Partial limits and sets of indices Let $\{a_n\}$, and $A_1...A_n \subseteq \mathbb{N}$, such that $A_1\cup A_2... \cup A_n= \mathbb{N}$. We denote $P_k$ as the set of all partial limits (= limit of a subsequence) with indices $\in$ $A_k$.  
Show that $P$, the set of all partial limits is:  $$P = P_1 \cup... \cup P_n$$
My thoughts:
First denote $L_i$ a partial limit of $A_i$, and $L_j$, a partial limit of $A_j$.
Lets assume by contradiction that there's a limit $L_k$ which involves indices from both $A_i$ and $A_j$. 
By definition of limit:
$$\forall \varepsilon  > 0\exists K \in N.\forall k > K:\left| {{a_{{n_k}}} - {L_k}} \right| < \varepsilon $$
We notice that $\left| {{L_i} - {L_k}} \right|$ is a fixed size.
The subsequences are infinite, and therefore we can choose $n_{{k_0}} \in A_i$ such that:
$$\left| {{a_{{n_{{k_0}}}}} - {L_k}} \right| < \varepsilon  \wedge \left| {{a_{{n_{{k_0}}}}} - {L_i}} \right| > \varepsilon $$
Which contradicts the assumption $L_i$ is a partial limit of $\{a_n\}$.
Can you criticize my work? Am I right? If not, what should I do different? 

You definitely seem to be misunderstanding the question, as you seem to be attempting that the $P_i$'s are disjoint which is untrue in general. 
Here is how to do it:
The inclusion $P_1 \cup \cdots \cup P_n \subseteq P $ is obvious. 
So let us show $P \subseteq P_1 \cup \cdots \cup P_n$. To this end consider $a = \lim_{k}a_{n_k}.$ Then clearly for some $i \in \{1,...n\}$ we have that $n_k \in A_i$ infinitely often (else the 'subsequence' $(a_{n_k})$ would contain only finitely many terms, a contradiction). Denote by $(n_{k_r})$ the subsequence of $(n_k)$ that lies in $A_i$, which we have just shown exists. Then we still have $a =\lim_{r}a_{n_{k_r}}$ and this shows that $a \in P_i \subseteq P_1 \cup \cdots \cup P_n .$
