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?

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?

Solution 1:

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.

Solution 2:

It seems difficult to be conjugation-variant. Any set that can be constructed from some set of entire functions (without using complex conjugation) has its conjugate constructible by conjugating the coefficients of the functions and all parameters in the construction. But we need $\bar{z}$ to be discontinuous and $z$ continuous.