Show that an arbitrary enumeration of rationals in interval $(0,1)$, represented as a sequence $(q_n)$, has $1$ as a subsequential limit.

Question

The rationals $\mathbb{Q}$ is countable, so rationals in the bounded interval $(0,1)$ can be enumerated as a sequence.

Let sequence $(q_n)_{n=1}^{\infty}$ be (a representation of) an enumeration of rationals in the bounded interval $(0,1)$. You probably have a better intuition than me that tells you the limit superior $\limsup_{n\to\infty} q_n=1$. (My first dumb guess is $0$.)

Now the question(s): How to write down a rigorous formal proof that $\limsup_{n\to\infty} q_n=1$? More importantly, what are some theorems or propositions enableing us to prove that $\limsup_{n\to\infty} q_n=1$?

I currently only thought of baby Rudin Theorem 3.17, rephrased as the following:

baby Rudin Theorem 3.17
Let $(s_n)$ be a sequence of real numbers. Then $s^{*}\in\mathbb{R}\cup\{\infty,-\infty\}$ is the limit superior of $(s_n)$, denoted by $\limsup_{n\to\infty} s_n = s^{*}$, if and only if $s^{*}$ has the following two properties:
(1.) $s^{*}$ is a subsequential limit of $(s_n)$.
(2.) If $x>s^{*}$ there is an integer $N$ such that $s_n<x$ for every $n\ge N$.

To apply baby Rudin Theorem 3.17 to prove that $\limsup_{n\to\infty} q_n=1$, we must show that $1$ has the two properties described in the theorem. To show that $1$ has property (2.) is trivial. The non-trivial part is to show that $1$ has property (1.), and I still haven't figured out how to do a convincing argument.

So in conclusion:

How to convincingly show that an arbitrary enumeration of rationals in interval $(0,1)$, represented as a sequence $(q_n)_{n=1}^{\infty}$, has $1$ as a subsequential limit?

Answer 1

We can construct a subsequence of $(q_k)$ inductively:

For the neighborhood $(1-\frac{1}{2},1)$, pick some natural number $n_1$ for which $q_{n_1}\in (1-\frac{1}{2},1)$.

For the neighborhood $(1-\frac{1}{2^2},1)$, pick some natural number $n_2>n_1$ for which $q_{n_2}\in (1-\frac{1}{2^2},1)$.

For the neighborhood $(1-\frac{1}{2^3},1)$, pick some natural number $n_3>n_2$ for which $q_{n_3}\in (1-\frac{1}{2^3},1)$.

$\dots$

Inductively we have a subsequence $(q_{n_k})$ of $(q_k)$ which converges to $1$.

Answer 2

For arbitrary $k < 1$ there is more rational numbers in an interval $(k, 1) $ than any arbitrary $N\in \mathbb N $.

So, for any $N$-term sequence of rationals let $k $ equal the greatest of those $N$ term. Then there is a (countably) infinite set remaining of rationals greater than any already included in the sequence.

Consequently, there exists some index $M$ greater than $N$, such that $q_M > k$. And as any $k \in (0,1) \cap\mathbb Q$ appears somewhere in the sequence, the $\limsup$ is greater than or equal to any such $k$.

The lowest such value is $1$, hence the answer.

Answer 3

Lemma Let $(q_n)_{n=1}^{\infty}$ be a real sequence that converges to $q_{\infty}\in\mathbb{R}$, then every rearrangement of $(q_n)_{n=1}^{\infty}$ also converges to $q_{\infty}$.
Proof Can rearranging a SEQUENCE (not a series) change the limit?

The rational sequence $(\frac{n}{n+1})_{n=1}^{\infty}$ converges to $1$ and is in interval $(0,1)$. Every arbitrary enumeration of rationals in interval $(0,1)$, by the definition of enumeration, must contains a rearrangement (sequence) of the rational sequence $(\frac{n}{n+1})_{n=1}^{\infty}$, which must also converges to $1$ by the lemma. Hence every arbitrary enumeration of rationals in interval $(0,1)$ must has $1$ as a subsequential limit.

P.S. Thanks very much to a friend on freenode IRC in channel ##math with nickname Z-module, who is the original person coming up with the idea of this proof by the lemma.