Cauchy sequences & topology on space of sequences

Question

Let us consider a metric space $(X, d)$, the space $X^\mathbb{N}$ of sequences of elements in $X$ and the metric $$ D : \begin{cases} X^\mathbb{N} \times X^\mathbb{N} \to \mathbb{R}_+ \\ (x, y) \mapsto \sum\limits_{n \in \mathbb{N}}{\min\left\{a_n, d\left(x\left(n\right), y\left(n\right)\right)\right\}} \end{cases} $$ where $\left(a_n\right)_{n \in \mathbb{N}}$ is a sequence of positive numbers such that $\sum\limits_{n \in \mathbb{N}}{a_n} < + \infty$.

I am having a hard time on two questions :

1. Prove that the sequence $\left(x_p\right)_{p \in \mathbb{N}} \subset X^\mathbb{N}$ is a Cauchy sequence if and only if $\forall n \in \mathbb{N}, \left(x_p \left(n\right)\right)_{p \in \mathbb{N}} \subset X$ is a Cauchy sequence.

2. Prove that $U$ is an open subset of $\left(X^\mathbb{N}, D\right)$ if and only if for every $x \in U$, there exists a finite subset $J$ of $\mathbb{N}$ and a positive number $\alpha$ such that if $y \in X^\mathbb{N}$ satisfies $\forall j \in J, (d\left(x\left(j\right), y\left(j\right)\right) < \alpha$, then $y \in U$.

My attempts so far

First question

$(\Rightarrow)$ If $\forall n \in \mathbb{N}$, $\left(x_p \left(n\right)\right)_{p \in \mathbb{N}}$ is a Cauchy sequence, then $\forall n \in \mathbb{N}, \forall \varepsilon > 0, \exists n_0 \in \mathbb{N}, \forall l \geq n_0, \forall m \geq n_0, d \left(x_l(n), x_m(n)\right) < \varepsilon$. Because of the assumption of finite sum of the $a_n$'s we have : $$ D(x_l, x_m) = \sum\limits_{n \in \mathbb{N}}{\min\left\{a_n, d\left(x_l\left(n\right), x_m\left(n\right)\right)\right\}} < + \infty $$ $$ \Rightarrow \forall \varepsilon > 0, \exists N_0 \in \mathbb{N}, \forall M \geq N_0, \forall N \geq N_0, D \left(x_M, x_N\right) < \varepsilon $$ so $\left(x_p\right)_{p \in \mathbb{N}}$ is Cauchy in $\left(X^\mathbb{N}, D\right)$.

Is it correct to use the finiteness of the infinite sum to justify that we can shrink it?

$(\Leftarrow)$ If $\left(x_p\right)_{p \in \mathbb{N}}$ is Cauchy in $\left(X^\mathbb{N}, D\right)$, then $\forall \varepsilon > 0, \exists N_0 \in \mathbb{N}, \forall M \geq n_0, \forall N \geq n_0, D \left(x_M, x_N\right) = \sum\limits_{n \in \mathbb{N}}{\min\left\{a_n, d\left(x_M\left(n\right), x_N\left(n\right)\right)\right\}} < \varepsilon$. Because the $a_n$'s are strictly positive, there exists $E > 0$ such that $\forall \varepsilon \in \left(0, E\right], \sum\limits_{n \in \mathbb{N}}{d\left(x_M\left(n\right), x_N\left(n\right)\right)} < \varepsilon \Rightarrow \forall n \in \mathbb{N}, d\left(x_M(n), x_N(n)\right) < \varepsilon$, thus $\forall n \in \mathbb{N}$, $\left(x_p(n)\right)_{p \in \mathbb{N}}$ is Cauchy. Is the reasoning rigorous?

My issue is that because of the $\min$ condition, $d$ can arbitrarily explode with $D$ remaining finite: how do we ensure that when $D$ is small, $d$ is small too? I would think that the positiveness of the $a_n$'s ensures that the infinite sum is growing (without exploding), unless the $d(x\left(n\right), y\left(n\right))$ are small enough, but I do not know how to write this.

Second question

$(\Leftarrow)$ Let $U$ be an open set of $X^\mathbb{N} : \forall x \in U, \exists \eta > 0, B_D(x, \eta) \subset U$, where $B_D(a, b)$ is the open ball with center $a$ and radius $b$ for metric $D$. Take $x \in U$. Let $\varepsilon > 0$ the radius of an open ball $B_D(x, \varepsilon) \subset U$ and take $N \in \mathbb{N}$ such that $\sum\limits_{k = N + 1}^{+ \infty}{a_k} < \frac{\varepsilon}{2}$, $J = {1, ..., N}$ and $\alpha = \frac{\varepsilon}{2N}$. Then $$ \forall y \in X^\mathbb{N}, \forall j \in J, \left(d\left(x\left(j\right), y\left(j\right)\right) < \alpha \Rightarrow D(x, y) < \varepsilon \Rightarrow y \in U \right) $$

$(\Rightarrow)$ Let us assume that there exists $J \subset \mathbb{N}$ finite with cardinality $N > 0$ and $\alpha > 0$ that satisfies the condition and take $x \in U$. If we pick some $\varepsilon < \alpha$, what guarantee do we have that $D(x, y) < \varepsilon \Rightarrow \forall j \in J, d\left(x\left(j\right), y\left(j\right)\right) < \alpha$ and thus $B_D(x, \varepsilon) \subset U$, proving that $U$ is an open set?

Answer 1

In 1, $\implies$ you don't take advantage of the sum and I don't understand your statement, it's just a mess of quantifiers.

If all sequences $(x_p(n))_p$ are Cauchy, let $\varepsilon >0$.

First pick $N_1$ such that $\sum_{n=N_1}^\infty a_n < \frac{\varepsilon}{2}$. This can be done by the convergence of the sum of $a_n$. It's then clear that in the formula for the distance $D$, a tail of the sum defining it is bounded by $\frac{\varepsilon}{2}$ already, whatever the points are.

So we only have to consider the sequences $(x_p(n))_p$ for $n < N_1$, so finitely many of them. For each $i < N_1$ we can find some $M_i$ such that when $p,q \ge M_i$ we have $d(x(i)_p, x(i)_q) < \frac{\varepsilon}{2N_1}$. So when we set $M=\max_i\{M_i: i < N_1\}$ we have that we can use all these finitely many estimates and that for all $p,q \ge M$ we have

$$\sum_{i< N_1} d(x(i)_p, x(i)_q) < N_1 \cdot \frac{\varepsilon}{2N_1} = \frac{\varepsilon}{2}$$

so that (with the tail estimate ) from earlier we know that $D(x_p, x_q) < \varepsilon$ for $p,q \ge M$ as required.

The reverse:Let $(x_p)_p)$ be $D$-Cauchy, and fix any $n \in \Bbb N$ and let $\varepsilon>0$. There is an $N_1$ such that $p,q \ge N_1$ implies $D(x_p,x_q) < a_n$. So then $\min(d_n(x(n)_p, x(n)_q), a_n) = d_n(x(n)_p, x(n)_q)$. But then another application of $D$-Cauchyness gives $N_2$ such that $D(x_p, x_q)< \varepsilon$ for $p,q \ge N_2$. But then for $p,q \ge N:= \max(N_1,N_2)$ we have:

$$d_n(x(n)_p, x(n)_q) = \min(d_n(x(n)_p, x(n)_q), a_n) \le D(x_p, x_q) < \varepsilon$$

which shows that $(x(n)_p)_p$ is $d$-Cauchy. And as $n$ was arbitrary, we're done.

The ideas for 2. are very similar to 1. and so hopefully you can figure it out from the proof I gave above.