(If the title is unclear, I'm looking at infinite cartesian product of $\mathbb{R}$ indexed by $\mathbb{R}$.)
I thought that I had reasoned this rather well, as follows:
$\mathbb{R}^\mathbb{R} = \{f\mid f:\mathbb{R}\rightarrow\mathbb{R}\}$. Note that this includes functions whose range is not $\mathbb{R}$, $\mathbb{R}$ here is just the codomain. Now consider an element $\textbf{x} \in \prod_{\mathbb{R}}\mathbb{R}$. This is an "uncountably infinite sequence," so to speak, so it does represent a function, where the index indicates the $x$ coordinate and the value at that index indicates the $y$ coordinate. (Consider a constant $x_i = 1 \;\forall\; i\in\mathbb{R}$, a straight line.) But this does not account for functions with vertical asymptotes, or functions which are not surjective (e.g. $f(x) = x^2$), so we conclude that $\prod_{\mathbb{R}}\mathbb{R} \subsetneq \mathbb{R}^\mathbb{R}$.
If we union in $\{+\infty, -\infty\}$, then we can get functions with vertical asymptotes, and if we allow coordinates $x_i = \emptyset$, then we get surjective functions as well. (If that even makes sense semantically).
This seems to make sense to me, and I find it a satisfying answer. However, I was reading this wikipedia article which states that:
An uncountable product of metric spaces need not be metrizable. For example, $\mathbb{R}^\mathbb{R}$ is not first-countable and thus isn't metrizable.
Which implicitly states that $\prod_{\mathbb{R}}\mathbb{R} = \mathbb{R}^\mathbb{R}$. So what am I missing? I feel like it must be something rather obvious.