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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?

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    Please use [MathJax](http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference) rather than images for mathematical text. – 2017-01-20

3 Answers 3

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By the triangle inequality, $$d(x_n,y_n)\leq d(x_n,x)+d(x,y)+d(y,y_n)\tag 1$$ and

$$d(x,y)\leq d(x,x_n)+d(x_n,y_n)+d(y_n,y)\tag 2$$

From $(1)$, we get $$d(x_n,y_n)-d(x,y)\leq d(x_n,x)+d(y_n,y)\tag 3$$ and from $(2)$, we get $$-[d(x_n,x)+d(y_n,y)]\leq d(x_n,y_n)-d(x,y)\tag 4$$

Combining $(4)$ and $(3)$, we get $$-[d(x_n,x)+d(y_n,y)]\leq d(x_n,y_n)-d(x,y)\leq d(x_n,x)+d(y_n,y)$$

and since $d(x_n,x)+d(y_n,y)\geq 0$, we get

$$\Big|d(x_n,y_n)-d(x,y)\Big|\leq d(x_n,x)+d(y_n,y).$$

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Hint:

By the triangle inequality,

$$ d(x,y) \leq d(x,u) + d(u,v) +d(v,y) $$

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    More generally, if you're looking for a property of metrics that doesn't look obvious, I find that it's a good strategy to check whether applying the triangle inequality (possibly more than once, as above) gives you what you need. – 2017-01-20
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Temporarily assume $d(x_n,y_n) \geq d(x,y)$.

Then using the triangle inequality for each step of this reasoning we have: $$ d(x_n, y) \leq d(x_n,x) + d(x,y)\\ d(x_n,y_n) \leq d(x_n, y) + d(y,y_n) \leq d(x_n,x) + d(x,y) + d(y,y_n) $$ And subtracting $d(x,y)$ from both sides $$ |d(x_n,y_n) -d(x,y)| = d(x_n,y_n) -d(x,y) \leq d(x_n,x) + d(y,y_n) \leq |d(x_n,x) + d(y,y_n)| $$ So the statement holds if $d(x_n,y_n) \geq d(x,y)$.

The same type of reasoning (about going around a quadrilateral in two ways) holds when $d(x_n,y_n) < d(x,y)$. In fact, if you just swap $x_n \leftrightarrow x$ and $y_n \leftrightarrow y$ in the above proof, it works for $d(x_n,y_n) \leq d(x,y)$.