I have been given the following problem to solve:
Let $M$ be the set of sequences $(x_n)_{n=0}^{\infty}$ where $x_n \in${$0,1$}. Define $d:M^2 \to \mathbb{R}$ by $d((x_n),(y_n)) = \sum_{n≥0} \frac{|x_n - y_n|}{2^n}$. Note that $d$ is a metric on $M$.
Is $M$ complete?
Intuitively, I feel like it must be complete: if there is a Cauchy sequence of sequences consisting of only $1$s and $0$s then surely the limit sequence must also consist of only $1$s and $0$s, and therefore be in $M$. However, I am finding it difficult to come up with a formal proof of this.
Just write down your idea: Let $(x^k)$ be a Cauchy sequence in $M$. We want to prove it converges, as you suggested, the first point is to find a "limit". Note, that it converges entrywise: For every $n \in \def\N{\mathbf N}\def\abs#1{\left|#1\right|}\N$ we have $$ \abs{x^k_n - x^\ell_n} = 2^n \cdot 2^{-n}\abs{x^k_n - x^\ell_n} \le 2^n \cdot d(x^k, x^\ell) $$ hence, as the latter becomes small for big $k$, $\ell$, each component $(x^k_n)_k$ is a Cauchy sequence in $\{0,1\}$. As $\{0,1\}$ - as a closed subset of the complete space $\mathbf R$ is complete - $(x^k_n)_k$ converges for every $n$. Call the limit $x_n := \lim_k x^k_n$. Now $x := (x_n)$ is an element of $M$. We still have to show that $x$ is the limit of $(x^k)$, that is that $d(x, x^k) \to 0$ for $k \to \infty$. To show that we will use that $\{0,1\}$ is bounded and $\sum 2^{-n}$ converges. Let $\epsilon > 0$. Choose $N \in \N$ such that $\sum_{n=N}^\infty 2^{-n} \le \frac{\epsilon}2$, as for $n < N$ we know that $(x^k_n)_k$ converges and we are looking at only finitely many sequences, there is an $K \in \mathbf N$, such that $$ \abs{x^k_n - x_n} < 2^n \cdot \frac{\epsilon}{2N}, \quad \text{all $k \ge K$, $n < N$} $$ For $k \ge K$ we have \begin{align*} d(x^k, x) &= \sum_n 2^{-n} \abs{x^k_n - x_n}\\ &\le \sum_{n < N} 2^{-n} \abs{x^k_n - x_n} + \sum_{n \ge N} 2^{-n}\\ &< \sum_{n < N} 2^{-n} \cdot 2^n \frac{\epsilon}{2N} + \frac \epsilon 2\\ &\le \epsilon. \end{align*} Hence, $x^k \to x$ in $M$.