Exercise 1.5.2

Question

Review the proof of Theorem 1.5.6, part (ii) showing that $\mathbf{R}$ is uncountable, and then find the flaw in the following erroneous proof that $\mathbf{Q}$ is uncountable:

Assume, for contradiction, that $\mathbf{Q}$ is countable. Thus we can write $\mathbf{Q}=$ $\left\{r_{1}, r_{2}, r_{3}, \ldotsight\}$ and, as before, construct a nested sequence of closed intervals with $r_{n} 
otin I_{n}$. Our construction implies $\bigcap_{n=1}^{\infty} I_{n}=\emptyset$ while NIP implies $\bigcap_{n=1}^{\infty} I_{n} 
eq$ $\emptyset$. This contradiction implies Q must therefore be uncountable.

Ulisse Mini · Accepted Answer

The nested interval property is not true for $\mathbf{Q}$. Consider $I_n$ being rational bounds for $\sqrt2$ with $n$ decimal places, then $\bigcap_{n=1}^\infty I_n = \emptyset$ since $\sqrt2 
otin \mathbf{Q}$.

Exercise 1.5.2

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Exercise 1.5.2

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