I have this question that has been on my head all day and I could not find a solution that can convince me.
Let $(X,d)$ be a metric space. Let $(p_n)_{n\in\mathbb{N}}$ be bounded sequence in $X$ with bound $R>0$. Assume that the sequence converges with limit $p$. Then is the set $\{p\}\subset X$ bounded with bound $R>0$.
I have two equivalent definitions for a bounded subset of a metric space:
\begin{equation}
V\subset X\ \text{is bounded} \Leftrightarrow \exists q\in X,R>0:v\in V \Rightarrow d(v,q)<R
\end{equation}
and
\begin{equation}
V\subset X\ \text{is bounded} \Leftrightarrow \forall q\in X:\exists R>0:v\in V \Rightarrow d(v,q)<R
\end{equation}
A sequence $(p_n)_{n\in\mathbb{N}}$ is bounded if the set $\{p_n| n\in\mathbb{N}\}\subset X$ is bounded.
I can prove this if $X=\mathbb{R}$ with standard metric and with the second definition of bounded set with $q=0$.
I don't even know if it make sense for a general metric space.
Best Answer
Since $\{ p_n \} $ is bounded with bound $R$ there exists $q \in X$ such that $d(p_n,q) <R$ for all $n$. Since $p_n \to p$ as $n \to \infty$, the second triangle inequality implies $$\vert d(p_m,q)- d(p,q) \vert\leqslant d(p_n,p) \to 0 $$ as $n \to \infty$. Hence, $$\lim_{n \to \infty} d(p_m,q) = d(p,q) . $$ It follows that $$ d(p,q) = \lim_{n \to \infty} d(p_n,q) \leqslant R.$$