# Is this space a complete generalized metric space?

In this paper, in the proof of Theorem 2.1, the authors use the following result. We recall that, in this context, a "generalized metric space" is a metric for which the metric may assume infinite values.

Let $$G$$ be a non-empty set and $$Y$$ a Banach space. Define $$S:=\{f:G\longrightarrow Y\}$$ and the map $$d:S\times S \longrightarrow [0,+\infty]$$ as

$$d(f,g):=\inf\{\alpha\geq 0: \|f(x)-g(x)\|\leq \alpha \psi(x) \quad \textrm{for all } x\in G\},$$ where $$\psi:G\longrightarrow [0,+\infty)$$ is a given function.

Then, the authors state that $$(S,d)$$ is a complete generalized metric space. It is easy to show that $$d$$ satisfies all of the axioms of the metric definition. But, How can we prove that $$d$$ is complete?

Let $$(f_n)_n$$ be a Cauchy sequence in $$S.$$ Let $$a_n=\sup \{d(f_m, f_{m'}):n\le m So $$\lim_{n\to \infty}a_n=0.$$

For any $$x\in G$$ we have $$\|f_m(x)-f_{m'}(x)\|\le 2a_n\psi(x)$$ whenever $$n\le m and $$\lim_{n\to \infty}2a_n\psi(x)=0.$$

So $$(f_n(x))_n$$ is a Cauchy sequence in $$Y.$$ Since $$Y$$ is complete, $$f_n(x)$$ converges to some $$f(x)\in Y.$$

Given $$\epsilon\in \Bbb R^+,$$ choose $$n$$ large enough that $$\forall m\ge n\,(d(f_n,f_m)\le\epsilon).$$ For each $$x\in G$$ the set $$\{f_m(x):m\ge n\}$$ is contained in the closed ball $$C(x)=\overline {B(f_n(x),2\epsilon \cdot \psi(x))}$$ of $$Y,$$ so $$f_m(x)$$ converges in $$Y$$ to a member of $$C(x)$$. That is, $$\|f(x)-f_n(x)\|\le 2\epsilon \cdot \psi(x)\in C(x).$$ Therefore $$\forall m\ge n\,\forall x\in G\,(\|f(x)-f_n(x)\|\le 2\epsilon \cdot \psi(x)\,).$$ That is,$$\forall m\ge n\,(\,d(f,f_n)\le 2\epsilon).$$

So $$\lim_{n\to \infty}d(f,f_n)=0.$$

• Many thanks for your reply DanielWainfleet. I do not understand at all the number "2" in the proof....For instance, as $d(f_{m},f_{m'})\leq a_{n}$ for all $m'>m\geq n$, then $\|f_{m}(z)-f_{m'}(x)\|\leq a_{n}\psi(x)$, it is correct? – user123043 Apr 28 at 16:52
• Just being cautious about the $\inf$ in the def'n of $d$, but that $\inf$ is actually a $\min$ so the $2$ could be dropped. – DanielWainfleet Apr 29 at 1:59
• Ok, thank you very much. – user123043 Apr 29 at 16:09