A new proof of Dirichlet's theorem concerning prime numbers in arithmetic progressions

It is known that there are infinitely-many prime numbers which take the form of a polynomial of degree one with integer coefficients, this is Dirichlet’s theorem. We use an elementary sieving argument together with bounds on the prime number counting function to provide a new proof of Dirichlet’s theorem. We show that if $a\in \mathbb{N},k\in \mathbb{N},a_k=(a,a+1,\mathrm{.}\mathrm{.}\mathrm{.},a+k-1)$ and $A=\{p_1,p_2,\mathrm{.}\mathrm{.}\mathrm{.},p_n\}$, a finite set of primes. Then the number of components of $a_k$ that are divisible by some prime in $A$ is less than or equal to $$\sum _{\begin{array}{c}d\mathrm{\mid }P(A)\\ d>1\end{array}}(-1{)}^{\omega \left(d\right)+1}\lfloor \frac{k}{d}\rfloor +2n$$ where $\omega \left( d\right)$isthenumberofdistinctprimedivisorsof$d$and$P(A)=\prod_{p\in A}p$\mathrm{.}Weclaimthatthe$+2n$intheboundcanbereplacedwith$n$, the $\mathtt{\text{best}}\mathtt{\text{possible}}\mathtt{\text{bound}}$. However, we did not demonstrate our claim in this paper since the $+2n$(bound) is enough for the new proof of Dirichlet’s theorem. This result effectively means that given $[1,x],x\in \mathbb{R}$; if the primes in $A$ divide $h$ integers in $[1,x]$ then for every $g>0$, they will divide at most $h+2\mathrm{\mid }A\mathrm{\mid }$ integers in $[1+g,x+g]\mathrm{.}$