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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?

hot_queen
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    I doubt. Gluing continuous functions is usual in general topology, but is forbidden for analytic functions. – Berci May 05 '13 at 23:21
  • A topology for which $f(z)=|z|$ is not continuous, then... – Julien May 05 '13 at 23:47
  • What happens if you look at the weak topology generated by the holomorphic functions? – ncmathsadist May 06 '13 at 00:30
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    a quick comment: If such a T exists then, observing bicontinuity of az +b. The translations, dialations and rotations of an open sets in T will be open. – rohit May 06 '13 at 05:26
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    Assume that $S \in T$ and S is bounded.

    Take unit disk D , and a point $p \in D$. Let r be such $B_{3r}(p)$ fits inside D. Translate S to origin s.t. a point in S matches with origin .Crunch S to fit in $B_{r}(0)$. Then translate by p.We get a U open in T s.t. $p \in U$ s.t. U fits in D. This we can do for every $p \in D$. Thus $D \in T$

    It easily follows by translations and dilations applied to unit disc D , T contains our open sets in archimedian topology.

     – rohit May 06 '13 at 12:28
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    Since complex functions take values in $\mathbb C$, "T-continuous" can be understood in two ways: continuous from T-topology to T-topology, or from T-topology to the standard topology. Which one did you mean? – 75064 May 06 '13 at 21:11
  • T-T-continuous. – hot_queen May 06 '13 at 21:19
  • Of course, any non trivial observations, about similar questions (e.g., replace analytic on C by infinitely differentiable functions on R etc.) are appreciated. – hot_queen May 06 '13 at 21:22
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    analytic or entire? –  May 23 '13 at 15:40
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    @rohit, you make one assumption: there exists $S \in T$, such that $S$ is bounded. I fail to see why this should be true. –  May 23 '13 at 17:09
  • Related: http://math.stackexchange.com/questions/427634/a-topology-such-that-the-continuous-functions-are-exactly-the-polynomials#427634 – Dominik Jun 23 '13 at 16:35

2 Answers2

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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.

Community
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    What your proof for the complex case seems to show, is that, under certain assumptions, there is a T-continuous map on a T-closed subspace ($A\cup B$) that has no T-continuous extension to all of $\mathbb{C}$. That does not look like a contradiction to me. Am I missing something? – Niels J. Diepeveen May 29 '13 at 11:55
  • @NielsDiepeveen yes, you're right. I was thinking about Tietze's extension theorem, but forgot that the topology on the other side is $T$. (Also, I didn't mention normality). Have removed that part until I figure out a way to salvage it. –  May 30 '13 at 13:44
  • Why is $f^{-1}(V)$ bounded? – dfeuer Jul 04 '13 at 20:32
  • @dfeuer: $f^{-1}(V)$ is a subset of the (compact) support of $f$, because $0 \notin V$ – Niels J. Diepeveen Jul 10 '13 at 11:43
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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.

zyx
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