Let $\mathbb{N}^{\mathbb{N}}=\{(x_n)_{n=1}^{\infty}: x_n \in \mathbb{N}\}$ with the Fréchet metric $d(x,y)=\sum_{n=1}^{\infty} 2^{-n}\dfrac{|x_n-y_n|}{1+|x_n-y_n|}$
a) Is the space $\mathbb{N}^{\mathbb{N}}$ complete? If not, find its completion.
b) What is the category of $\mathbb{N}^{\mathbb{N}}$ in $\omega$? Here $\omega$ is the space of all real sequences equipped with $d$.
Here is my attempt:
The space $\mathbb{N}^{\mathbb{N}}$ with the Fréchet metric is complete. That is, for $\forall \varepsilon \enspace k \in \mathbb{N} \enspace \exists N :m,n\ge N$
\begin{align} d(x_{m},x_n)&<{\varepsilon\over2^k(\varepsilon+1)}\\ \sum_{j=1}^\infty{|x_{m,j}-x_{n,j}|\over2^{j}[1+|x_{m,j}-x_{n,j}|]}&<{\varepsilon\over2^k(\varepsilon+1)}\\ {|x_{m,k}-x_{n,k}|\over2^{k}[1+|x_{m,k}-x_{n,k}|]}&<{\varepsilon\over2^k(\varepsilon+1)}\\ |x_{m,k}-x_{n,k}|&<\varepsilon \end{align}
Therefore we conclude that $x_n$ is a Cauchy sequence in $\omega$ and $(\mathbb{N}^{\mathbb{N}},d)$ is a complete metric space.
b) By Baire's theorem, if $(\mathbb{N}^{\mathbb{N}},d)$ is a complete metric space, it implies that if of the $2^{\text{nd}}$ category in itself, hence it is also of $2^{\text{nd}}$ category in $\omega$.
Is my reasoning correct? I am not quite sure about the category part, especially. Thanks.
The space $\omega$ ($\mathbb{R}^\mathbb{N}$) is complete in this Fréchet metric. This is well-known. $\mathbb{N}^\mathbb{N}$ is closed in it (product of closed sets is closed; the Fréchet metric induces the product topology), and hence it is also complete in that metric.
It's also clear that $\mathbb{N}^\mathbb{N}$ is nowhere dense in $\omega=\mathbb{R}^\mathbb{N}$ (as it contains no (product) open set at all, so has empty interior), so is first category in the ambient space, despite being second category in itself (cf. a line in the plane, which is a nowhere dense closed copy of a complete metric space too; it happens a lot).