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Someone asked a version of this 10 years ago, but no satisfactory answer was given, so I want to try again. I have a class of functions F from the reals to a set S (the specifics are complicated and don't matter). I'm curious to know whether one can find a metric topology or any topology T such that F is exactly the set of continuous functions from the reals to S relative to T. So I'd be interested in either a strong set of necessary conditions (to prove that there isn't any such topology) or a weak set of sufficient conditions (to prove that there is).

Here are some necessary conditions of the sort I'm thinking of:
Constant: Every constant function is in F.
Composition with continuous functions: If f is in F and g:R->R is continuous then f(g(t)) is in F.
Splicing: If f and g are in F and f(0) = g(0), then the function h which is equal to f on negative numbers and equal to g on non-negative numbers is also in F.
Separation: (for Hausdorff topologies). If f is in F, u and v are in S, u!=v and f(0)=u then there is some open interval I around 0 such that v is not in f(I).

Can any one suggest any good theorems of this kind?

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    $\begingroup$ how about if $f(t)=g(t)$, for $t\in A$, then $f(t)=g(t)$, for $t\in \overline{A}$? $\endgroup$
    – erz
    Commented Jul 4, 2020 at 9:30

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