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.