# Homeomorphism between two metric spaces via identity

Suppose $$(X,d)$$ is a complete metric space with $$U_1,U_2,...$$ nonempty open subsets, with none equal to $$X.$$ Let $$U= \bigcap_{n=1}^{\infty } U_n \neq \emptyset$$ and define $$d_n$$ on $$U_n$$ as $$d_{n}(x,y) =\text{min} (D_{n} (x,y),1)$$ where $$D_{n}(x,y) =d(x,y)+\lvert \frac{1}{d(x,U_n^c) } - \frac{1}{d(y,U_n^c)} \rvert.$$ Define $$D(x,y)=\sum_{n=1}^{\infty } \frac{1}{2^n} d_{n} (x,y).$$

If I want to show that $$(U,d)$$ is homeomorphic to $$(U,D)$$ by the identity function, then I can show that $$(U,d)$$ and $$(U,D)$$ have the same open sets.

$$\implies$$ Let $$V$$ be an open set in $$(U,d)$$. Let $$x \in V$$ and $$r<1.$$ If $$d(x,y), then $$y \in V.$$ $$f(x)=x$$ so $$x \in f ^{-1} (V).$$ If $$D(x,y) then $$f(y)=y \in V$$ so $$y \in f ^{-1} (V).$$ Thefore $$f ^{-1} (V)$$ is an open set.

$$\impliedby$$ Let $$x \in f ^{-1} (V)$$ open set. Let $$r<1.$$ $$D(x,y) $$f(x)=x \in V.$$ If $$d(x,y) then $$y \in f ^{-1} (V)$$ so $$f(y)=y \in V.$$ Thus $$V$$ is an open set so $$(U,d)$$ is homeomorphic to $$(U,D).$$

Is it correct to use a radius of less than 1? I did this because $$D(x,y)$$ is less than or equal to 1.

• Sorry to say that this nowhere near a proof. How could you ignore the second term in the definition of $D_n(x,y)$? Nov 23 '20 at 5:31
• Can you elaborate?
– 000
Nov 23 '20 at 6:01