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Consider the product topology on $\mathbb{R}^{\mathbb{R}}$. My textbook (Willard) describes the basic neighborhood of the zero function, which they label as $g$, as the set

$$U(g) = \{h \in \mathbb{R}^{\mathbb{R}} \mid |h(y) - g(y)| < \epsilon, y \in F\},$$

for some finite set $F \subset \mathbb{R}$ and some $\epsilon > 0$. I'm not quite sure where this comes from. I guess it intuitively makes sense that you'd want the inequality part with the $\epsilon$, but why a finite set $F$? What, in general, does a basic neighborhood of any given function $f: \mathbb{R} \rightarrow \mathbb{R}$ look like in this topology?

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    Define, for each $\alpha\in\mathbb R$, the evaluation function: $e_{\alpha}:\mathbb R^{\mathbb R}\to\mathbb R$ defined as $e_{\alpha}(f)=f(\alpha)$. Then the product topology is the smallest topology such that each of the $e_{\alpha}$ is continuous. – 2017-01-24

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Okay I've thought about it a little bit more. From the definition of the product topology the basic nbhds of a point will be a finite product of open sets crossed with the remaining topological spaces. I think this is where $F$ comes from above. And the $\epsilon$ part of the definition comes from the fact that basic nbhds of $\mathbb{R}$ are the sets $(a, b)$ for $a < b \in \mathbb{R}$. So I think I can work out the missing details.