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

Question

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?

Answer 1

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

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

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

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 $:

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

Answer 2

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$

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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*}