Let $X$ be separable and normal. Suppose $A\subset X$ is discrete in the relative topology. I'd like to show that $A$ is countable. By separability, it suffices to exhibit a mutually disjoint collection of non-empty open sets indexed by $A$.
The fact that $A$ is discrete means that, for each $a\in A$, there exists an open set $U_a\subset X$ such that $U_a\cap A=\{a\}$. The problem is that the $U_a$ need not be pairwise disjoint.
We know that $a\not\in U_a^c$, which is closed, so by Urysohn's lemma we can choose $f_a\in C(X,[0,1])$ such that $f_a(a)=1$ and $f_a(U_a^c)=\{0\}$. My idea was to use these $f_a$ to somehow shrink the sets $U_a$ and obtain a collection that is pairwise disjoint, but I wasn't able to come up with something that works. Any hints?
Best Answer
If $X=[0,1]^\mathbb{R}$ in the product topology, then $X$ is compact Hausdorff (so normal) and separable (Hewitt-Marczewski-Pondiczery theorem) while $C = \{f \in X: \exists p \in \mathbb{R}: (f(p) = 1 \land \forall x \neq p: f(x) = 0) \}$ is discrete in the subspace topology and of uncountable size $\mathfrak{c}$. So you'd have to at least assume that your discrete subspace $C$ is closed as well.
Assuming $C$ is closed, the standard Urysohn function argument from Jones' lemma immediately implies that $2^{|C|} \le 2^{\aleph_0} = \mathfrak{c}$. That last fact does not necessarily imply that $C$ is at most countable: there are models of ZFC where $2^{\aleph_1} = 2^{\aleph_0}$, e.g. and under MA there are spaces $X$ that are normal and separable and have a closed discrete subspace of size $\aleph_1$.