TITLE: Is there a topology $T$ on the set of complex numbers such that the class of $T$-continuous functions and the class of analytic functions coincide?
QUESTION [46 upvotes]: Is there a topology $T$ on the set of complex numbers such that the class of $T$-continuous functions and the class of analytic functions coincide?

REPLY [13 votes]: For infinitely differentiable functions on $\mathbb{R}$, there is no such topology $T$:
If there is, as rohit has noted, due to the bicontinuity of $ax+b$, translations and dilations of open sets will be open. If $U\in T$ such $U$ is bounded, then we can construct the usual topology on $\mathbb{R}$ from $U$ (see rohit's comment)
Now, let $V \in T$, such that $V\neq \mathbb{R}$ (wlog, assume $0\notin V$). Then, take any $f \in C_c^{\infty}(\mathbb{R})$.  Now, $f^{-1} (V)$ is bounded.
From this, it follows that any such topology $T$ is finer than the usual topology. But, as Berci said, in the usual topology, pasting doesn't work in general for differentiable functions. (Take $x$ on $[0,\infty)$ and $-x$ on $(-\infty,0]$)

Edit: As NielsDiepeveen pointed out, my earlier answer (about the complex case) was wrong.