If $\lim_{n\to\infty}x_n=x$ and $\lim_{n\to\infty}y_n=y$ then $\lim_{n\to\infty}d(x_n,y_n)=d(x,y)$

I want to prove the following (Exercise 4.7 Apostol's Mathematical Analysis).

If $\lim_{n\to\infty}x_n=x$ and $\lim_{n\to\infty}y_n=y$ then $\lim_{n\to\infty}d(x_n,y_n)=d(x,y)$

Here is my solution. Would you please check that to see if it is OK. Also, any suggestion for improvement is appreciated. Can we make a geometric interpretation for this?

The assumptions imply that

\begin{align*} &\forall \varepsilon_1>0\,\,\exists N_1\in\mathbb{Z}^+:n\ge N_1\implies d(x,x_n)\lt\varepsilon_1 \\ &\forall \varepsilon_2>0\,\,\exists N_2\in\mathbb{Z}^+:n\ge N_2\implies d(y,y_n)\lt\varepsilon_2 \tag{1} \end{align*}

and we are to prove

\begin{align*} \forall \varepsilon>0\,\,\exists N\in\mathbb{Z}^+:n\ge N\implies |d(x_n,n_n)-d(x,y)|\lt\varepsilon. \tag{2} \end{align*}

Let us set $N=\max\{N_1,N_2\}$ so the inequalities in $(1)$ hold. By the triangle inequality, we also have

\begin{align*} |d(x_n,y_n)-d(x,y)|&\le d(x_n,y_n)+d(x,y)\le d(x_n,x)+d(x,y_n)+d(x,y)\\ &\le d(x_n,x)+d(x,y)+d(y,y_n)+d(x,y) \\ &=d(x,x_n)+d(y,y_n)+2d(x,y). \tag{3} \end{align*}

Choosing $\varepsilon_1=\varepsilon_2=\frac{\varepsilon}{2}-d(x,y)$, $(1)$ and $(3)$ lead us to

\begin{align*} |d(x_n,y_n)-d(x,y)|\lt\Big(\frac{\varepsilon}{2}-d(x,y)\Big)+\Big(\frac{\varepsilon}{2}-d(x,y)\Big)+2d(x,y)=\varepsilon \end{align*}

which completes the proof.

• your epsilons are not necessarily positive, this is a problem... why not keeping the absolute values in $(3)$ ? – zwim Nov 17 '17 at 14:26
• How we know that $\frac{\varepsilon}{2}-d(x,y)>0$ ? – Hector Blandin Nov 17 '17 at 14:27
• @zwim: Ah! I totally missed that part. I knew something was wrong! :) Thanks. Should rethink about it. Because $d$'s are always non-negative. – Hosein Rahnama Nov 17 '17 at 14:37

$d(x_n,y_n)\le d(x_n,x)+d(x,y)+d(y,y_n)$
$d(x,y)\le d(x,x_n)+d(x_n,y_n)+d(y_n,y)$
This gives $\bigg|d(x,y)-d(x_n,y_n)\bigg|\le d(x_n,x)+d(y_n,y)\le \epsilon_1+\epsilon_2$