, not converted, is no longer a subset of $I_3$... –  Willie Wong Nov 18 '11 at 13:42
For changing infinitely many: switch $\{ [0,1/n]: n=1,2,3,\ldots\}$ to $[0,1], (0,1/2), (0,1/3),\ldots$. –  David Mitra Nov 18 '11 at 13:45
@WillieWong what is meant by holding trivially? math.stackexchange.com/questions/79395/…. I hope you will also like to share your views on this question. –  user16186 Nov 18 '11 at 13:54