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 $$D_{n} =d(x,y)+\lvert \frac{1}{d(x,U_n^c) } - \frac{1}{d(y,U_n^c)} \rvert. $$ Define $$D(x,y)=\sum_{n=1}^{\infty } \frac{1}{2^n} d_{j} (x,y). $$

  1. I have a question about showing that it's a metric under the condition that $D(x,y)=0 \iff x=y.$ I'm trying to prove it in the $\implies $ direction. Since $\frac{1}{2^n} $ is a geometric series and $d_n(x,y)$ is less than or equal to 1, we know that the series converges. This implies that at some point, $\frac{1}{2^{n} } d_n(x,y)$ goes to 0, but this is only possible if $d_n(x,y)=0,$ which means that $x=y.$ Is this correct?

  2. I also want to show that if $\{ x_{n}\} $ is a Cauchy sequence in $(U,D),$ then it is a Cauchy sequence in each $(U_n,D_n).$ My attempt is as follows. Since $\{ x_{n}\} $ is Cauchy in $U,$ we have for any $\epsilon >0,\exists N,\forall n,m \geq N,$ $$D(x_n,x_{m} )=\sum_{n=1}^{\infty } \frac{d_n(x_n,x_m)}{2^n}< \epsilon. $$ We can rewrite this as $$\sum_{k=1}^{n-1 } \frac{d_k(x_n,x_m)}{2^k}+\frac{d_n(x_n,x_m)}{2^n}+\sum_{k=n+1}^{\infty } \frac{d_k(x_n,x_m)}{2^k}< \epsilon $$ $$\implies d_n(x_n,x_m)<(\epsilon-\sum_{k=1}^{n-1 } \frac{d_k(x_n,x_m)}{2^k} -\sum_{k=n+1}^{\infty } \frac{d_k(x_n,x_m)}{2^k})2^n$$ Set $$\epsilon = \frac{\epsilon}{2^n}+ \sum_{k=1}^{n-1 } \frac{d_k(x_n,x_m)}{2^k} +\sum_{k=n+1}^{\infty } \frac{d_k(x_n,x_m)}{2^k}>0. $$ After substituting and canceling terms we get $d_n(x_n,x_{m} )< \epsilon .$ Since $\epsilon $ is arbitrary, $D_n(x_n,x_m)<\epsilon .$ Thus $\{ x_{n}\} $ is Cauchy in each $(U_n,D_n).$ Is this argument correct?

  3. I want to show that $(U,D)$ is complete. Suppose $\{ x_{n}\} $ is a Cauchy sequence in $(U,D)$ that converges to $L.$ If $L \in U^c,$ then $$ \forall \epsilon >0 \lim_{n \to \infty} d(x_n,U^c) \leq \lim_{n \to \infty} d(x_n,L) < \epsilon.$$ By the contrapositive, we have if $$ \exists \epsilon >0 \lim_{n \to \infty} d(x_n,U^c) > \lim_{n \to \infty} d(x_n,L) \geq \epsilon ,$$ then $L \in U.$ So we want to show that there is an $\epsilon >0$ such that $$\lim_{n \to \infty} \frac{1}{d(x_n,U^c)} \leq \frac{1}{\epsilon } .$$ Since $$\frac{1}{d(x_n,U^c)} \to \frac{1}{d(L,U^c)},$$ we have that $\frac{1}{d(x_n,U^c)}$ is bounded. I'm not sure how to incorporate any other details about $D(x,y)$ other than the $d(x_n,U^c)$ term.


1 Answer 1


Your reasoning in $(1)$ isn’t at all clear. I would simply prove the contrapositive. If $x,y\in U$ and $x\ne y$, then $d(x,y)>0$, so $D_n(x,y)\ge d(x,y)>0$, and therefore $d_n(x,y)>0$. Clearly this implies that $D(x,y)>0$.

In $(2)$ you can’t set


First, it makes no sense: it’s like saying ‘let $x=x+1$’. You need to use a different name for the new value. Next, you’re trying to show that $d_k(x_n,x_m)$ is less than the original $\epsilon$, not some new $\epsilon$. Finally, this sentence makes no sense:

Since $\epsilon$ is arbitrary, $D_n(x_n,x_m)<\epsilon$.

Unless $D_n(x_n,x_m)=0$, it is not less than $\epsilon$ for arbitrary $\epsilon$. And you probably mean $D_k$, not $D_n$. You appear to have made a similar error in the displayed line

$$D(x_n,x_{m} )=\sum_{n=1}^{\infty } \frac{d_n(x_n,x_m)}{2^n}< \epsilon\,:$$

the index of summation has to be different from the subscripts on $x$.

What you need to do here is let $\epsilon>0$ and $k\in\Bbb Z^+$. Since $\langle x_n:n\in\Bbb Z^+\rangle$ is $D$-Cauchy, there is an $n_0\in\Bbb Z^+$ such that $D(x_n,x_m)<\frac{\epsilon}{2^k}$ whenever $n,m\ge n_0$. But then for all $n,m\ge n_0$ we have

$$\frac{d_k(x_n,x_m)}{2^k}\le D(x_n,x_m)<\frac{\epsilon}{2^k}$$

and therefore $d_k(x_n,x_m)<\epsilon$. Thus, $\langle x_n:n\in\Bbb Z^+\rangle$ is $d_k$-Cauchy and therefore $D_k$-Cauchy. (You should have no trouble proving that a sequence is $d_k$-Cauchy if and only if it is $D_k$-Cauchy.)

The that you start $(3)$ makes no sense. To show that $\langle U,D\rangle$ is complete, you need to start with an arbitrary $D$-Cauchy sequence in $U$ and show that it converges; you cannot simply assume that it converges.

HINT: The sequence $\langle x_n:n\in\Bbb Z^+\rangle$ is $d_k$-Cauchy for each $k\in\Bbb Z^+$, and each $\langle U_k,d_k\rangle$ is complete, so for each $k\in\Bbb Z^+$ there is a $y_k\in U_k$ such that $\langle x_n:n\in\Bbb Z^+\rangle$ converges to $y_k$ in $\langle U_k,d_k\rangle$. Show that there is a $y\in U$ such that $y_k=y$ for each $k\in\Bbb Z^+$, and show that $\langle x_n:n\in\Bbb Z^+\rangle$ converges to $y$ in the metric $D$.

  • $\begingroup$ Suppose $\{ x_{n}\} $ is a D-Cauchy sequence in $U.$ Since $(X,d) $ is complete, $\{ x_{n}\}$ has a limit $L \in (X,d).$ Since $(U_k,d_k)$ is complete $\forall k,$ there are $L_k \in U_k$ that $\{ x_{n}\} $ converges to in $(U_{k},d_k ), \forall k.$ We proved in math.stackexchange.com/questions/3918418/… that if $\{ x_{n}\} $ is a Cauchy sequence in $(U_k,d_k) ,$ then $L_k=L,$ and since $L_k=L \in U_k,\forall k,$ $L \in U$ so $U$ is complete. How is this? Also can you check the link in this comment? I use a result from there for this proof. $\endgroup$
    – 000
    Nov 23, 2020 at 2:51
  • 1
    $\begingroup$ @000: You really should say a bit more to justify the assertion that $L_k=L$ for all $k$; this is because all of these metrics induce the same topology (or relative topology, in the case of the metrics $d_k$), and a sequence in a Hausdorff space (and hence in any metric space) can converge to at most one point; otherwise it’s okay. I have the earlier question set aside to look at, but I’ve not managed to get to it yet. $\endgroup$ Nov 23, 2020 at 2:58
  • $\begingroup$ Ok thank you. I also have a question about show that (U,D) and (U,d) are homeomorphisms via the identity. I did this by showing they have the same open sets, however is it correct to set the radius of the distance function to less than one? This question is here math.stackexchange.com/questions/3918982/… $\endgroup$
    – 000
    Nov 23, 2020 at 3:27
  • $\begingroup$ @000: You’re welcome. I’ll take a look, though I may not get to it until tomorrow. $\endgroup$ Nov 23, 2020 at 3:48

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .