Irreducible factors in an arithmetic sequence The question is 
Let $T=\{1, 4, 7, 10, 13, 16, 19,...\}$.  An element of $T$ is called irreducible if it is not $1$ and its only factors within $T$ are 1 and itself.  Demonstrate that every element of T can be factored as a product of irreducible elements of $T$.
My answer:
Suppose we had a set $B=\{b\in T : d\mid b,  d\in T \ \Rightarrow d=1 \vee d=b\}$ of all of the irreducibles of $T$.  Let $X=\{a \in T : \forall p \in B, p$ does not divide $a\} $
Assume $X \neq \emptyset$
By the well ordering principle, $$\exists a_0\ \in X, \forall x \in X,  a_0 \leq x$$
Suppose $a_0$ is irreducible.  However, since $a_0 \mid  a_0$, $a_0 \notin X$. Hence we have a contradiction.  
Suppose $a_0$ is reducible.  Then $$\exists n \in T, n>1, n \neq a_0, n\mid a_0 \Rightarrow \exists q \in \mathbb{Z}, nq=a_0$$
Now $q \in T$ by part a).  Since $$ 1< q < a_0 \Rightarrow q \notin X \Rightarrow \exists p \in B, p\mid q \Rightarrow p\mid a_0$$
So $a_0 \notin X$ and this contradicts the fact that $a_0 \in X$.  Thus in both cases where $a_0$ is reducible and where $a_0$ is not irreducible, we have a contradiction.  Thus, $$X=\emptyset \Rightarrow  \forall a \in T, \exists p \in B, p\mid a \Rightarrow \exists m \in \mathbb{Z}, pm =a$$  So $m \in T$ by part a)
$$\exists p_1 \in B, p_1\mid m \Rightarrow \exists m_1 \in \mathbb{Z}, m=p_1m_1 \Rightarrow p_1\mid a$$
$$\exists p_2 \in B, p_2\mid m_1 \Rightarrow \exists m_2 \in \mathbb{Z}, m_1=p_2m_3 \Rightarrow p_2\mid a$$
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$$\exists p_n \in B, p_n, p_{n-1},p_{n-2},...,p_1\mid a$$
$a=p_1p_2p_3...p_n$ where $p_1, p_2, p_3,..., p_n$ are all irreducible factors of $a$
I am having trouble on this question and I wonder if there is a simpler way to do it.  I understand it is almost similar to prime numbers in that the the irreducible numbers in $T$ are basically like the prime numbers in $\mathbb{Z}$
Part a of the problem was that if $$a=bc, a \in T, b \in T \Rightarrow c \in T$$
 A: The key to this problem is proving that $T$ is divisor-closed in $\mathbb{N}$; that is, if $rs=t$ and $r,t\in T$, then also $s\in T$.  
With this property, we can proceed by induction.  Base case trivial.  Assume that all of the elements of $T$ up to some $n$ can be factored into irreducibles.  Let $n'$ be the next largest element of $T$.  $n'$ either is irreducible (and is a product of one irreducible), or it has an irreducible divisor $m$.  By the divisor-closed property, $n'/m$ is in $T$, and so can be factored into irreducibles by the inductive hypothesis.  Combining $m$ with that factorization of $n'/m$, gives a factorization of $n'$.
Unfortunately, OP's proof does not prove that $T$ is divisor-closed, because it doesn't appear to use properties of $T$ at all, apart from being a subset of $\mathbb{N}$.

We have $$T=\{1+3k:k\in\mathbb{N}_0\}$$
Suppose now that $rs=t$, and $r,t\in T$.  Then $rs\equiv t\equiv 1\pmod{3}$, and since $r\equiv 1\pmod{3}$, also $s\equiv 1\pmod{3}$.  Thus $s\in T$.  This proves that $T$ is divisor-closed.

To demonstrate the need to prove divisor-closed, consider instead the monoid $M=\langle 2,6\rangle=\{1,2,4,6,8,12,16,24,32,36,\ldots\}$.  Note that $6$ is not an irreducible, nor is it the product of irreducibles.
A: Likely the exercise is meant to guide you into discovering nonunique factorization. So let's proceed along that path. Given  $t\in T$ consider its prime factorization in $\Bbb N.\,$ Since $\,t\equiv 1\pmod 3\,$ it must contain (counting multiplicity)  an even number of primes $\,p\equiv -1\pmod{3},\,$ so we can pair them all up $\ p,q \mapsto pq\,$ to obtain irreducibles  $\,pq\in T,\,$ by $\,pq\equiv 1\pmod{\!3}\,$ so $\,pq\in T\,$ and it is clearly irreducible in $T$ since its only nontrivial factorization $\,p\cdot q$ in $\,\Bbb Z\,$ has both factors $\,\not\in T$. 
So this pairing process yields a factorization in $T$. But this pairing up process is generally nonunique if there is more than one pair, $ $ e.g. $\,pq,\, p'q'$ vs. $\,pp'\!,\,qq',\,$  e.g. $\, 2\cdot 2,\, 5\cdot 5\,$ vs. $\,2\cdot 5,\,2\cdot 5,\,$ hence $\, 4\cdot 25 = 10\cdot 10\,$ is a nonunique factorization into irreducibles in $\,T.$
