About $\Sigma=\{p_2\to p_1, p_3\to p_2,\, \dots\,\}$ . . . 
Suppose $$\Sigma=\{p_2\to p_1, p_3\to p_2,\, \dots\,\}.$$
Which of the following is true? Explain your answer.
  
  
*
  
*For any $n$, $$\Sigma\cup\{p_n, \neg p_{n+1}\}$$ is complete and consistent.
  
*It's not possible to complete $\Sigma$ with (a) finite sentence(s). 

I think (2) is correct. Any ideas, descriptions, and excellent answers would be appreciated. (Pn is Atomic Sentence).
 A: I will show that 1 is true and therefore 2 is false. Note that since $\Sigma=\{p_2\to p_1, p_3\to p_2,\, \dots\,\}$, we have that $\Sigma\equiv\{\neg p_1\to \neg p_2, \neg p_2\to \neg  p_3,\, \dots\,\}$. In particular, we have $\Sigma \equiv \bigcup_{1 \leq n<\omega}\{\neg p_n \to \neg p_{n+1}\}$. 
(1) Claim: for all $n<\omega$ such that $\Sigma \cup \{p_n, \neg p_{n+1}\}$ is consistent and complete. 
Proof: Let $n\geq 1$. We are considering $\Sigma \cup \{p_n, \neg p_{n+1}\}$ Since $\Sigma \equiv \{\neg p_1\to \neg p_2, \neg p_2\to \neg  p_3,\, \dots\,\}$, we know that, as consequence, $\Sigma \models \{\neg p_m\}$ for $m\geq n+1$. $\Sigma \models p_n$ by assumption. Note that by the definition of $\Sigma$, $\Sigma \models p_s$ for $s\leq n$. Therefore, since $\Sigma$ makes a judgement on every atomic sentence, $\Sigma$ is complete. 
It should be pretty obvious that $\Sigma$ is also consistent by observing the fact that every finite set of axioms is clearly consistent, and therefore, by the compactness theorem, $\Sigma$ also is consistent. 
Therefore, any pair $\{p_n,\neg p_{n+1}\}$ added to $\Sigma$ make $\Sigma$ both complete and consistent. 
