Farey Sequence has Convergent Subsequences for all x in Closed Unit Interval

Theorem
Consider the Farey sequence:
 * $\sequence {a_n} = \dfrac 1 2, \dfrac 1 3, \dfrac 2 3, \dfrac 1 4, \dfrac 2 4, \dfrac 3 4, \dfrac 1 5, \dfrac 2 5, \dfrac 3 5, \dfrac 4 5, \dfrac 1 6, \ldots$

Every element of the closed real interval $\closedint 0 1$ is the limit of a subsequence of $\sequence {a_n}$.

Proof
We have that every rational number $\dfrac p q$ between $0$ and $1$ occurs infinitely often in $\sequence {a_n}$:


 * $\dfrac p q, \dfrac {2 p} {2 q}, \dfrac {3 p} {3 q}, \ldots$

Let $x \in \closedint 0 1$.

From Between two Real Numbers exists Rational Number, a term $a_{n_1}$ of $\sequence {a_n}$ can be found such that:


 * $x - 1 < a_{n_1} < x + 1$

Then a term $a_{n_2}$ of $\sequence {a_n}$ such that $n_2 > n_1$ can be found such that:


 * $x - \dfrac 1 2 < a_{n_2} < x + \dfrac 1 2$

Hence we can create a subsequence $\sequence {a_{n_k} }$ such that:


 * $x - \dfrac 1 k < a_{n_k} < x + \dfrac 1 k$

and so by the Squeeze Theorem for Real Sequences:


 * $r_{n_k} \to x$ as $k \to \infty$