Prove that $(\mathbb{N}^\mathbb{N},d)$ is complete.

Question

Baire space $\mathbb{N}^\mathbb{N}$ is the set of sequences with natural number as entries. For example, $(n)_{n=1}^\infty \in \mathbb{N}^\mathbb{N}.$

Equip Baire space with a metric $d$ given by $$d(f,g) = \frac{1}{2^{n+1}}$$ where $n$ is the least element of $\mathbb{N}$ such that $f(n)\neq g(n)$ for all $f,g\in\mathbb{N}^\mathbb{N}.$ If $f=g,$ then define $d(f,g)=0.$

Question: Prove that $(\mathbb{N}^\mathbb{N},d)$ is complete.

My attempt: Let $(f_n)_{n=1}^\infty$ be a Cauchy sequence in $\mathbb{N}^\mathbb{N}.$ Fix $\varepsilon>0.$ Then there exists $N\in\mathbb{N}$ such that for all $m>n\geq N,$ we have $$d(f_n,f_m)<\varepsilon.$$ Fix $k\in\mathbb{N}.$ Let $k$ be the smallest natural number such that $$\frac{1}{2^{k+1}}<\varepsilon.$$ I stuck here.

My intention is to construct a limit $f\in\mathbb{N}^\mathbb{N}$ by letting its $k$ entries to be the same as $f_n$ when $n$ is large. However, I fail to express it using mathematics.

Any hint would be appreciated.

Answer 1

Hint: for all $k \in \mathbb{N}$ the sequence $\left< f_n(k) : n \geqslant 1 \right>$ is fundamental in $\mathbb{N}$, so it stabilises eventually (i.e. there is $f(k)$ such that $f_n(k) = f(k)$ for large enough $n$). Show that $f_n \to f$ in $(\mathbb{N}^{\mathbb{N}}, d)$.

Answer 2

Instead of the integer k you are considering look at the largest k for which $\frac {2} {k} > \varepsilon$. Then $f_n$ and $f_m$ have the same first k coordinates. It is now clear how to get the limit of $\{f_n\}$.