Infinite series of metric spaces is metric space

cauchy-sequencesconvergence-divergencegeneral-topologymetric-spacesreal-analysis

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.

Best Answer

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

$$\epsilon=\frac{\epsilon}{2^n}+\sum_{k=1}^{n-1}\frac{d_k(x_n,x_m)}{2^k}+\sum_{k=n+1}^\infty\frac{d_k(x_n,x_m)}{2^k}\,.$$

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 $D$.

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