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 $D( \frac{1}{d(x_n,U^c)},\frac{1}{d(L,U^c)} )<r,\;\forall n.$ 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?


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