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.

  • 1
    $\begingroup$ your epsilons are not necessarily positive, this is a problem... why not keeping the absolute values in $(3)$ ? $\endgroup$ – zwim Nov 17 '17 at 14:26
  • $\begingroup$ How we know that $\frac{\varepsilon}{2}-d(x,y)>0$ ? $\endgroup$ – Hector Blandin Nov 17 '17 at 14:27
  • $\begingroup$ @zwim: Ah! I totally missed that part. I knew something was wrong! :) Thanks. Should rethink about it. Because $d$'s are always non-negative. $\endgroup$ – Hosein Rahnama Nov 17 '17 at 14:37

Apply two times the triangular inequality allows to have the same quantity but with reversed sign, thus we get a bound for the absolute value.

$d(x_n,y_n)\le d(x_n,x)+d(x,y)+d(y,y_n)$

$d(x,y)\le d(x,x_n)+d(x_n,y_n)+d(y_n,y)$

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.