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Background Suppose $(x_i)$ and $(y_j)$ are two sequences in a metric space $(X,d)$ that converge respectively to $x$ and $y$. I am trying to show that the sequence $(d( x_i, y_j))$ converges to $d(x,y)$

Thoughts I think probably the first thing to do is to show that the metric is itself a continuous function. I believe this is so because if $V$ is an open subset of $\mathbb{R}$ in the range of $d$, say $V := (a, b)$ then the preimage of $V$ is the set $$ d^{-1}(V) = \{(x,y) \in X \times X: a < d(x,y) < b \} $$ which is an open ball of radius $|b-a|$. Assuming this part is OK, since $d$ is continuous, it commutes with the limit operation but in this case, there are actually two limits: one as $i \rightarrow \infty$ and one as $j \rightarrow \infty$ I think maybe there is a couple of ways around this; one, by fixing $x := x_0$ and taking the limit as $j \rightarrow \infty$ and then by similarly fixing $y := y_0$ and take the limit as $i \rightarrow \infty$. Perhaps another way is to simply use the same sequence index and consider each point $(x_i, y_i)$ as a term in the sequence.

Question I think I understand intuitively what's going on, but I'm not sure of how to make the argument precise. Is the approach I outlined promising or is there a cleaner approach that I'm overlooking?

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Your inverse image of $V$ is incorrect. $d^{-1}(V)=\{(x,y)\in X\times X\mid a\lt d(x,y)\lt b\}$. –  Arturo Magidin Feb 15 '12 at 19:55
    
What ordering are you placing on $\mathbb{N}\times\mathbb{N}$ in order to consider $(d(x_i,y_j))$ a sequence? –  Arturo Magidin Feb 15 '12 at 19:57
    
@Arturo I'll fix V; with regard to ordering, well, perhaps I'm not stating the problem clearly. I'm trying to say that as $x_i$ approaches $x$ and $y_i$ approaches y then the distance between $x_i$ and $y_i$ approaches $d(x,y)$ –  ItsNotObvious Feb 15 '12 at 20:01
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in other words: you want to prove that for all $\epsilon\gt 0$ there exist $N,M\gt 0$ such that if $i\geq N$ and $j\geq M$, then $|d(x,y) - d(x_i,y_j)|\lt\epsilon$. –  Arturo Magidin Feb 15 '12 at 20:06

3 Answers 3

up vote 2 down vote accepted

Assume that $\{x_n:n\in\mathbb{N}\}\subset X$ connverges to $x\in X$ and $\{y_n:n\in\mathbb{N}\}\subset X$ connverges to $y\in X$. Fix $\varepsilon >0$, then there exist $K\in\mathbb{N}$, $L\in\mathbb{N}$ such that for all $k>K$ and $l>L$ we have $$ d(x_k,x_0)<\varepsilon/2\qquad\qquad d(y_l,y_0)<\varepsilon/2 $$ Consider $N=\max(M,L)$, then for all $k>N$ and $l>L$ we have $$ |d(x_k,y_l)-d(x_0,y_0)|\leq d(x_k,x_0)+d(y,y_0)<\varepsilon $$ Since $\varepsilon >0$ is arbitrary we conclude $$ \lim\limits_{k,l\to\infty}d(x_k,y_l)=d(x_0,y_0). $$

P.S. In my proof I've used the following inequality: $$ |d(p,q)-d(r,s)|\leq d(p,r)+d(q,s) $$ which holds for all $p,q,r,s\in X$. It can be proved by the following way $$ d(p,q)\leq d(p,r)+d(r,q)\leq d(p,r)+d(r,s)+d(s,q)\Longrightarrow $$ $$ d(p,q)-d(r,s)\leq d(p,r)+d(q,s) $$ Similarly, $$ d(r,s)\leq d(r,p)+d(p,q)\leq d(r,p)+d(p,q)+d(q,s)\Longrightarrow $$ $$ d(r,s)-d(p,q)\leq d(r,p)+d(q,s) $$ Then $$ |d(p,q)-d(r,s)|\leq d(p,r)+d(q,s) $$

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I was not familiar with the $|d(p,q)−d(r,s)| \leq d(p,r)+d(q,s)$ inequality that you used; this makes everything very clear. Thanks. –  ItsNotObvious Feb 15 '12 at 20:22
    
So, using this result, I believe we can say the following: $\lim_{i \to \infty} \lim_{j \to \infty} d(x_i, y_j) = \lim_{k \to \infty}d(x_k, y_k)$, correct? –  ItsNotObvious Feb 16 '12 at 16:30
    
Yes, if limit of the type $\lim\limits_{k,l\to\infty}$ exists, then exists repeared limits like $\lim\limits_{i\to\infty}\lim\limits_{j\to\infty}$, and limit over subsequences like $\lim\limits_{l=k,k\to\infty}$ –  userNaN Feb 16 '12 at 16:39

It is true that $d$ is a continuous function on $X\times X$, so if you show $(x_i,y_i)\to (x,y)$ in $X\times X$ the result follows. However, I would use a different approach. First note that $d(x_i,y_i)\leq d(x_i,x)+d(x,y)+d(y,y_i)$ by the triangle inequality. Furthermore, since $d(x_i,x)+d(x_i,y_i)\geq d(x,y_i)$ and $d(x,y_i)+d(y,y_i)\geq d(x,y)$ by the triangle inequality, we get $$d(x_i,y_i)\geq d(x,y_i)-d(x_i,x) \geq d(x,y)-d(x_i,x)-d(y,y_i)$$ so putting our two inequalities together $$d(x,y)-d(x_i,x)-d(y,y_i)\leq d(x_i,y_i)\leq d(x_i,x)+d(x,y)+d(y,y_i)$$ and the left and right expressions go to $d(x,y)$ because $x_i\to x$ and $y_i\to y$ (and by the definition of convergence this means $d(x_i,x),d(y,y_i)\to 0$). Thus $d(x_i,y_i)\to 0$.

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$$ \color{red}{0\leqslant|d(x_i,y_j)-d(x,y)|\leqslant d(x_i,x)+d(y_j,y)\longrightarrow0} $$

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