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I have found somewhere the following statement:

"Every computable real function has to be continuous,"

but I'm not able to prove it and the "proofs" that I found in some blog posts don't seem rigorous enough to me. Could you provide a formal proof of the statement?

Note: I have some knowledge of Turing machines.

Dal
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    http://blog.sigfpe.com/2008/01/what-does-topology-have-to-do-with.html – Kyle Gannon Jan 02 '15 at 01:43
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    Is a step function computable? – Bernard Jan 02 '15 at 01:43
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    @KyleGannon Could you point out a more academic presentation? – Dal Jan 02 '15 at 02:15
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    In intuitionistic mathematics, Brouwer's Continuity Theorem boldly states that all total real functions are (uniformly) continuous on the unit interval. Somewhat more elaborate/general: any (total real) function which is defined everywhere at an interval of real numbers is also continuous at the same interval. With other words: For real valued functions, being defined is very much the same as being continuous. In view of the above, is this a coincidence? I think not. Sad remark: it's difficult to find a reference for the theorem nowadays. – Han de Bruijn Feb 21 '15 at 10:37
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    @Bernard: A step function is not computable where it steps :-( – Han de Bruijn Feb 21 '15 at 10:58
  • @HandeBruijn, how can that possibly be true? if we have the step function $f(x)=0 \text { if } x<0, f(x)=1 \text { if } x\leq 0$, then it is discontinuous at $0$, but I can compute it: $f(0)=1$. There I've just computed it. So the step function is computable. I just computed it. Is this a joke? I checked whether the question was posted on April 1, but no. I must either significantly misunderstand something, or the world has gone mad. – user56834 Dec 08 '17 at 18:45
  • @Programmer2134: According to LEJ Brouwer, such a "function" is not a function, because in intuitionistic mathematics it is "multiple valued" at $x=0$. If you take a look at an oscilloscope with the Heaviside on it, then this point of view makes more sense than you might think at first sight. Sorry for the late response. – Han de Bruijn Dec 17 '17 at 11:41
  • @HandeBruijn, Could you elaborate? I don't understand the oscilloscope example. – user56834 Dec 17 '17 at 11:45
  • @Programmer2134: Consider instead the function: $$ H(x) = \lim_{\sigma\to 0}\left[\frac{1}{\pi}\arctan\left(\frac{x}{\sigma}\right)+\frac{1}{2}\right] $$ – Han de Bruijn Dec 18 '17 at 12:53

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It is proved on the book Computable Analysis: An Introduction written by Klaus Weihrauch

Nicolás Ippolito
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