2

Every separable metric space $E$ has cardinality less or equal than the cardinality of continuum.

As $E$ is Hausdorff; $\{x\}$ is closed, so $E-\{x\}$ is open in $E$; as $E$ is second-countable (let $(B_n)_{n \in \mathbb{N}}$ be the countable basis) , then $E-\{x \}= \bigcup_{j \in J_x \subset \mathbb{N}}B_j$

Consider the mapping:

\begin{align*} \varphi:\ &E \mapsto \mathcal{P}(\mathbb{N})\\ &x \longrightarrow J_x \end{align*}

Then, if $\varphi(x)=\varphi(y) \implies J_x=J_y \implies \{x\}=\{y\}$; so, $\varphi$ is one-to-one and $|E| \leq 2^{|\mathbb{N}|}$.

Is this a valid argument? Should I use the separability as well?

J P
  • 1,152
  • 2
    https://math.stackexchange.com/questions/723948/how-big-can-a-separable-hausdorff-space-be#:~:text=The%20maximum%20possible%20cardinality%20of,is%2022%E2%84%B50. – Evangelopoulos Foivos Oct 19 '24 at 19:22
  • 2
    You use separability when you say it's 2nd countable. – spaceisdarkgreen Oct 19 '24 at 19:24
  • 3
    Simpler (also doesn't require knowing separability is same as 2nd countability for metric spaces) is to note that, for a given countable dense subset, there are at most continuum many sequences, all of whose terms belong to that subset, because ${\aleph_0}^{\aleph_0} = \mathfrak c.$ Incidentally, if the countable dense subset is actually finite, but at least $2,$ then still get $\mathfrak c.$ Now note that there cannot be MORE (in the sense of cardinality) elements in the metric space than there are sequences of this type. – Dave L. Renfro Oct 19 '24 at 19:38
  • 2
    Yes, your proof is correct. In fact, you have shown, that every second-countable $T_1$ space has cardinality less than or equal to continuum size. This implies the mentioned statement, since separable metric spaces are second-countable, but covers many more spaces. – Ulli Oct 20 '24 at 11:42

0 Answers0