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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.

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

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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 $$x_n \to x \iff \forall \epsilon > 0 \exists N \in \mathbb{N} \forall n \geq N: d(x_n, x) < \epsilon \iff \forall \epsilon > 0 \exists N \in \mathbb{N} \forall n \geq N: |d(x_n, x) - 0| < \epsilon \iff d(x_n,x) \to 0.$$

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