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Possible Duplicate:
Cantor's completeness principle

Cantor's completeness principle states that the intersection of a nest of closed intervals is non empty and is a point. Now let us consider a nest of open intervals. I1 contains I2 contains in I3........ Now The intersection of I1 and I2 is nothing but the interval I2 so is I3 for I1,I2 and I3. So if N tends to infinity then is the intersection not going to be a point? Can anyone show me with rigor and reason why is does cantor principle fail in case of open intervals,i.e. the intersection is empty.?

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  • $\begingroup$ @TonyK - Could you please tell me where I am wrong in the statement? Thanks a ton for pointing. $\endgroup$
    – user16186
    Oct 30, 2011 at 16:31
  • $\begingroup$ "and is a point". See the duplicate question for more info. $\endgroup$
    – TonyK
    Oct 31, 2011 at 0:35

3 Answers 3

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Take $I_n = (0,\dfrac{1}{n})$. For every finite intersection you can find some $k$ large enough such that $\dfrac{1}{k}$ is in the intersection.

However for every $x>0$ there is some $k$ such that $x<\dfrac{1}{k}$, therefore $x\notin\bigcap_n I_n$, thus the intersection is empty.

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Asaf's example is clearly right. Let me add just a comment about the reason why Cantor's principle holds for closed interval but not for open ones. It is simply because closed intervals contain the limits of the sequences. Indeed, the unique way to prove the existence of an element in your intersection $\bigcap I_n$ is picking a point in $I_1$, one in $I_2\cap I_1$ and so on. In this way you construct a sequence, which admit a convergent subsequence, whose limit belongs to the closure of the intersection (which is the intersection itself, because the intersection is still closed). You can't do this when the sets are open, because the intersection of open sets need not to be close, so you have a limit, but it can be outside the intersection.

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$$ \bigcap_{n=1}^\infty \left(0,\frac1n\right) = \varnothing. $$

And $$ \bigcap_{x>0} (0,x) = \varnothing. $$

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