# Open sets and Complete Metric Spaces

Let $$(X,d)$$ be a metric space with $$A$$ a subset. For each $$x \in X$$, define $$d'(x,A)=\text{inf}_{a \in A}d(x,a)$$.

Let $$U$$ be an non-empty open subset of $$X$$ and $$U\ne A$$. For $$x,y \in U$$, define $$D(x,y)=d(x,y)+\lvert \frac{1}{d'(x,U^c)} - \frac{1}{d'(y,U^c} \rvert.$$

Show that D is complete.

My attempt at a solution. Let $$\{ x_{n}\}$$ be a $$D$$-Cauchy sequence in $$U.$$ Since $$(X,d)$$ is complete $$\{ x_{n}\}$$ has a limit $$L$$ in $$X.$$ If $$L \notin U,$$ then $$\forall \epsilon>0, \;d(L,U^c)\leq d(x_{n} ,L)<\epsilon.$$ By the contrapositive, if $$\exists \epsilon >0, \;d(L,U^c)>d(x_{n} ,L)\geq \epsilon ,$$ then $$L \in U.$$ $$\frac{1}{d(x_{n} ,U^c)} \to \frac{1}{d(L,U^c)}$$ so $$\frac{1}{d(x_{n} ,U^c)}$$ is bounded. Therefore there is an $$\epsilon >0$$ such that $$\frac{1}{d(x_n,U^c)} \leq \epsilon$$ for $$n \geq N.$$ Thus $$L \in U$$ so $$(U,D)$$ is complete.

There is another definition of bounded which is that there is $$r$$ such that $$D( \frac{1}{d(x_n,U^c)},\frac{1}{d(L,U^c)} ) Let $$\alpha =\frac{1}{d(x_n,U^c)}, \beta =\frac{1}{d(L,U^c)}.$$ Using this definition I get $$d(\alpha ,\beta ) + \lvert \frac{1}{d(\alpha , U^c )} - \frac{1}{d(\beta ,U^c)} \rvert.$$ How can I proceed from here to show that $$\alpha$$ is bounded? And is the second definition a generalization of boundedness?