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Let $X:=\{f: \mathbb{C}\to \mathbb{C}\}$ be a class of total functions on $\mathbb{C}$ closed under composition, addition, multiplication, and scalar multiplication. Does there exist a topology on $\mathbb{C}$ making these functions and only these functions continuous?

If it's not true in general (it probably isn't), are there any interesting known cases where it is true?

Note: I emphasize total functions because we want them to be everywhere defined. This avoids functions with bad singularities.

Edit: Obviously, continuous functions in the standard topology fit this bill, but this is tautological and not in the spirit of the problem.

Edit 2: Apparently the way I asked this question made it seem like I was looking for an answer to the "general case" which seems pretty untrue although I haven't actually worked it out. Rather, the real question was interesting cases where it is true.

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    $\begingroup$ Example machine: Take X to be the set of all continuous function for a topology on C with respect to which addition and multiplication are continuous... $\endgroup$
    – François G. Dorais
    Commented Feb 13, 2010 at 6:11
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    $\begingroup$ OK, here are 2^{2^{\aleph_0}} examples: Let s be any field automorphism of C, and consider {s \circ f \circ s^{-1} : f is continuous in the usual sense}! $\endgroup$
    – Bjorn Poonen
    Commented Feb 13, 2010 at 6:15
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    $\begingroup$ Could you guys close this? I know at least Joel and Pete can vote to close. $\endgroup$
    – Harry Gindi
    Commented Feb 13, 2010 at 6:15
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    $\begingroup$ I guess I didn't put enough thought into the question, as Pete revealed to me via the Socratic method. $\endgroup$
    – Harry Gindi
    Commented Feb 13, 2010 at 6:21
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    $\begingroup$ @JDH: We are all finding our way in this new social environment, so what I say is tentative, but: it seems to me that if the questioner wants a question to be closed, it should be -- and stay -- closed. I think Harry is right when he says that anyone who is interested in the question can easily ask a new (and, one hopes, improved) version of the question. Please go ahead and ask a new question if you like: the topic seems interesting to me too. $\endgroup$
    – Pete L. Clark
    Commented Feb 14, 2010 at 1:57

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If X is the set of all functions, then it has your closure properties, and the indiscrete topology {emptyset, C} makes exactly those functions continuous.

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