1
$\begingroup$

I want to show

In a metric space $(X,d)$ we have $\lim\limits_{n\to\infty}x_n=a\implies\lim\limits_{n\to\infty}d(x_k,x_n)=d(x_k,a)$

My proof: by the triangle inequality $$d(x_k,a)\leq d(x_k,x_n)+ d(x_n,a)$$ And if we take $n\to\infty$ we have $x_n\to a$ $$d(x_k,a)\leq \lim\limits_{n\to\infty}d(x_k,x_n)$$

Now I want to Show that

$$\lim\limits_{n\to\infty} d(x_k,x_n)\leq d(x_n,x_n)+d(x_k,\lim\limits_{n\to \infty} x_n)=d(x_k,a)$$ Is it correct?

$\endgroup$
  • 1
    $\begingroup$ All that the triangle inequality says is that the sum of two sides of a triangle is bigger than the third side. It is up to you to choose which one the third side should be. $\endgroup$ – NickD May 4 '17 at 18:11
1
$\begingroup$

Does domething like:

Fix $\varepsilon>0$, it exists $N\in\mathbb{N}$ s.t. $\forall n>N$, $d(x_n,a)<\varepsilon$. It follows $$\vert d(x_k,x_n)-d(x_k,a)\vert\leq\vert d(x_k,a)+d(a,x_n)-d(x_k,a)\vert=\vert d(a,x_n)\vert<\varepsilon,\ \ \forall n>N.$$

convince you?

$\endgroup$
  • $\begingroup$ Yes it does. Thank you very much! $\endgroup$ – WaldoRozir May 4 '17 at 19:11
1
$\begingroup$

The result follows immediately by the fact that the distance function is continuous.

$\endgroup$

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.