Is $\text{Cont}(\mathbb{R},\mathbb{R})$ dense in $\mathbb{R}^\mathbb{R}$? Let $\text{Cont}(\mathbb{R},\mathbb{R})$ denote the set of continuous self-maps of $\mathbb{R}$ and let $\mathbb{R}^\mathbb{R}$ denote the set of all self-maps of $\mathbb{R}$, endowed with the product topology. Is $\text{Cont}(\mathbb{R},\mathbb{R})$ dense in $\mathbb{R}^\mathbb{R}$?
 A: This question seems to be posed too quickly (without substantial preliminary thinking) and has an (almost trivial) affirmative answer: We should prove that $\mathrm{Cont}(\mathbb R,\mathbb R)$ intersects each non-empty open set $U\subset \mathbb R^{\mathbb R}$. We can assume that $U$ is of basic form: $U=\prod_{r\in\mathbb R}U_r$, where for each $r\in\mathbb R$ the set $U_r$ is open in $\mathbb R$ and the set $F=\{r\in\mathbb R:U_r\ne\mathbb R\}$ is finite. Now take any (piecewise linear) continuous function $f:\mathbb R\to\mathbb R$ such that $f(r)\in U_r$ for any $r\in F$. Then $f\in \mathrm{Cont}(\mathbb R,\mathbb R)\cap U$. So, $\mathrm{Cont}(\mathbb R,\mathbb R)$ is dense in $\mathbb R^{\mathbb R}$ (even for the Tychonoff product topology of the real lines endowed with the discrete topology).
A: Yes.
Let $g: \mathbb{R} \to \mathbb{R}$ be an abritrary function.
Let $\mathcal{F}$ denote the set of all finite subsets of $\mathbb{R}$. We endow $\mathcal{F}$ with the order $\subseteq$, which renders it a directed set.
For each $F \in \mathcal{F}$, choose a continuous function $f_F$ which fulfils $f_F(x) = g(x)$ for all $x \in F$. Then the net $(f_F)_{F \in \mathcal{F}}$ converges pointwise to $g$.
