In a metric space $(S,d)$, assume that $x_n\,\to\,x$ and $y_n\,\to\,y$. Prove that $d\left(x_n,y_n\right)\,\to\,d(x,y)$.
$\textbf{Proof:}$ Since $x_n\,\to\,x$ and $y_n\,\to\,y$, given $\epsilon>0$ there exists a positive integer $N$ such that as $n\ge N$, we have
\begin{equation}
d\left(x_n,x\right)<\epsilon/2 \text{ and }d\left(y_n,y\right)<\epsilon/2
\end{equation}
Hence, as $n\ge N$, we have
\begin{eqnarray}
\left\lvert\,d\left(x_n,y_n\right)-d\left(x,y\right)\,\right\rvert&\le&\left\lvert\,d\left(x_n,x\right)+d\left(y_n,y\right)\,\right\rvert\\
&=&d\left(x_n,x\right)+d\left(y_n,y\right)\\
&<&\epsilon/2+\epsilon/2\\
&=&\epsilon
\end{eqnarray}
I cannot understand why this is true:
\begin{equation}
\left\lvert\,d\left(x_n,y_n\right)-d\left(x,y\right)\,\right\rvert\le\left\lvert\,d\left(x_n,x\right)+d\left(y_n,y\right)\,\right\rvert
\end{equation}
Can you please explain me that?
Best Answer
Hint:
By the triangle inequality,
$$ d(x,y) \leq d(x,u) + d(u,v) +d(v,y) $$