# Show $\lim\limits_{n\to\infty}x_n=a\implies\lim\limits_{n\to\infty}d(x_k,x_n)=d(x_k,a)$

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?

• 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. – NickD May 4 '17 at 18:11

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?

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

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