An open question on MSE, http://math.stackexchange.com/questions/427634/a-topology-such-that-the-continuous-functions-are-exactly-the-polynomials , asks whether there is an infinite field and a topology over that field such that the continuous functions from the field to itself are precisely the polynomials (and whether such a topology exists for $\Bbb Q$, $\Bbb R$, or $\Bbb C$). Another question, linked from there, asks the same about holomorphic functions. The common notion is this:

Given a set $X$ and a set $\mathscr C$ of functions from $X$ to $X$, what are necessary and/or sufficient conditions under which there is a topology $\mathscr T$ on $X$ such that a function $f\colon X\to X$ is $\mathscr T$-$\mathscr T$ continuous iff $f\in \mathscr C$.

It is immediately clear that $\mathscr C$ must contain the identity function and all constant functions, and must be closed under composition. Nothing else seems terribly obvious. What is known about this?

Edit: for the purpose of answering the specific questions on MSE, even results limited to connected, bihomogeneous $T_1$ spaces would be very helpful.

muchdiscussion there. – dfeuer Jul 8 '13 at 19:13