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Suppose $(X,d)$ is a complete metric space and $U$ is a countable collection of open subsets of $X$. Show that $\bigcap U$ can be endowed with a metric topologically equivalent to $d$ with respect to which $\bigcap U$ is complete.

Attempt: Some Background:

(i) Theorem: Open subsets of a complete metric space can be made into complete metric spaces by judiciously altering the metric. Let $U$ be an open set in $X$. Define $f(x) = \dfrac {1}{dist_d(x,U^c)}$ where $dist$ refers to the distance function i.e. $\inf \{d(x,w)~|~w \in U^c \}$. Then, the metric $e:(a,b) \rightarrow d(a,b)+|f(a)-f(b)|$ converts $U$ into a complete space. Also,$d,e$ are topologically equivalent

EDIT: I found a solution below:enter image description here

I couldn't understand why $e(a,b) \le d(a,b) + |f_,m(a) - f_m(b)|$. Is it an error?

Also, is there really a need for the summation? Thanks

MathMan
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1 Answers1

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You could considere the metric on $U$ :

$$\rho(a,b)=\sum\limits_{n=1}^\infty \frac{1\wedge e_n(a,b)}{2^n}$$

It is known as Alexandroff-Mazurkiewicz theorem : Proof of Mazurkiewicz theorem

  • I have edited and added a solution image. Could you please have a look and suggest me if the solution is correct? Thanks! – MathMan Apr 13 '18 at 18:09
  • and thank you for the link! – MathMan Apr 13 '18 at 18:29
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    I don't know where does the inequality $e\leq d+|f_m(a)-f_m(b)|$ come from. But you can see that $d$ and $e$ are equivalent metrics showing they have the same convergent sequences. –  Apr 13 '18 at 19:50
  • yep got it. I have one more query though. Do you think we could have done away with the summation sign.? What specific use did it serve? – MathMan Apr 14 '18 at 05:50