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Prove that there does not exist a sequence $I_n=[a_n,b_n], n=1,2,\ldots $ of nonempty, pairwise disjoint intervals such that $\cup_{n}I_n=[0,1]$.

My solution -- is it correct?

My idea is that taking an open cover $\{A_{\alpha_i}\}$ of $[0,1]$ then $(\cup_{i}A_{\alpha_i})\cap (\cup_i I_i)=\cup_{i}(A_{\alpha_i}\cap I_i)$ is a open cover of $[0,1]$ as well (correspondence to induced topology).

Moreover, $(A_{\alpha_i}\cap I_i)\cap (A_{\alpha_j}\cap I_j)=\emptyset$ with $i\ne j$.

Since $[0,1]$ is compact, there exists $N$ such that $ \cup_{i=1}^N(A_{\alpha_i}\cap I_i) \supset[0,1]$.

However, $[0,1] \supsetneq \cup_{i=1}^N(A_{\alpha_i}\cap I_i)$ since $\cup_{i=1}^N(A_{\alpha_i}\cap I_i)$ and $I_n=[a_n,b_n], n=1,2,\ldots $ of nonempty, pairwise disjoint intervals.

So, we get contradition $[0,1] \subsetneq [0,1]$.

Could you please give your idea about my solution? Thank you very much for your help.

Chill2Macht
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user52523
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    The sentence beginning with "Moreover" seems to be incomplete: something about that intersection... – DonAntonio Aug 10 '16 at 22:21
  • You wrote : "$(\cup_{i}A_{\alpha_i})\cap (\cup_i I_i)=\cup_{i}(A_{\alpha_i}\cap I_i)$" You need $i,j$ on the right, – zhw. Aug 10 '16 at 22:27
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    This looks like a Baire category theorem problem to me. – carmichael561 Aug 10 '16 at 22:28
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    http://math.stackexchange.com/questions/6314/is-0-1-a-countable-disjoint-union-of-closed-sets/216595#216595 – Sargera Aug 10 '16 at 22:29
  • @DonAntonio: I just edited.Thank Taylor very much. I know there exists other way to solve. Could you please give your idea about my method? – user52523 Aug 10 '16 at 22:33
  • You can only have a countable union: There are at most countably many (closed/open) balls of positive radius in $R^n$. – Behnam Esmayli Aug 11 '16 at 00:02
  • @user52523 u said pairwise disjoint then this is trivial. – ibnAbu Aug 11 '16 at 00:04
  • Just take an arbitrary such sequence and describe its union. Note it is not as desired. – Jacob Wakem Aug 11 '16 at 01:46
  • Note these closed sets are all internal. Thus it is equivalent to a countable closed partition on the reals. – Jacob Wakem Aug 11 '16 at 02:26
  • @Behnam moreover their centers cannot be dense. – Jacob Wakem Aug 11 '16 at 04:59
  • I now have a completely correct solution! – Jacob Wakem Aug 11 '16 at 05:53
  • @Behnam "There are at most countably many (closed/open) balls of positive radius in Rn" Already with a given center, there are uncountably many. – Did Aug 11 '16 at 14:55
  • I have meant "pairwise disjoint" collection. – Behnam Esmayli Aug 11 '16 at 14:57
  • @Behnam So, you are just stating the desired conclusion? (Unrelated: Please use @.) – Did Aug 11 '16 at 15:02
  • But the proof of my claim is so simple and straightforward. Take increasing sequence of balls of radius $1,2,3,...$ all centered at origin. One of them must contain infinitly many of our collection. Now, If there are uncountably many disjoint positive radius balls within a huge bounded ball, then their union will have a measure equal to infinity ("sum" of uncountably many positive numbers is never bounded.) However, the union is contained in a ball with finite measure. Contradiction. – Behnam Esmayli Aug 12 '16 at 17:03

1 Answers1

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I think the equation $(\cup_{i}A_{\alpha_i})\cap (\cup_i I_i)=\cup_{i}(A_{\alpha_i}\cap I_i)$ does not make sense. $A_\alpha's$ has any union and $I_n's$ has countable union. So how do you define the union in the right side? There is a very easy way of solving this if you like to try. To the contrary assume such sequence exists. Then the set (Say $E$) that contains $a_n's$ and $b_n's$ is countable. Now show that $E$ is a perfect set. A perfect set in $\mathbb{R}$ is uncountable. Thus we get a contradiction.

Extremal
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  • The indexing set of the $A_\alpha$ can be chosen so that it is finite, given that the closed unit interval is compact. – user259242 Aug 10 '16 at 23:19
  • Ok. But then how do we have intersection for the same index on the RHS? I mean it should be corrected as $A_{\alpha_i}\cap I_j$ – Extremal Aug 10 '16 at 23:21