Munkres has the following exercise:
Show that the product space $R^I$, where $I=[0,1]$, has a countable dense subset. If $J$ has cardinality greater than $2^\mathbb{N}$, then the product space $\mathbb R^J$ does not have a countable dense subset.
There is already examples of countable dense subsets and a more general theorem stating that a product of $\leq 2^\kappa$ spaces of density $\leq \kappa$ has density $\leq \kappa$ for an infinite cardinal $\kappa$. By using the fact that the index set is Hausdorff, we can conclude that the product space $\mathbb R^J$ is separable.
Let us return to the case $I=J=[0,1]$. Now I may have found an other, slightly different, countable dense subset of $\mathbb R^{[0,1]}$.
Consider the set $$A:= \left\{f\in\text{Hom}\left([0,1], \mathbb R\right) : f = \displaystyle\sum_1^n \lambda_k H_{q_k}\right\}$$ of finite linear combinations of Heaviside functions $H_{q_k}$ in rational points $q_k\in\mathbb Q$ and rational coefficients $\lambda_k\in \mathbb Q$.
Now, $|A| =\left|\displaystyle\prod_{n\geq 1}\mathbb (\mathbb Q\times\mathbb Q)^n\right| = \left|\displaystyle\prod_{n\geq 1}\mathbb Q^{2n}\right|$, so $A$ should be countable.
Moreover; since the topology is generated by $\prod U_\alpha$ with only finitely many non-trivial $U_\alpha$, we should be able to find a finite linear combination of Heavisides (i.e. some element of $A$) such that it intersects a basis element $U=\prod U_\alpha.$
Does this example hold or have I overlooked something?
