False proofs claiming that $\mathbb{Q}$ is uncountable.

Question

Here are two false proofs of the fact that $\Bbb Q$ is uncountable. From Stephen Abbott's Understanding Analysis.

Proof 01

Lemma:(Nested Interval Property - NIP). For each $n ∈ N$, assume we are given a closed interval $I_n = [a_n, b_n]$. Assume also that each I_n contains I_{n+1}. Then, the resulting nested sequence of closed intervals $$I_1 ⊇ I_2 ⊇ I_3 ⊇ I_4 ⊇ · · ·$$ has a nonempty intersection.

Assume, for contradiction, that $\Bbb Q$ is countable. Thus we can write $\Bbb Q =\{r_1, r_2, r_3, . . .\}$ and, construct a nested sequence of closed intervals with $r_n \not \in I_n$. Our construction implies $\bigcap_{n=1}^{\infty} I_n = ∅$ while NIP implies $\bigcap_{n=1}^{\infty} I_n \neq ∅$ . This contradiction implies $\Bbb Q$ must therefore be uncountable.

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Proof 02

Theorem : The open interval $(0,1)$ is uncountable.

Proof: Lets assume that $(0,1)$ is countable and thus let $f:\Bbb N\to (0,1)$ be a bijective function. Then let $f(m) = 0.a_{m1}a_{m2}a_{m3} . . . .$ (decimal representation). Let $x=0.b_1b_2...$ with $b_i=3$ if $a_{ii}=2$ else $b_i=2$. Its not difficult to see x is a different number from the counted set a contradiction.

Now the question is: Every rational number has a decimal expansion, so we could apply this same argument to show that the set of rational numbers between $0$ and $1$ is uncountable. However, because we know that any subset of $\Bbb Q$ must be countable, the proof of Theorem.

I can't figure out the flaws in these two arguments.

Answer 1

In proof $01$ you have not proved that $\bigcap\limits_{i=1}^n=\varnothing$, it could contain only one irrational number.

In proof $02$ you can't apply the same argument. Because the resulting number might be irrational.

Answer 2

The NIP doesn't hold in general metric spaces. For instance, let $a_1 = 3.1, a_2 = 3.14, a_3 = 3.141, \ldots$ (so $a_n$ is the first $n$ digits of the decimal expansion of $\pi$, and $b_n = a_n + 10^{-n}$. Then in $\mathbb{R}$, we have that $\bigcap_{n = 1}^{\infty} [a_n , b_n] = \{ \pi \}$. But if we limit ourselves to $\mathbb{Q}$, we have empty intersection. This relates to a property of $\mathbb{R}$ called completeness.

EDIT: Also, in 02 you are indeed constructing a number, but it needn't be rational (and if $f : \mathbb{N} \to \mathbb{Q}$ is a bijection, it necessarily won't be).

Answer 3

I'd say both of these make the same error: that in only considering the rationals, we forget that the irrationals still exist.

In case 1 we conclude the non-empty intersection contains no rationals.

In case 2 we construct a number that is different from all rationals.

To which, if we remember that irrationals still exist, we should conclude the intersection contains only rationals (only one actually) and the constructed number is irrational.

Answer 4

In the first case, you've ignored the possibility that your sequence of intervals is, say, $I_n = [\pi - 1/n, \pi + 1/n]$. Then $\bigcap_nI_n$ is nonempty, but the only element it includes is $\pi$, which is not rational. Importantly, the NIP holds only within the reals, not the rationals.

In the second proof, observe that the number you construct (0.3233222232333... or some such) is going to be nonterminating - and it would be very easy to concoct a listing of rational numbers so that the decimal was also nonrepeating. For example, say our list was 0.2, 0.03, 0.002, 0.0002, 0.00003, 0.000003, ...; repeating in increasing "patches". Then the "diagonal" number we construct is a nonrepeating decimal - which is necessarily irrational. Thus the diagonal argument only tells us that our list of rationals is missing an irrational - which we already knew, because that was what it was supposed to do.