# 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.

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1) no, it is homeomorphic to the set of irrational numbers. 2) yes, it contains $\{0,1\}^{\mathbb{N}}$, as you say 3) yes, a countable product of metrizable spaces is metrizable. –  t.b. Jul 6 '11 at 15:19
@Theo Buehler: without using the homeomorphism you mention, how would you prove it is not discrete? –  user10 Jul 6 '11 at 15:21
Your argument is fine. Note that $(0,\cdots,0,1,0,\cdots)$ ($1$ at entry $n$) converges to zero as $n \to \infty$ by what you say. Alternatively, prove 3) by writing down an explicit metric and check that this sequence converges to zero. –  t.b. Jul 6 '11 at 15:24
@user10: just exhibit a nontrivial convergent sequence. –  Qiaochu Yuan Jul 6 '11 at 15:24

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:

1. 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.

2. 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.

3. 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.

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There is also Moschovakis' book about descriptive set theory, which is highly recommended by many people as well. –  Asaf Karagila Jul 6 '11 at 15:49
@Asaf: Yes, true. Moschovakis is a very fine book, too; but I never really worked with it, so I can't recommend it with clear conscience. I find Kechris very interesting because of all its interesting applications to other parts of mathematics. –  t.b. Jul 6 '11 at 15:51
Just bringing it up, I have the both open (I have an exam on the subject in a few days) and my teacher praised Moschovakis' book often (although he worked with both, and on some portion he came up with his own proofs and ideas). –  Asaf Karagila Jul 6 '11 at 15:59
@Asaf: Well, actually I think Moschovakis is a person, not a book, but you got my meaning in my previous comment :) Good luck at your exam! (I'm glad that I have all my exams way behind me). As a person working a lot with functional analysis, Kechris is especially appealing because it contains many things that are extremely hard to locate elsewhere (in a language I can understand without much effort, that is). –  t.b. Jul 6 '11 at 16:04
@Theo Buehler: thanks a lot! –  user10 Jul 6 '11 at 16:05