# Thread: limits in a metric space question

1. ## limits in a metric space question

The Question:

Show that if $\displaystyle x_n -> x$ in a metric space (M,d), then for any $\displaystyle a$ in M, $\displaystyle d(x_n,a) -> d(x,a)$.

Where I am struggling:

I understand that $\displaystyle x_n->x => d(x_n,x) -> 0$

Then I tried relating this to the triangle inequality:

$\displaystyle d(x_n,a) <= d(x_n,x) + d(x,a)$
$\displaystyle lim[ d(x_n,a)] <= 0 + d(x,a)$
$\displaystyle d(x_n,a) -> c <= d(x,a$)

I don't know how to show that $\displaystyle c$ is strictly equal to $\displaystyle d(x,a)$ and not less than $\displaystyle d(x,a)$.

Thank you for any help that can be offered.

2. Originally Posted by rgriss1
The Question: Show that if $\displaystyle x_n -> x$ in a metric space (M,d), then for any $\displaystyle a$ in M, $\displaystyle d(x_n,a) -> d(x,a)$.
There is a wonderfully useful theorem: If $\displaystyle \left\{ {a,b,x} \right\} \subseteq M$ then $\displaystyle \left| {d(a,x) - d(x,b)} \right| \leqslant d(a,b)$
Here we can say that $\displaystyle \left| {d(x_n ,a) - d(a,x),} \right| \leqslant d(x_n ,x)$.
We are almost done.

3. Haha, yes. Thank you so much.

I actually started messing around with the inequality after I posted this and remembered that theorem.

But, again, thank you for confirming that I eventually found the right direction. :-)