$(x_n)^{\infty}_{n=m}$ converges to x if and only if $lim_{n \rightarrow \infty} d(x_n,x)=0$

Let $$(x_n)^{\infty}_{n=m}$$ be a sequence of real numbers, and let x be another real number. Then $$(x_n)^{\infty}_{n=m}$$ converges to x if and only if $$lim_{n \rightarrow \infty} d(x_n,x)=0$$

How do I begin with this problem? All I can think of is this definition: let $$(x_N)$$ be a sequence of real numbers. The sequence $$(x_n)$$ is said to converge to a real number a if for all $$\epsilon>0$$, there exists N in $$\mathbb{N}$$ such that $$|x_n-a|<\epsilon$$ for all $$n \geq N$$. A hint is helpful.

• Just recall that $d(x_n,x)=|x_n-x|$. Apr 18, 2020 at 1:50

Defintion: $$x_n \to x$$ in a metric space $$(X,d)$$ if $$\forall \epsilon > 0 \exists N \in \mathbb{N} \forall n \geq N: d(x_n, x) < \epsilon$$.
Then $$\mathbb{R}$$ is just a metric space with metric $$d(a,b) = |a-b|$$. So $$x_n \to x \iff \forall \epsilon > 0 \exists N \in \mathbb{N} \forall n \geq N: d(x_n, x) < \epsilon \iff \forall \epsilon > 0 \exists N \in \mathbb{N} \forall n \geq N: |d(x_n, x) - 0| < \epsilon \iff d(x_n,x) \to 0.$$