If $a$ is irrational, there are infinitely many $\frac pq$ s.t $|a-\frac pq|<\frac1{q^2}\tag1$
I have the proof but don't understand it:
Take a finite set of rationals $S$ then for sufficiently large $Q$ the result of Dirichlet approximation theorem$^1$ does not hold for any $s\in S$ (this is clear), so the set of rational numbers satisfying $(1)$ is infinite
My Questions:
$\bullet$ Does that mean the following ?
Due the Dirichlet theorem for a given $Q$ you find $p/q$ satisfying the condition below, then you choose another $\tilde Q$ (much greater than $Q$) for which we cannot find a $q$ in the set $\{q:q\le Q\}$, but the Dirichlet theorem must hold, so there must be another $\tilde q$, so different from $q$ (actually $q<\tilde q)$ and proceeding in that manner there are infinitely many rationals.
$\bullet$ Why ''for sufficiently large $Q$ the result of Dirichlet approximation theorem does not hold for any $s\in S$''
($^1$Dirichlet approximation theorem states: for any $r\in\mathbb R$ and $Q\in \mathbb Z_{+}$ there are $p,q$ with $1\le q\le Q$ s.t. $|r-\frac pq|<\frac1{Qq}$)
Dirichlet approximation: for any $\alpha$, any $N$, there exists $p,q$ integers $1\leq q\leq N$ such that:
$\mid q\alpha-p\mid <{1\over N}$ this implies $q\mid \alpha-{p\over q}\mid <{1\over N}$, thus $\mid \alpha-{p\over q}\mid <{1\over {Nq}}\leq {1\over{q^2}}$， since $q\leq N$. In particular this applies if $\alpha$ is irrational, you can take a sequence $N_n$ going to infinity such that
$\mid q_n\alpha-p_n\mid <{1\over {N_n}}$. The sequence of $q_n$ has to go towards infinity also, so you have an infinite numbers of $q_n$ such that
$\mid \alpha-{{p_n}\over {q_n}}\mid <{1\over {q_n^2}}$.