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 ?
2 Answers
$\begingroup$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:
Symmetry: $$d(x,y) = d(y,x)$$
Non-negativity $$d(x,y)\geq 0\qquad d(x,y)=0\iff x = y$$
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").
$\endgroup$ 1 $\begingroup$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 $| d (x,y) -d (x_n,y_n)|= \pm[d (x,y) -d (x_n,y_n) ]\le d (x,x_n)+d (y_n,y)$
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