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Theorem 1.7 (Cantor) No interval $[a, b]$ is countable.

Proof. Suppose not. Then the elements of $[a, b]$ can be arranged into a sequence $c_1, c_2, c_3, \ldots$.

Select an interval $[a_1, b_1] \subset [a, b]$ so that $c_1\in[a_1, b_1]$ and so that $b_1 − a_1 < \dfrac12$.

Continuing inductively, we find a nested sequence of intervals $\{[a_i, b_i]\}$ with lengths $$b_i − a_i < 2^{−i}\to 0$$ and with $c_i\in [a_i, b_i]$ for each $i$.

By Theorem 1.2 (variant of Cantor's intersection theorem), there is a unique point $c \in [a, b]$ common to each of the intervals. This point cannot be equal to any $c_i$ and this is a contradiction, since the sequence $c_1, c_2, c_3,\ldots$ was to contain every point of the interval $[a, b]$.

can someone clarify why and how we $c_i$ are not in the $[a_i,b_i]$.

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    $\begingroup$ Oh yes! The famous theorem 1.2. I can't imagine how our lives would be without theorem 1.2. $\endgroup$ – José Carlos Santos Mar 10 at 7:24
  • $\begingroup$ At each iteration you choose $a_i$ and $b_i$ such that $c_i \not \in [a_i,b_i]$ and $[a_i,b_i] \subset [a_{i-1},b_{i-1}]$ and $b_i-a_i < 2^{-i}$ $\endgroup$ – Henry Mar 10 at 7:26
  • $\begingroup$ if our sequence of c was increasing I could accept it but we have no information on the sequence $\endgroup$ – Nima Balesini Mar 10 at 7:32
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The way we can make sure $c_j\notin [a_j, b_j]$ because if $c_j$ is not in the previous interval we can select any subinterval less than half the size. Otherwise $c_j$ is either in the upper or lower half of the previuous subinterval and we can select the lower third or upper third of the previous interval that would not contain $c_j$.

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