Same limit iff $d(x_n,y_n) = 0 $

Question

If $(M, d)$ is complete, prove that two Cauchy sequences $(x_n)_n$ and $(y_n)_n$ have the same limit if and only if $d(x_n,y_n) \rightarrow 0.$

My Method :

For forward implication ,

$$ d(x_n,a) < \frac{\varepsilon}{2},$$ $$ d(y_n,a) < \frac{\varepsilon}{2}.$$ Using the triangle inequality, we have $$d(x_n,y_n) < \varepsilon,$$ and therefore $d(x_n , y_n ) \rightarrow 0$.

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Is this right?

But, for the reverse implication, I used the triangle inequality and cannot proceed further .

Answer 1

Forward :

Let $x_n$ and $y_n$ be two Cauchy Sequences in $(M,d)$. Also, let $x_n \to L$ and $y_n \to L$ with $L \in M$. This indeed means that : $$d(x_n,L) \to 0 \quad \text{and} \quad d(y_n,L) \to 0 $$ This can equally be expressed as that $\exists \varepsilon >0$ such that $$d(x_n,L) < \varepsilon/2 \quad \text{and} \quad d(y_n,L) < \varepsilon/2$$ as $\varepsilon$ can become arbitrary small.

But $d$ is a metric in the space $M$ and thus the Triangle Inequality holds : $$d(x_n,y_n) \leq d(x_n,L) + d(y_n,L) < \varepsilon \implies d(x_n,y_n) \to 0$$

Reverse :

Let $x_n$ and $y_n$ be two Cauchy Sequences in $(M,d)$, such that $d(x_n,y_n) \to 0$. Now, let $x_n \to x$ and $y_n \to y$ with $x,y \in M$. We are interested to see if their limits coincide, aka if $x$ and $y$ can be brought infinitely close. For this, we will work over their distance under the metric $d$, namingly $d(x,y)$. But then, it is : $$d(x,y) \leq d(x,x_n) + d(x_n,y_n) + d(y_n,y) $$ Now, note that since $x_n \to x$ it is $d(x_n,x) \to 0$ and the same holds for $d(y_n,y)$. We have also let that $d(x_n,y_n) \to 0$. Thus, it is $d(x,y) \to 0$ and that essentialy means that $x \equiv y$.

Reference/Note : A nice post on why the limit exists.

Answer 2

Suppose $x_n \rightarrow x$ and $y_n \rightarrow y$ as $n \rightarrow \infty.$ Then $d(x,y) \leq d(x,x_n) + d(x_n,y_n) + d(y_n,y) \rightarrow 0$ as $n \rightarrow \infty$ if $d(x_n,y_n) \rightarrow 0$ as $n \rightarrow \infty.$ So the distance between $x$ and $y$ can be made less than any positive number. This proves that $d(x,y)= 0 \implies x=y,$ as claimed.

Answer 3

Your forward implication is basically correct, but would benefit from a bit more words. What is $a$, what is $n$, what is $\epsilon$, and how exactly does $d(x_n,y_n)<\epsilon$ imply $d(x_n, y_n)\to 0$?

For instance:

Assume both sequences converge to $a$. Given an $\epsilon>0$, let $M\in \Bbb N$ be such that $d(x_m, a)<\epsilon/2$ for all $m>M$, and let $N\in\Bbb N$ be such that $d(y_n, a)>\epsilon/2$ for all $n>N$.

Let $K=\max(M, N)$. Then by the triangle inequality we have $d(x_k, y_k)<\epsilon$ for all $k>K$, showing that $d(x_k, y_k)$ converges to $0$.

As for the other direction, I like the contrapositive. Assume they converge to different limits, and show that there is some $\epsilon$ that $d(x_n, y_n)$ after some $N$ doesn't go below.