Show that the product space \(\mathbb{R}^I\), where \(I = [0, 1]\), has a countable dense subset.
Show that if \(J\) has cardinality greater than \(\mathcal{P}(\mathbb{Z}_+)\), then the product space \(\mathbb{R}^J\) does not have a countable dense subset. [Hint: If \(D\) is dense in \(\mathbb{R}^J\), define \(f: J \to \mathcal{P}(D)\) by the equation \(f(\alpha) = D \cap \pi_\alpha^{-1}((a, b))\), where \((a, b)\) is a fixed interval in \(\mathbb{R}\).]
Solution
- We construct a countable dense subset of \(\mathbb R^I\), \(I=[0,1]\). Choose an injection \[ \iota:I\hookrightarrow\mathcal P(\mathbb Z_+). \] For every finite set \(F\subset\mathbb Z_+\) and every function \[ \phi:\mathcal P(F)\to\mathbb Q, \] define \(d_{F,\phi}\in\mathbb R^I\) by \[ d_{F,\phi}(t)=\phi(\iota(t)\cap F). \] There are only countably many such pairs \((F,\phi)\), so \[ D=\{d_{F,\phi}\} \] is countable.
To see \(D\) is dense, let a basic open set prescribe nonempty open intervals \(U_1,\dots,U_n\) at distinct coordinates \(t_1,\dots,t_n\). Since the sets \(\iota(t_i)\) are distinct, there is a finite \(F\subset\mathbb Z_+\) such that their traces \(\iota(t_i)\cap F\) are pairwise distinct: for each pair choose one integer where the two subsets differ, and take the finite union of these witnesses. Choose rationals \(q_i\in U_i\), and define \(\phi\) on the distinct traces by \(\phi(\iota(t_i)\cap F)=q_i\), arbitrarily elsewhere. Then \(d_{F,\phi}\in D\) lies in the given basic open set. Hence \(D\) is dense.
- Let \(D\subset\mathbb R^J\) be countable and dense. Fix a nonempty proper open interval \(U=(0,1)\). For each \(\alpha\in J\), define \[ F(\alpha)=D\cap\pi_\alpha^{-1}(U)\subset D. \] We claim \(F:J\to\mathcal P(D)\) is injective. If \(\alpha\ne\beta\), the basic open set \[ \pi_\alpha^{-1}(U)\cap\pi_\beta^{-1}((2,3)) \] is nonempty, so density of \(D\) gives \(d\in D\) with \(d_\alpha\in U\) and \(d_\beta\notin U\). Hence \(d\in F(\alpha)\setminus F(\beta)\). Thus \[ |J|\le |\mathcal P(D)|=|\mathcal P(\mathbb Z_+)|. \] Therefore if \(J\) has larger cardinality than \(\mathcal P(\mathbb Z_+)\), \(\mathbb R^J\) cannot be separable.