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$1)$ If $(x_n)$ is Cauchy in $(X,d)$, then $x_n \longrightarrow x_0$. Or, equivalently, $\forall \epsilon > 0, \exists N \in \mathbb{N}$ s.t: $d(x_n,x_0) < \epsilon , \forall n \geq N$.

$2)$ Defn: A function $f: X \longrightarrow Y$ is called uniformly continuous if, $\forall \epsilon > 0, \exists \delta > 0$ s.t: $\rho(f(x),f(y)) < \epsilon, \forall x,y \in X$ satisfying $d(x,y) < \delta$.

Proof: It suffices to show that $(f(x_n))$ converges to some $f(x_0)$ in $(Y,\rho)$. Indeed, since $(x_n)$ is Cauchy, it converges in $(X,d)$ and so $\forall \delta > 0, \exists N \in \mathbb{N}$ s.t: $d(x_n,x_0) < \delta$, whenever $n \geq N$. So fix $\delta > 0$. Then, in view of the uniform continuity of $f$, we have that $\forall \epsilon > 0, \exists \delta > 0$ s.t: $\rho(f(x_n),f(x_0)) < \epsilon, \forall x_n,x_0 \in X$ satisfying $d(x_n,x_0) < \delta$.

In other words, $(f(x_n)) \longrightarrow f(x_0)$, and so $(f(x_n))$ is a Cauchy sequence in $(Y,\rho)$ since it converges. QED.

How did I do?

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  • $\begingroup$ Why $(x_{n})$ converges in $(X.d)$? $\endgroup$
    – Rafael
    Aug 13, 2016 at 21:19
  • $\begingroup$ Fuck. I thought $(x_n)$ converges because it's Cauchy... But I just remembered a Cauchy sequence does not necessarily converge -_- $\endgroup$
    – Javier
    Aug 13, 2016 at 21:22
  • $\begingroup$ Just use the definitions, there is not much possibilty to do anything wrong. $\endgroup$
    – Lukas Betz
    Aug 13, 2016 at 21:26

1 Answer 1

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Let $\epsilon$ be a positive real number. Thus exist $\delta>0$ such that $$d(x_{n},x_{m})<\delta\Rightarrow\rho(f(x_{n})f(x_{m}))<\epsilon.$$ Since $(x_{n})$ is Cauchy, there is $n_{0}\in\mathbb{N}$ such that $$n,m>n_{0}\Rightarrow d(x_{n},d_{m})<\delta.$$ It follow that $$n,m>n_{o}\Rightarrow\rho(f(x_{n}),f(x_{m}))<\epsilon$$ that is, $(f(x_{n}))$ is Cauchy.

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