convergence of a metric sequence I need help proving a homework problem and I have no idea how to start or how approach it, I'd appreciate some hints or advice instead of a solution:
Let $(X,d)$ be a metric space, $\{ p_{n} \} \subset X$ a convergent sequence with $p_{n} \rightarrow p$, and $\{q_{n}\} \subset X$ a convergent sequence with $q_{n} \rightarrow q$. 
Prove that $d(p_{n},q_{n}) \rightarrow d(p,q)$ where convergence takes place in $\mathbb{R}$.
One thing that I first did was to use the definition of convergence. So if  $d(p_{n},q_{n}) \rightarrow d(p,q)$, then for any $\varepsilon > 0$, there is an integer $N$ such that for all $n \geq N$ implies $d_{\mathbb{R}}(d(p,q),d(p_{n},q_{n})) < \varepsilon.$
However, I don't know what to do after this, and then there is also the question of which metric $d_{\mathbb{R}}$ is being used in $\mathbb{R}$ in the last inequality (after the word implies). I am assuming it is the Euclidean metric $d(x,y) = |x-y|$, but it isn't specified. I tried to negate the statement in the last paragraph which essentially says: $d(p_n,q_n) \rightarrow d(p,q)$ and $\exists \varepsilon>0,  \hspace{2mm}\forall N \hspace{1mm} \text{s.t. } [n\geq N \text{ and } d_{\mathbb{R}}(d(p,q),d(p_n,q_n)\hspace{.5mm}) \geq \varepsilon]$, but I don't think this is the correct approach and the last bit doesn't make sense to me. Any help would be appreciated, thank you.
 A: The metric on $\mathbb{R}$ is indeed the Euclidean metric.
Hint: $d(p,q) \leq d(p,p_n)+d(p_n,q_n)+d(q_n,q)$ and $d(p_n,q_n) \leq d(p_n,p)+d(p,q)+d(q,q_n)$. What does that tell you about $|d(p,q)-d(p_n,q_n)|$?
A: 
So if  $d(p_{n},q_{n}) \rightarrow d(p,q)$, then for any $\varepsilon > 0$, there is an integer $N$ such that for all $n \geq N$ implies $d_{\mathbb{R}}(d(p,q),d(p_{n},q_{n})) < \varepsilon.$

That is what you're trying to prove, not what you've already established.
What you know is
\begin{align}
\forall \varepsilon>0\  \exists N_1\  \forall n\ge N_1\  d(p,p_n)<\varepsilon/2 \\
\forall \varepsilon>0\  \exists N_2\  \forall n\ge N_2\  d(q,q_n)<\varepsilon/2 
\end{align}
where the reason for dividing by $2$ will appear below.
Let $N = \max\{N_1,N_2\}.$ Then you have
\begin{align}
d(p_n,q_n) & \le d(p_n,p) + d(p,q) + d(q,q_n) \text{ by the triangle inequality} \\
\text{and therefore } d(p_n,q_n) - d(p,q) & \le d(p_n,p) + d(q,q_n) \le \frac\varepsilon 2 + \frac\varepsilon 2 = \varepsilon \\[15pt]
\text{and } d(p,q) & \le d(p_n,p) + d(p_n,q_n) + d(q,q_n)  \text{ by the triangle inequality} \\
\text{and therefore } d(p,q) - d(p_n,q_n) & \le d(p_n,p) + d(q_n,q)  \le \frac\varepsilon 2 + \frac\varepsilon 2 = \varepsilon. \\[10pt]
\text{So } |d(p,q)-d(p_n,q_n)| & \le \varepsilon.
\end{align}
A: Suppose the distance between $p$ and $q$ is some finite positive quantity. If it is true that $p_n$ approaches $p$ as $n$ gets large then we can make the distance between $p_n$ and $p$ as small as we want for sufficiently large $n$. The same will be true for the distance between $q_n$ and $q$ albeit the $n$ chosen for $p_n$ might not work for $q_n$. So we will need to choose whichever $n$ was bigger. Then we are guaranteed that the distances $d(p_n,p)$ and $d(q_n,q)$ will be as small as we want. Say $\epsilon/2$. Then if you invoke the triangle inequality on $d(p_n,q_n)$ and $d(p,q)$ you should find the result you need. 
