# can someone clarify one line of the proof (non-countability of real interval)

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 $$[a_i,b_i]$$.

• Oh yes! The famous theorem 1.2. I can't imagine how our lives would be without theorem 1.2. – José Carlos Santos Mar 10 at 7:24
• 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 $b_i-a_i < 2^{-i}$ – Henry Mar 10 at 7:26
• if our sequence of c was increasing I could accept it but we have no information on the sequence – Nima Balesini Mar 10 at 7:32

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 $$c_j$$.