Edit: Prove that $(U,\delta)$ is complete where $\delta(x,y)=d(x,y)+\Big|\frac{1}{d(x,F)}-\frac{1}{d(y,F)}\Big|,\;\;\forall\; x,y\in U$

Question

Edit: Let $(E,d)$ be a complete metric space such that U is open in $E$, $U\neq E.$ Let $F=U^c$. We define

$$\delta(x,y)=d(x,y)+\Big|\frac{1}{d(x,F)}-\frac{1}{d(y,F)}\Big|,\;\;\forall\; x,y\in U$$

1.) Prove that $\delta$ is a distance in $U$

2.) Prove that $(U,\delta)$ is complete.

Number 1 is clear to me. However, to show 2, we can let $\{x_n\}\subset U$ be Cauchy, then $\delta(\{x_n\},\{x_m\})\to 0.$

Question: How do I show that $x_n\to x$ and $x\in U$ using metric $\delta$? Thanks for your time and help!

Answer 1

Let $\{x_n\}$ be a Cauchy sequence in $(U,\delta)$. Then $d(x_n,x_m)\to 0$ so there exists $x \in E$ such that $d(x_n,x) \to 0$. By triangle inequality $|d(x_n,F)-d(x,F)|\leq d(x_n,x) \to 0$ as $n \to \infty$. Since $\frac 1 {d(x_n,F)}-\frac 1 {d(x_m,F)} \to 0$ it follows that $\{\frac 1 {d(x_n,F)}\}$ is also a Cauchy sequence of real numbers. This implies that this sequence is bounded. It follows that $d(x,F)$ cannot be $0$. Thus $\frac 1 {d(x_n,F)} \to \frac 1 {d(x,F)}$. It is now clear that $\delta (x_n,x) \to 0$. Hence $(U,\delta)$ is complete.

Answer 2

The theorem as stated is false. Let $F$ be the origin in the plane and let $U$ be an open half-circumference of the unit circle. Consider this space with the metric inherited from the plane, which is clearly not complete. Then for every pair of points in $U$, the second term of your new metric will be constantly zero. Thus this new metric induces the same (non-complete) metric as the planar one.

Do you want to assume that $E$ or $U$ are complete, maybe?

Edit: a new assumption, namely that $U$ is complete in the original space, has been added. I have outlined the proof in a comment below.