I want to prove that every countable metrizable spaces can be embedded in reals, using that the irrational numbers $\mathbb{P}$ are homeomorphic to the Baire space $\omega^\omega$. Let $(X, d)$ be a countable metric space. Fix an enumeration $(x_n)$ of $X$. Since $A := \{d(x, y) \mid x, y \in X\} \cup \mathbb{Q}$ and $\mathbb{Q} + \pi \subset \mathbb{P}$ are unbounded countable dense linearly ordered sets, we can find an order isomorphism $f\colon A \to \mathbb{Q} + \pi$ which can be extended to a unique homeomorphism $F\colon \mathbb{R} \to \mathbb{R}$. Let $g_n\colon X \to \mathbb{I}$ be defined as $g_n(x) = F(d(x, x_n))$. $\{g_n\}$ separates points from closed sets and $X$ is $T_0$, so $g\colon X \to \mathbb{P}^\omega$ defined as $[g(x)](n) = g_n(x)$ is a homeomorphic embedding. Since $\mathbb{P}^\omega \sim (\omega^\omega)^\omega \sim \omega^\omega \sim \mathbb{P} \subset \mathbb{R}$, we are done. Is this right?
Every countable metrizable spaces can be embedded in reals
general-topologymetric-spacessolution-verification
Related Solutions
I believe that you are looking for ideas from the Cantor Bendixson theorem.
The main idea of the proof is the Cantor-Bendixson derivative. Given a closed set $X$, the derived set $X'$ consists of all limit points of $X$. That is, one simply throws out the isolated points. Continuing in a transfinite sequence, one constructs $X_\alpha$ as follows:
- $X_0=X$, the original set.
- $X_{\alpha+1}=(X_\alpha)'$, the set of limit points of $X_\alpha$.
- $X_\lambda=\bigcap_{\alpha\lt\lambda}X_\alpha$, for limit ordinals $\lambda$.
Thus, $X_1$ consists of the limit points of $X$, and $X_2$ consists of the limits-of-limits, and so on. The set $X_\omega$ consists of points that are $n$-fold limits for any particular finite $n$, and $X_{\omega+1}$ consists of limits of those kind of points, and so on. The process continues transfinitely until a set is reached which has no isolated points; that is, until a perfect set is reached. The Cantor Bendixon rank of a set is the smallest ordinal $\alpha$ such that $X_\alpha$ is perfect.
The concept is quite interesting historically, since Cantor had undertaken this derivative before he developed his set theory and the ordinal concept. Arguably, it is this derivative concept that led Cantor to his transfinite ordinal concept.
It is easy to see that the ordinal $\omega^\alpha+1$ under the order topology has rank $\alpha+1$, and one can use this to prove a version of your desired theorem.
The crucial ingredients you need are the Cantor Bendixson rank of your space and the number of elements in the last nonempty derived set. From this, you can constuct the ordinal $(\omega^\alpha+1)\cdot n$ to which your space is homeomorphic. Meanwhile, every countable ordinal is homeomorphic to a subspace of $\mathbb{Q}$, and is metrizable. The compact ordinals are precisely the successor ordinals (plus 0).
Update 5/11/2011. This brief article by Cedric Milliet contains a proof of the Mazurkiewicz-Sierpiński theorem (see Stefan Mazurkiewicz and Wacław Sierpiński, Contribution à la topologie des ensembles dénombrables, Fundamenta Mathematicae 1, 17–27, 1920), as follows:
Theorem 4. Every countable compact Hausdorff space is homeomorphic to some well-ordered set with the order topology.
The article proves more generally that any two countable locally compact Hausdorff spaces $X$ and $Y$ of same Cantor-Bendixson rank and degree are homeomorphic. This is proved by transfinite induction on the rank, and the proof is given on page 4 of the linked article.
$F$ injective is quite obvious as $x \neq y$ implies that there is some basic element $B_n$that contains $x$ and misses $y$ (we include $T_1$ in regularity for this theorem). By regularity we have some $x \in \overline{B_m} \subseteq B_n$ and then when $k$ is the index corresponding to this $(m,n)$ pair, $f_k(x) = 1$ and $f_k(y) = 0$, ensuring that $F(x) \neq F(y)$, because the images differ at least in one coordinate.
$F$ is continuous as $\pi_k \circ F = f_k$ is continuous for all $k$, by the universal property the product topology.
In the $F$ being open proof, the step where we chose the $f_N$ is where the "magic" happens later; we have $x_0 \in U$ with $F(x_0) = z_0$. Then we find basic elements $(n,m)$ such that $x_0 \in \overline{B_n} \subseteq B_m \subseteq U$. This is possible because the $B_n$ form a base for $X$ and $X$ is regular (like in the 1-1-ness proof), these fact guarantee us the existince of enough pairs and thus enough functions defined for these pairs: we have some index $N$ coresponding to this $(n,m)$ pair (this is the re-indexing mentioned in the beginning) and we know that $f_N = g_{n,m}$ which was chosen in the beginning to obey $g_{n,m}[\overline{B_n}] = \{1\}$ and $g_{n,m}[X \setminus B_n] = \{0\}$, so as $x_0 \in B_n$ in particular $1=g_{n,m}(x_0) = f_N(x_0)$, and so $\pi_N(F(x))= f_N(x) = 1>0$ so that later in the proof we can conclude that $z_0 = F(x_0) \in \pi_N^{-1}[(0,\infty)]=V$, and so indeed $z_0 \in V \cap Z$
This is just by "construction" (how the $f_N = g_{n,m}$ were chosen) and the definitions of inverse images and projections and $F$ itself.
The proof (Munkres' I believe) is fine, really, but could do with a bit more explicitness for people not used to arguments like this with the product map. In my class at the time we first proved a more general embedding theorem for products and deduced the Urysohn metrisation theorem from that; all we need is that a second countable regular $T_1$ space is normal, and this implies (with the countable base again) that $X$ has a countable family of continuous $[0,1]$-valued functions that separates points, and points and closed sets. The embeddability of $X$ into $[0,1]^\omega$ follows then by the general theorem.
In the posted proof, these ideas are intermixed. IMHO it might be better to separate the embedding idea in general from the specific details for a regular second countable space. I've always found it clarifying (and the embedding theorem I mentioned also has other applications, so it makes sense to specify it separately. Munkres does that later in an exercise.
Best Answer
A countable metric space embeds into $2^\omega=\{0,1\}^\omega$ via a countable clopen base (use the characteristic functions of the countable clopen base). And $2^\omega$ trivially embeds into $\Bbb P \simeq \omega^\omega$ and $\Bbb R$ (middle third set).