Why is $\mathbb{C}_p$ isomorphic to $\mathbb{C}$?

I know that two closed fields of caracteristic $0$ and uncountable are isomorphic iff they have the same cardinality. But I don't know why $\mathbb{C}_p$ has the same cardinality as $\mathbb{C}$. Can anyone give me some reference or hint ?

• By $C$, do you mean $\mathbb{C}$? – Thomas Andrews May 23 '11 at 7:56

Step 1: Any complete metric space without isolated points has cardinality at least $\mathfrak{c}$ (continuum cardinality). I got a nice explanation of this here. In particular $\mathbb{C}_p$ has cardinality at least $\mathfrak{c}$.

Step 2: As wikipedia knows, a topological space which is Hausdorff, first countable and separable (so in particular a separable metric space) has cardinality at most $\mathfrak{c}$: indeed, every point is the limit of a sequence from a countable set, and $\aleph_0^{\aleph_0} = \mathfrak{c}$. Now $\overline{\mathbb{Q}}$ is dense in $\overline{\mathbb{Q}_p}$ (a consequence of Krasner's Lemma: see e.g. $\S 3.5$ of these notes) and hence also in its completion $\mathbb{C}_p$. So $\mathbb{C}_p$ has cardinality at most $\mathfrak{c}$.

[Added: Here is alternate -- less elegant but more elementary -- argument for Step 2:
(i) For any infinite field $K$, the cardinality of $K$ is equal to the cardinality of its algebraic closure.
(ii) For any metric space $X$, the cardinality of the completion of $X$ is at most $\# X^{\aleph_0}$ (the number of sequences with values in $X$). By standard facts on cardinal exponentation, we have $\# X \leq \mathfrak{c} \iff \# X^{\aleph_0} \leq \mathfrak{c}$.]

Thus by the Schroder-Bernstein Theorem, $\mathbb{C}_p$ has cardinality $\mathfrak{c}$.

• You need a separation assumption (Hausdorff, say) in step 2. Of course, we have that in this case. – Chris Eagle May 23 '11 at 8:13
• @Chris: I sure do, thanks. – Pete L. Clark May 23 '11 at 8:15
• Step 1 also has an elementary version : using $p$-adic Hensel expansion, we see that $\mathbb{Z}_p$ is uncountable (which is very similar to what can be done in $\mathbb{R}$). – Joel Cohen May 23 '11 at 14:59
• @Joel: that's definitely another nice way to see Step 1, thanks. (I'm not sure whether it's more or less elementary than what I said; it depends on the person, I suppose.) – Pete L. Clark May 24 '11 at 5:14