Let $(M,d)$ be a metric space and $q \in M$. If $(p_n)$ be a bounded sequence in $M$, show that $(d(p_n,q))$ has a convergent subsequence in $\Bbb R$. Let $(M,d)$ be a metric space and $q \in M$. Let $(p_n)$ be a bounded sequence in $M$. Show that $(d(p_n,q))$ has a convergent subsequence in $\Bbb R$.
Monotone Subsequence Theorem. If $(x_n)$ is a any sequence of real numbers, then there exists a subsequence in $(x_n)$ that is monotone.
Here's what I tried so far:
Since $(p_n)$ is bounded, then by the above theorem, it follows that $(p_n)$ has a subsequence $(p_{n_k})$ that is monotone. Clearly, $(p_{n_k})$ is bounded. Hence, by the Monotone Convergence Theorem, we obtained $(p_{n_k})$ is convergent.
The goal is to show that $d(p_{n_k},q) \to d(p,q)$ for some $p \in M$. But, I got stucked there.
How to approach it correctly? Thanks for your help in advanced.
 A: $(M, d) $ metric space.
$(p_n) \subset M $ bounded implies $\exists q\in M, r>0 $ such that for all $n\in \Bbb{N}$ , $d(p_n, q) <r$
Implies, $\{d(p_n, q) \} \subset \Bbb{R}$ is a bounded sequence and by the Bolzano–Weierstrass theorem $\{d(p_n, q) \}$ has a convergent subsequence.
Edit: For any $p\in M $ ,
\begin{align*}
d(p_n, p) &\le d(p_n, q) +d(q, p) \\ 
&\le r+d(q, p) \\
&= r'. 
\end{align*}
Hence, $\{d(p_n,p) \}$ is again a bounded sequence of Real numbers and by B-W theorem, it has a convergent subsequence.
A: You are on the right track, sort of!  However, remember $p_n$ aren't real numbers, they don't even have to come from an ordered space in which monotonicity would make sense!  (For instance, they could be complex numbers).   The sequence of real numbers is the DISTANCE sequence you are working with.  $d(p_n,q)$.   That is a sequence of real numbers, so has a monotone subsequence.
Now, we need to show that that sequence of real numbers is also bounded. Recall that $p_n$ being bounded in a metric space means there is some radius $r$ and some center $c$ such that for all $n$,   $d(p_n,c)\leq r$  Now use the triangle inequality to show that there is some other real number that bounds the values of $d(p_n,q)$.
Once you have that, you have a monotone bounded sequence of real numbers,  which since the reals are complete,  converges.
A: Since $(p_{n_k})$ convergent in $M$, by letting $p_{n_k} \to p \in M$ with $p \ne q$, we have $d(p_{n_k}, p) \to 0$.
Now, it's easy to check that
$$|d(p_{n_k},q) - d(p,q)| \le d(p_{n_k},p). \qquad (1)$$
Since $d(p_{n_k},p) \to 0$, from $(1)$, we must have $d(p_{n_k},q) \to d(p,q) \in \Bbb R$. Hence, $(d(p_n,q))$ has a convergent subsequence, namely, $(p_{n_k})$, as desired.
