Basic questions about $\mathbb{Z}^{\mathbb{N}}$ with the product topology can someone please let me know if the following is correct:
1) Let $\mathbb{Z}$ be the integers endowed with the discrete topology and $\mathbb{N}$ the natural numbers. Is $\mathbb{Z}^{\mathbb{N}}$ a discrete space with the product topoogy?
2) Does $\mathbb{Z}^{\mathbb{N}}$ contain a compact infinite set?
3) Is $\mathbb{Z}^{\mathbb{N}}$ metrizable?
My work:
1) I think this is false, let $A= \{0\} \times \{0\} \times ...$ Suppose $A$ is open in $\mathbb{Z}^{\mathbb{N}}$ then we can find a basic open set $U=\prod_{n \in \mathbb{N}} U_{n}$ such that $(0,0,0,...) \in U \subset A$. By definition of product topology there exists a natural number $J$ such that if $n>J$ then $U_{n} = \mathbb{Z}$. This in turn implies that:
$U_{1} \times U_{2} ...\times U_{J} \times \mathbb{Z} \times \mathbb{Z} ... \subset  \{0\} \times \{0\} \times ...$
which is not true since we can pick $z \in \mathbb{Z} \setminus \{0\}$ then $(0,0,...0,z,z,z...)$ is the LHS while not in the RHS. 
2) Can we simply say, take $\{0,1\}$ endowed with the discrete topolgy then $\{0,1\}$  is compact since it is finite. But then by Tychonoff theorem $\{0,1\}^{\mathbb{N}}$ is compact and clearly infinite.
3) I think this one is true right? $\mathbb{Z}$ is metrizable (e.g discrete metric) and the countable product of metrizable spaces is metrizable.
 A: The space $\mathbb{Z}^{\mathbb{N}}$, which is of course homeomorphic to $\mathbb{N}^{\mathbb{N}}$, is often called the Baire space — not to be confused with spaces satisfying the Baire category theorem, which are also called Baire spaces. Using continued fractions one can show that $\mathbb{N}^{\mathbb{N}}$ is homeomorphic to the subspace of $\mathbb{R}$ consisting of the irrational numbers. If you believe this last fact, all your questions become rather obviuous. Here are some comments on your approach:


*

*Your argument in 1) shows that the sequence $x_{n} = (0,\ldots,0,1,0,\ldots)$ (the $1$ appears at the $n$th entry) converges to $0$. Therefore $\mathbb{Z}^{\mathbb{N}}$ is not discrete, and as it is a topological group, it contains no isolated points at all.

*Yes, you can embed the space $C = \{0,1\}^{\mathbb{N}}$ into $\mathbb{\mathbb{Z}}^{\mathbb{N}}$. As a product of compact spaces, $C$ is compact (alternatively, show that $C$ is homeomorphic to the usual Cantor ternary set). The obvious map $C \to \mathbb{Z^N}$ (coming from the inclusion ${0,1} \to \mathbb{Z}$) is continuous and injective, hence a homeomorphism onto its image, as $C$ is compact and $\mathbb{Z^N}$ is Hausdorff. Clearly $C$ is uncountable.

*It is not hard to check that a countable product $X = \prod X_i$ of metric spaces is metrizable, for instance $\sum 2^{-i} \frac{d_i}{1+d_i}$ furnishes a compatible metric. If each $X_n$ is separable/completely metrizable then so is $X$.
As a last remark, the Baire space is a separable, completely metrizable space (such spaces are usually called Polish spaces). In fact, it is universal in the the following sense: every other Polish space is a quotient space of $\mathbb{N}^{\mathbb{N}}$. This, together with the fact that one can encode a basis of open sets handily using paths in the tree $\omega^{\lt\omega}$ makes the space $\mathbb{N}^{\mathbb{N}}$ an extremely important tool in descriptive set theory. A very nice book on all I'm claiming here and very much more is Kechris's Classical descriptive set theory, which I heartily recommend.
