# Definition of convergence in metric spaces

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?

• Do you know the definition of convergence in $\Bbb{R}$ ? – Maxime Ramzi Jul 28 '17 at 8:46
• It shows that convergence in $(X,d)$ can be reduced to a question about convergence in the reals. – Henno Brandsma Jul 28 '17 at 9:00

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).