Space of Katetov-Functions with sup-metric is complete metric space Let (X,d) be a metric space. $f: X\longrightarrow\mathbb{R}$ is called "Katetov map" iff : $$∀,,∈:|()−()|≤(,)≤()+()$$
The set of all Katetov-maps on X is denoted by $E(X)$.
Prove that $E(X)$ turns into a complete metric space if endowed with the  sup-metric.
Since Katetov-maps don't necessarily have to be bounded, I have no idea how to prove the statement. Can anyone help me please?
NB: I tried to use that if $f_n$ is a Cauchy-sequence in $E(X)$, then $f_n(x)$ is a Cauchy-sequence in $\mathbb{R}$ for every fixed x and thus converges. But how do I show that $f_n$ uniformly converges to $f$?
Many thanks!
 A: First, we show that d is indeed a metric on E(X). We use a different approach to the (supposed) metric d.
Note that we can embed X in E(X) via $x \mapsto \delta_x$ with $\delta_x: X \to \mathbb{R}: y \mapsto d(x,y)$. For a Katetov-map f we extend the metric on X to a metric on $X \cup \{f\}$ by defining $d(x,f)=f(x)$ (which is a metric, since (i) $d(f,x)=f(x)=0 \Rightarrow f(y) = d(x,y) \forall y \in Y \Rightarrow f = \delta_x \Rightarrow f=x$ and (ii)Katetov-maps are exactly those that fulfil the triangle inequality).
We now show that being a Katetov-map is equivalent to fulfilling $|f(x)-d(x,y)| \leq f(y) \ \forall y \in X$. We define -- as proposed in the question -- the metric d on E(X) as $d(f,g) = \sup |f(x)-g(x)|$ which is always finite, since $|f(x)-d(x,y)| \leq f(y)$ and $|g(x)-d(x,y)|\leq g(y)$ hence $|f(x)-g(x)| \leq f(y)+g(y)$. Therefore $f-g$ is a bounded function and d is a well defined metric on E(X).
We define f as the pointwise limit of the $f_n$.
Since $f_n$ is Cauchy in $E(X)$, we know that $\forall \epsilon > 0 \exists N_0 \in \mathbb{N} \ st \ \forall n,m \geq N_0 : d(f_n,f_m) \leq \epsilon$. Now suppose we find $x$ and $n \geq N_0$ with $d(f_n(x), f(x)) = \epsilon + \delta, \delta > 0$. But then for $ m \geq \max \{N_0, N_x \}$ ($N_x$ being the number st $\forall n \geq N_x: d(f_n(x), f(x)) < \epsilon$) it holds that $d(f_n(x), f_m(x)) \geq d(f_n(x), f(x)) - d(f_m(x), f(x)) \geq \delta >0$ hence $(f_n)$ is not Cauchy. Therefore $d(f_n, f) \rightarrow 0$ and f still fulfils the inequalities necessary to be a Katetov-map.
Edited: Proof that d is a metric on $E(X)$.
