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For $X$ a closed subset of $\mathbb{R}$, let $C(X)$ denote the family of all continuous functions $f:X\to\mathbb{R}$ equipped with the topology of uniform convergence. Let $\mathcal{F}$ be the family of all sets $F\subseteq C(\mathbb{R})$ such that for any closed set $X\subseteq\mathbb{R}$, $\{f\upharpoonright X\colon f\in F\}$ is a closed subset of $C(X)$.

Question: Is $\mathcal{F}$ the family of all closed sets in some topology on the set $C(\mathbb{R})$?

My guess is that $\mathcal{F}$ is not closed under intersections, but I could not find an example.

Edit: It is rather straightforward that $\emptyset,C(\mathbb{R})\in\mathcal{F}$. Since $\big(\bigcup\mathcal{F}'\big)\upharpoonright X= \bigcup\{F\upharpoonright X\colon F\in\mathcal{F}'\}$ for any $\mathcal{F}'\subseteq\mathcal{F}$, and a finite union of closed sets is closed, one obtains that $\mathcal{F}$ is closed under finite unions. However, $(F\cap G)\upharpoonright X$ can be a proper subset of $(F\upharpoonright X)\cap(G\upharpoonright X)$, so it is not clear whether it has to be closed if $F\upharpoonright X$ and $G\upharpoonright X$ are. Here we denote $F\upharpoonright X=\{f\upharpoonright X\colon f\in F\}$.

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    Not clear what you mean by "in some topology" because (1) if "in some topology on $C(R)$ means being a member of a topology on $C(R),$ then $\mathbb F$ is not, because it is not a subset of $C(R).$...(2) If you mean "is it a topology on $C(R)$?" then no, as it is not closed under unions. Let $f_n$ be the constant function with value $1/n \;(n\in \mathbb N). $ Then $\{f_n\}\in \mathbb F$ but $\cup_n\{f_n\}$ is not in $\mathbb F.$ – 2017-01-18
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    I meant (2). Your argument shows that $\mathbb{F}$ is not closed under countable unions. However, a family of closed sets of a topology has to be closed only under finite unions, which is true in our case. – 2017-01-22

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Here's a counterexample. Let $e_n(x)=nx+\frac{1}{n}$ for $n\in\mathbb{Z}_+$, and let $f,g\in C(\mathbb{R})$ be two distinct functions such that $f(0)=g(0)=0$. Let $F=\{e_n:n\in\mathbb{Z}_+\}\cup\{f\}$ and $G=\{e_n:n\in\mathbb{Z}_+\}\cup\{g\}$. Note that $F$ and $G$ are both closed: if a sequence of distinct elements of $F$ (or $G$) converges uniformly on a set $X$, that set can contain no elements besides $0$, and so since $F\upharpoonright\{0\}$ and $G\upharpoonright\{0\}$ are closed, $F$ and $G$ are closed. But $F\cap G=\{e_n:n\in\mathbb{Z}_+\}$ is not closed, since the sequence $(e_n)$ converges uniformly to the zero function $0$ on $\{0\}$ but no element of $F\cap G$ restricts to the zero function.

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    Thank you for the example. I am still interested in some characterization of sets from family $\mathcal{F}$. One question is whether the topology generated by $\mathcal{F}$ coincides with the uniform one. By other words, can be any closed set $F\subseteq C(\mathbb{R})$ (closed in the uniform topology) expressed as an intersection of sets from $\mathcal{F}$? Let me also note that one can equivalently define family $\mathcal{F}$ as consisting of all $F\subseteq C(\mathbb{R})$ such that $F\upharpoonright X$ is closed, for any $X\subseteq\mathbb{R}$ (not just for $X$ closed). – 2017-01-22