Is $\prod_{\mathbb{R}}\mathbb{R} = \mathbb{R}^\mathbb{R}$? (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.
 A: Your argument about the product starts well, but then it goes off the rails. Why do you say that "it doesn't account for functions that are not surjective"? Or "functions with vertical asymptotes"? It certainly doesn't include functions that are not defined on all of $\mathbb{R}$, but then neither does $\mathbb{R}^{\mathbb{R}}$. As for surjectivity, nothing in the definition implies surjectivity: the "constant tuple" that is $1$ everywhere is an element of $\prod \mathbb{R}$; what makes you think this is surjective?
Remember the very definition of a direct product: if $\{X_i\}_{i\in I}$ is a family of sets, then
$$\prod_{i\in I}X_i = \Bigl\{ f\colon I\to \cup X_i \,\Bigm|\, f(i)\in X_i\text{ for each }i\in I\Bigr\}.$$
And if $X$ and $Y$ are sets, then
$$X^Y = \{f\colon Y\to X\}$$
that is, all functions with domain all of $Y$ and with $f(y)\in X$ for all $y\in Y$. 
So
$$\prod_{r\in\mathbb{R}}\mathbb{R} = \Bigl\{ f\colon \mathbb{R}\to\mathbb{R}\,\Bigm|\, f(i)\in\mathbb{R}\text{ for each }i\in\mathbb{R}\Bigr\} = \mathbb{R}^{\mathbb{R}}.$$
