Sequences that converge to the same point

Question

In $(X,d)$ metric space, If $\{x_n\}_{n\in\Bbb N}$ and $\{y_n\}_{n\in\Bbb N}$ are two sequences which converge to the same point, then prove that

$$\lim_{n\to\infty} d(x_n,y_n)=0$$

This question is found in the chapter of completeness, but I could not figure out where to use completeness for this question.

Answer 1

let $X_n\rightarrow X$ thus $Y_n$ converges to Y as well. Given $\epsilon>0$, $\exists m\in N$ such that $\forall n>m$ $d(X_n,X)<\epsilon /2$ and $d(Y_n,X)<\epsilon /2$ Then $\forall n>m $ we have $d(X_n,Y_n)<d(X-n,X)+d(Y_n,X)<\epsilon$. Thus $d(X_n,Y_n)\rightarrow 0$

Answer 2

Hint:

If $x$ is the limit of $x_n$ and $y$ is the limit of $y_n$, then

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

You can use this to prove (using the $\epsilon$ -definition) that $$\lim_{n\to\infty} d(x_n, y_n) = 0$$

Answer 3

Let $l$ be the limit of the sequences $(x_{n}), (y_{n})$ by definition Note that $d(x_{n},y_{n}) \leq d(x_{n},l) + d(l, y_{n})$ for all $n$. If $\varepsilon > 0$, then there are $N_{1},N_{2}$, by assumption, giving $d(x_{n},l), d(l,y_{n}) < \varepsilon/2$ for all $n \geq N_{1}$ and $\geq N_{2}$. So $d(x_{n},l) + d(l,y_{n}) < \varepsilon$ for all $n \geq \max \{ N_{1},N_{2} \}$; so $d(x_{n},y_{n}) < \varepsilon$ for all $n \geq \max \{ N_{1},N_{2} \}$.