Proving completeness of a metric space Let $\mathbb{R}^{\infty}$ be the infinite Cartesian product of $\mathbb{R}$ with itself. Define the metric $d(x,y) = \sum_{t \in \mathbb{N} \cup \{0\}} \beta^{-t} \rho(x_t, y_t)$ where $\rho(x_t, y_t) = \min\{|x_t - y_t|, 1\}$ for some fixed $\beta>1$ and for all $x, y \in \mathbb{R}^{\infty}$ where $x = (x_0, x_1, \cdots)$ and $y=(y_0, y_1, \cdots)$. Prove that $(\mathbb{R}^{\infty}, d)$ is a complete metric space.
My attempt: I know that I need to show that every Cauchy sequence $\{x_n\} \subseteq \mathbb{R}^{\infty}$ converges to a limit $x \in \mathbb{R}^{\infty}$. So let $\{x_n\}$ be a Cauchy sequence contained in $\mathbb{R}^{\infty}$. Then, $\forall \varepsilon>0$ $\exists N$ such that $n,m > N$ implies $d(x_n, x_m) < \varepsilon$. 
Now $d(x_n, x_m) = \sum_{t \in \mathbb{N} \cup \{0\}}  \beta^{-t} \rho(x_{n, t}, x_{m, t})$. I need to somehow show that $\{x_n\}$ is actually converging and converges to a value in $\mathbb{R}^{\infty}$. How can I finish the proof?
 A: You should describe a sequence $x$, and then show that $x_n \to x$.
A reasonable first guess is to define $x$ by setting $x_t = \lim_{n \to \infty} x_{n,t}$. It's not too hard to show that these $x_t$'s all exist. (Brief sketch: Let $\epsilon > 0$. For large enough $n$ and $m$, $d(x_n,x_m) < \epsilon \beta^{-t}$. We have $\beta^{-t} \rho(x_{n,t},x_{m,t}) \leq d(x_n,x_m)$. So, for large enough $n$ and $m$, $\rho(x_{n,t},x_{m,t}) < \epsilon$. This shows $(x_{n,t})_n$ is Cauchy.)
Next, showing $x_n \to x$, i.e., $d(x_n,x) \to 0$, is a little more involved but should be also possible. For $\epsilon > 0$, let $T$ be large enough so that the tail of the geometric sum $\beta^{-T} + \beta^{-T-1} + \dotsb < \epsilon$. Let $B = \beta^{-0}+\beta^{-1}+\dotsb+\beta^{-T+1}$. Let $N$ be large enough so that for $n \geq N$, $\rho(x_{n,t},x_t) < \epsilon/B$, for each $t=0,1,\dotsc,T-1$. (Why is this possible to say?) Now what is $d(x_n,x)$ for $n \geq N$? I'll leave the rest (and filling in details) to you.
A: Here is my attempt after reading through the answers.
The claim is that $(\mathbb{R}^{\infty}, d)$ is a complete metric space. Let $\{x_n\}$ be a Cauchy sequence contained in $\mathbb{R}^{\infty}$. We want to show that $\{x_n\}$ converges to $x \in \mathbb{R}^{\infty}$. First, we must find a candidate for $x$. Denote $x_n = (x_{n, 0}, x_{n, 1}, \cdots )$ and $x = (x_0, x_1, \cdots)$. We claim that $\lim_{n \rightarrow \infty} x_{n, t} = x_t \in \mathbb{R}$ for $t = 0, 1, \cdots$. Fix $t$, we will show that the sequence of real numbers $(x_{n, t})_n$ is Cauchy. Since $\{x_n\}$ is Cauchy, we can pick large enough $n$ and $m$ such that $d(x_n, x_m) < \varepsilon \beta^{-t}$. For a fixed $t$, we have $\beta^{-t} \rho(x_{n, t}, x_{m, t}) \le d(x_n, x_m) < \varepsilon \beta^{-t}$ which implies that $\rho(x_{n, t}, x_{m, t}) < \varepsilon$. Thus, for large enough $n$ and $m$, we can make $|x_{n, t} - x_{m, t}|$ arbitrarily small, hence $(x_{n, t})_n$ is Cauchy. Since $(x_{n, t})_n$ is just a sequence of real numbers, by the completeness of $\mathbb{R}$, we know that $(x_{n, t})_n$ converges to some value $x_t \in \mathbb{R}$. 
Now we show that $\{x_n\}$ converges to $x$. Let $\varepsilon >0$. Since $(x_{n, t})_n$ converges to $x_t$, we know that there exists an $N_t$ such that for $n \ge N_t$, we can make $|x_{n, t} - x_t|$ arbitrarily small for $t = 0, 1, \cdots$. Note that
\begin{align*}
d(x_n, x) & = \sum_{t \in \mathbb{Z}^+} \beta^{-t} \rho(x_{n, t}, x_t) \\
& = \beta^{-0} \rho(x_{n,0}, x_0) + \beta^{-1}\rho(x_{n,1}, x_1) + \cdots + \beta^{-T} \rho(x_{n,T}, x_T) \\
& + \beta^{-(T+1)} \rho(x_{n,T+1}, x_{T+1}) + \beta^{-(T+2)} \rho(x_{n,T+2}, x_{T+2})  + \cdots \\
& < \beta^{-0} \rho(x_{n,0}, x_0) + \beta^{-1}\rho(x_{n,1}, x_1) + \cdots + \beta^{-T} \rho(x_{n,T}, x_T) \\
& + \beta^{-(T+1)} + \beta^{-(T+2)} + \cdots
\end{align*}
We can make $\beta^{-(T+1)} + \beta^{-(T+2)} + \cdots$ arbitrarily small since it is the tail of a geometric series. Thus, let $B = \beta^{-0} + \beta^{-1} + \cdots + \beta^{-T}$ and set $N = \max\{N_1, N_2, \cdots, N_{T}\}$. Thus, for $n \ge N$, we have $\rho(x_{n, t}, x_t) <  \varepsilon/B$. So, for $n \ge N$, we have $d(x_n, x) = \sum_{t \in \mathbb{Z}^+} \beta^{-t} \rho(x_{n, t}, x_t) <  B \cdot \varepsilon/B = \varepsilon$. Therefore, since $\{x_n\}$ converges to $x \in \mathbb{R}^{\infty}$, the metric space $(\mathbb{R}^{\infty}, d)$ is complete. 
A: First show each $(x_{n,t})_n$ is Cauchy. This is pretty clear from the definition of the metric. Since $\mathbb{R}$ is complete, each $x_{n,t}$ will converge as $n \to \infty$ to some $x_t$. Let $x = (x_1,x_2,\dots)$ be the formed sequence. You must show $x_n \to x$ in the metric given. But this pretty much follows from the definition of the metric as well. Dominated convergence theorem is helpful.
A: You should first find the point where $\{x_n\}$ should converge, which you might be able to guess is the point $x$ so that $\pi_i(x) = \lim_{n\rightarrow \infty} \pi_i(x_n)$ (where $\pi_i$ is the projection onto the $i^{th}$ coordinate). You need to show that this point exists, and that the $x_n$ actually converge to this point.
