I was reading the following theorem in Kechris' Classical Descriptive Set Theory:

(6.2) Theorem. Let $X$ be a nonempty perfect Polish space. Then there is an embedding of $\mathcal C$ onto $X$.

Proof. We will define a Cantor scheme $(U_n)_{s\in2^{<\mathbb N}}$ on $X$ so that
i) $U_s$ is open nonempty;
ii) $\operatorname{diam}(U_s)\le2^{-\rm{length}(s)}$;
iii) $\overline{U_{s^-i}}\subseteq U_s$, for $s\in2^{<\mathbb N}$, $i\in\{0,1\}$.

Then for $x\in\mathcal C$, $\bigcap_nU_{x|n}=\bigcap_n\overline{U_{x|n}}$ is a singleton (by the completeness of $X$), say $\{f(x)\}$. Clearly, $f:\mathcal C\to X$ is injective and continuous, and therefore an embedding.

We define $U_s$ by induction on $\rm{length}(s)$. Let $U_\emptyset$ be arbitrary satisfying i), ii) for $s\ne\emptyset$. Given $U_s$, we define $U_{s^-0}$, $U_{s^-1}$ by choosing $x\ne y$ in $U_s$ (which is possible since $X$ is perfect) and letting $U_{s^-0}$, $U_{s^-1}$ be small enough open balls around $x,y$, respectively. $\square$

(6.3) Corollary. If $X$ is a nonempty perfect Polish space, then $\rm{card}(X)=2^{\aleph_0}$. In particular, a nonempty perfect subset of a Polish space has the cardinality of the continuum.

And I'm not really getting the corollary. We've just proved that we can embed the Cantor space (which has the cardinality of the continuum) into any non-empty perfect polish space. But why does this imply that any non-empty perfect polish space has the cardinality of the continuum? I mean, it should have at least the cardinality of the continuum, since the mapping is injective, but why are we saying that it has exactly that cardinality, acting like we've found a bijection between the two spaces? Thanks


Any separable metric space has size at most continuum (the set of sequences from a countable dense set has size continuum, and every point of the space is a limit of at least one such sequence). This does not need completeness.

The theorem you quoted shows it's at least continuum too. So we get the exact size continuum. (Cantor-Bernstein theorem).

  • $\begingroup$ You are absolutly right, thanks $\endgroup$ – Lorenzo Jul 6 at 7:52
  • $\begingroup$ @Lorenzo de nada. $\endgroup$ – Henno Brandsma Jul 6 at 7:53

Other argument is that every separable and metrizable space can be embedding in Hilbert's cube and Hilbert's cube has $2^\omega$ points.

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