# Show that $d(x_n,y_n) \longrightarrow d(x,y)$ if $x_n \longrightarrow x$ and $y_n \longrightarrow y$

Let $$(X,d)$$ be a metric space and $$(x_n),(y_n)$$ be two sequences in $$X$$ such that $$x_n \longrightarrow x$$ and $$y_n \longrightarrow y$$, with $$x,y \in X$$. Show that $$d(x_n,y_n) \longrightarrow d(x,y)$$.

To begin with, for any $$x,y,u,v \in X$$ we have: $$|d(x,y) -d(u,v)|\leq d(x,u) +d(y,v)$$ which ensures that the metric is continuous in topology generated by itself and any other weaker topology. Given that, according to the Heine definition of continuity, $$d$$ is continuous at $$(x,y)$$ if and only if for any sequences $$x_n,y_n$$ with $$x_n \longrightarrow x$$ and $$y_n \longrightarrow y$$: $$d(x_n,y_n) \longrightarrow d(x,y)$$ Is my approach correct?

• I think so, but I'm guessing this is an exercise where you're expected to write an $\epsilon-N$ proof, using the triangle inequality. – Robert Shore Mar 12 at 20:43

$$d(x,y)\le d(x,x_n)+d(x_n,y) \le$$

$$d(x,x_n)+d(x_n,y_n)+d(y_n,y).$$

$$d(x,y) -d(x_n,y_n) \le d(x,x_n) +d(y,y_n)$$.

Similarly:

$$d(x_n,y_n)-d(x,y) \le d(x_n,x)+d(y_n,y).$$

$$|d(x_n,y_n)-d(x,y)| \le d(x_n,x)+d(y_n,y).$$

Let $$\epsilon >0$$.

For $$\epsilon/2$$ there is a $$n_1$$ s.t.

for $$n \ge n_1$$

$$d(x,x_n) < \epsilon/2.$$

For $$\epsilon/2$$ there is a $$n_2$$ s.t. for $$n \ge n_2$$

$$d(y,y_n) < \epsilon/2.$$

Let $$N=\max (n_1,n_2)$$ .

For $$n \ge N$$:

$$|d(x,y)-d(x_n,y_n)| < d(x,x_n)+d(y,y_n) < \epsilon.$$

It follows from the following simple property about metric $$d$$:

For any $$x,y,a\in X$$, we have $$|d(x,a)-d(y,a)|\leq d(x,y)$$.

Proof: By the triangular inequality, $$d(x,a)\leq d(x,y)+d(y,a)$$, so $$d(x,a)-d(y,a)\leq d(x,y)$$.

Again, $$d(y,a)\leq d(y,x)+d(x,a)$$, so $$d(y,a)-d(x,a)\leq d(y,x)$$. Hence, we obtain $$-d(x,y)\leq d(x,a)-d(y,a)\leq d(x,y)$$ which is equivalent to $$|d(x,a)-d(y,a)|\leq d(x,y)$$. $$\hspace{75mm} \square$$

Suppose that $$(x_{n})$$ and $$(y_{n})$$ are sequences in $$X$$ such that $$x_{n}\rightarrow x$$ and $$y_{n}\rightarrow y$$. Then $$\begin{eqnarray*} |d(x_{n},y_{n})-d(x,y)| & \leq & |d(x_{n},y_{n})-d(x_{n},y)|+|d(x_{n},y)-d(x,y)|\\ & \leq & |d(y_{n},y)|+|d(x_{n},x)|\\ & \rightarrow & 0 \ \ \text{as} \ \ n \rightarrow \infty. \end{eqnarray*}$$