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I am confused between two definitions for convergence of a sequence in metric spaces. The first definition is:

Let $(X, d)$ be a metric space. A sequence $\{x_n\}$ converges to $x$ (denoted $x_n \rightarrow x$) if $\forall \epsilon >0$ $\exists N$ such that if $n > N$ implies $d(x_n, x)< \varepsilon$.

But then, in working through some problems and questions, I see something along the lines of:

If $\{x_n\}$ is a sequence contained in a metric space $(X, d)$, then $x_n \rightarrow x$ if and only if $d(x_n, x) \rightarrow 0$.

I can see "intuitively" why the second definition "works", if $(x_n)$ converges to $x$ then obviously the distance between them must eventually tend to $0$, but I am having a hard time seeing why this is true formally using the first definition. In the second definition, are we defining convergence of a sequence ($(x_n) \rightarrow x$) using the convergence of another sequence? Can someone explain to me why these two definitions are the same?

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  • $\begingroup$ Do you know the definition of convergence in $\Bbb{R}$ ? $\endgroup$ – Maxime Ramzi Jul 28 '17 at 8:46
  • $\begingroup$ It shows that convergence in $(X,d)$ can be reduced to a question about convergence in the reals. $\endgroup$ – Henno Brandsma Jul 28 '17 at 9:00
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\begin{align*}(x_n)_{n\in\mathbb N}\text{ converges to }x&\iff(\forall\varepsilon>0)(\exists p\in\mathbb{N}):n\geqslant p\Longrightarrow d(x_n,x)<\varepsilon\\&\iff(\forall\varepsilon>0)(\exists p\in\mathbb{N}):n\geqslant p\Longrightarrow\bigl|d(x_n,x)-0\bigr|<\varepsilon\\&\iff\lim_{n\in\mathbb N}d(x_n,x)=0.\end{align*}

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The formal proof is gotten by directly applying the first definition of convergence to the statement $d(x_n, x)\to 0$, which is short for "The sequence $d(x_n, x)$ of real numbers converges to $0$ (under the standard metric)", along with knowing that $d(d(x_n, x),0) = d(x_n, x)$ (if you excuse my abuse of notation).

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