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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 $d(x,y)<r$, then $y \in V.$ $f(x)=x$ so $x \in f ^{-1} (V).$ If $D(x,y)<r,$ 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 $r<1.$ $D(x,y)<r \implies y \in f ^{-1} (V).$ $f(x)=x \in V.$ If $d(x,y)<r,$ 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.

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

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