Complete metric space of sequence of positive integers 
Let $(A,d)$ be the space $\mathbb{N}^{\mathbb{N}}$ of sequences of positive integers where $d((a_i)_i, (b_i)_i)= \frac{1}{n}$ where $n$ is the least coordinate at which $(x_i)_i$ and $(y_i)_i$ disagree (so $a_j = b_j$ for $j<n$).

*

*Show $(A,d)$ is complete metric space.


*Show $A$ is not a countable union of compact sets.

My progress: For (1) I was able to show two properties for $(A,d)$ to be a metric space, but for the inequality one, I don't see how this can be true from the given definition of the distance function $d((a_i)_i, (b_i)_i)$. About the completeness part, I also have the same problems when trying to prove any Cauchy sequence with $d((a^{(m)}_i)_i, (a^{(n)}_i)_i) < \epsilon$ for every $m, n > M$ for some $M>0$.
For (2), I couldn't make much progress.
 A: Since for me $\Bbb N$ includes $0$, which is inconsistent with the defintion of $d$, I’ve written $\Bbb Z^+$ instead.
For the triangle inequality, let $x=\langle x_k:k\in\Bbb Z^+\rangle$ and $y=\langle y_k:k\in\Bbb Z^+\rangle$ be distinct points of $A$, and suppose that $d(x,y)=\frac1n$. Let $z=\langle z_k:k\in\Bbb Z^+\rangle\in A$; we want to show that 
$$d(x,z)+d(z,y)\ge\frac1n\;.\tag{1}$$
If not, then $d(x,z)<\frac1n$ and $d(z,y)<\frac1n$, which means that $x_k=z_k=y_k$ for $k\le n$. But then $d(x,y)\le\frac1{n+1}$, contradicting the original assumption. Thus, at least one of $d(x,z)$ and $d(z,y)$ is $\frac1n$ of more, and $(1)$ holds.
For each $n\in\Bbb Z^+$ let $x^{(n)}=\langle x_k^{(n)}:k\in\Bbb Z^+\rangle\in A$, let $\sigma=\langle x^{(n)}:n\in\Bbb Z^+\rangle$, and suppose that $\sigma$ is a Cauchy sequence in $A$; we want to show that $\sigma$ converges to some point of $A$. Since $\sigma$ is Cauchy, for each $\epsilon>0$ there is an $r_\epsilon\in\Bbb Z^+$ such that $d(x^{(m)},x^{(n)})<\epsilon$ whenever $m,n\ge r_\epsilon$. In particular, for each $s\in\Bbb Z^+$ there is an $r_{1/s}\in\Bbb Z^+$ such that $d(x^{(m)},x^{(n)})<\frac1s$ whenever $m,n\ge r_{1/s}$. But $d(x^{(m)},x^{(n)})<\frac1s$ if and only if $x_k^{(m)}=x_k^{(n)}$ for each $k\le s$, so we’re saying that

for each $s\in\Bbb Z^+$ there is an $r_{1/s}\in\Bbb Z^+$ such that whenever $m,n\ge r_{1/s}$, then $x_k^{(m)}=x_k^{(n)}$ for each $k\le s$.



*

*Show that this implies that for each $k\in\Bbb Z^+$, the sequence $\langle x_k^{(n)}:n\in\Bbb Z^+\rangle$ is eventually constant: there are some $x_k\in\Bbb Z^+$ and some $m_k\in\Bbb Z^+$ such that $x_k^{(n)}=x_k$ for all $n\ge m_k$. 

*Then let $x=\langle x_k:k\in\Bbb Z^+\rangle$, and show that $\sigma$ converges to $x$.
For part (b), suppose that $A=\bigcup_{n\in\Bbb Z^+}K_n$, where each $K_n$ is closed. Because $A$ is a complete metric space, it’s a Baire space, so there must be an $n\in\Bbb Z^+$ such that $K_n$ has non-empty interior. 


*

*Show that there is a finite sequence $\langle a_1,\ldots,a_m\rangle$ of positive integers such that $$B=\{\langle x_k:k\in\Bbb Z^+\rangle\in A:x_k=a_k\text{ for }k\le m\}$$ is a clopen subset of $K_n$.  

*Show that $B$ is not compact. (In thinking about this you may or may not find it helpful to notice that $B$ is homeomorphic to $A$: it’s really just $\{\langle a_1,\ldots,a_m\rangle\}\times A$, where $\langle a_1,\ldots,a_m\rangle\in(\Bbb Z^+)^m$.)  

*Conclude that $K_n$ is not compact.
Remark: It’s hardly obvious, but as a topological space $A$ is homeomorphic to $\Bbb R\setminus\Bbb Q$ with the topology that it inherits from $\Bbb R$, though $d$ is very different from the Euclidean metric (since the latter is not a complete metric on $\Bbb R\setminus\Bbb Q$).
