Limit in metric space

Question

Suppose $x_n \to x$ and $y_n \to y$ in the metric space $\big( M,p \big)$. Prove that $lim_{n \to \infty} p(x_n, y_n) = p(x,y)$

Well, I am not sure how to get this from the properties of the metric space

Answer 1

EDIT: I believe I have now corrected the inept nonsense which I unfortunately posted earlier. My sincerest apologies and I hope your head doesn't hurt as much as mine..

If $$p(x_n,x)<{\epsilon\over 2}$$ and $$p(y_n,y)<{\epsilon\over 2}$$ then $$d(p(x_n,y_n),p(x,y))=|p(x_n,y_n)-p(x,y)|$$ Write $$|p(x_n,y_n)-p(x,y)|=|p(x_n,y_n)-p(x_n,y)+p(x_n,y)-p(x,y)|\le|p(x_n,y_n)-p(x_n,y)|+|p(x_n,y)-p(x,y)|\le|p(y,y_n)|+|p(x_n,x)|\le{\epsilon\over 2}+{\epsilon\over 2}=\epsilon$$

Answer 2

Hint: Prove that the distance function $p:X\times X\to \mathbb R$ is continuous. The easiest way to show that is to actually show that it is a non-expanding map (when $X\times X$ is given the metric structure $d((x,y),(x',y'))=d(x,x')+d(y,y')$ and $\mathbb R$ is given the usual metric). Any non-expanding mapping is claerly uniformly continuous, and thus continuous. Your claim then is just the preservation of limits under continuous functions.

Answer 3

just look at the two bellow inequalities: $$p(x_n,y_n)\leq p(x_n,x)+p(x,y)+p(y,y_n)$$ $$p(x,y)\leq p(x,x_n)+p(x_n,y_n)+p(y_n,y)$$