# Suppose $x_n \rightarrow x$ and $y_n \rightarrow y$, then $d(x_n,y_n) \rightarrow d(x,y)$

I am currently reading Erwin Kreyszig-Introductory functional analysis with applications. I don't understand the proof below in it. Why is it true that $d(x_n,y_n) \leq d(x_n,y) + d(x,y) + d(y,y_n)$ ? I also don't understand similiar inequality is given by interchanging $x_n$ and $x$ as well as $y_n$ and $y$ ? why is that true ? • It's the triangle inequality: d (xn,yn) <= d (xn,x)+d (x,yn) <= d (xn,x) + d (x,y) + d (y,yn). – fleablood Sep 4 '16 at 23:03
• You have a typo in your question -- the first term in the RHS is $d(x_n,x)$, not $d(x_n,y)$. – Clement C. Sep 4 '16 at 23:07

Recall that any metric space $M$ has a metric $d$ defined on it where: $$d:M\times M\to \mathbb R_{\geq 0}$$ and $d$ fulfills the following axioms:

1. Symmetry: $$d(x,y) = d(y,x)$$

2. Non-negativity $$d(x,y)\geq 0\qquad d(x,y)=0\iff x = y$$

3. Triangle inequality: $$d(x,y)\leq d(x,z)+d(z,y)$$ We can apply the triangle inequality twice to $d(x_n,y_n)$ as follows:

$$d(x_n,y_n) \leq d(x_n,y)+d(y,y_n)$$

We also have that $$d(x_n,y)\leq d(x_n,x)+d(x,y)$$

We can combine these to get that:

$$d(x_n,y_n)\leq d(x,x_n)+d(x,y)+d(y,y_n)$$ Here I implicitly used the symmetry condition to say that $d(x,x_n)=d(x_n,x)$. It wasn't required, but it's good to recognize that these are equal.

We can interchange $x_n$ and $x$ and $y_n$ and $y$ because throughout this argument we've been treating them as points, and "forgetting" that they are sequences. We could have used points called $a,b,c,d$ and still obtained a valid relation between them. So, interchange the points is just saying "this argument is true for any $4$ points, and as we don't need to repeat it, we'll just take the end result and modify it to how we want it").

• Sorry for late reply had problems with computer ! – user329017 Sep 8 '16 at 2:21

It's just the triangle inequality, $d (a,c) \le d (a,b) + d (b,c)$ applied twice.

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

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

Likewise:

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

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

So $| d (x,y) -d (x_n,y_n)|= \pm[d (x,y) -d (x_n,y_n) ]\le d (x,x_n)+d (y_n,y)$