0
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Robert Shore Mar 12 at 20:43
1
$\begingroup$

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

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

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

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 $:

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

$\endgroup$
1
$\begingroup$

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*}

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.