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What about the function of real number f(x) where 0 when x is an uncomputable real number(chaitins number) and 1 if its computable like π.

Its certainly not continuous anywhere What else can we know about this function?

In the spirit of Dirichlet function which can be construct using limits of sequence of functions, is there a way to approximate this?

multiple of 3
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  • In some sense this function is $0$ everywhere. "Almost all" real numbers are uncomputable. – Peter Nov 19 '22 at 16:05
  • Yes, "Almost all" real numbers are irrational, despite dirichlet function is much more managble, since we can detect if its rational or not, its much harder to detect if its computable or not – multiple of 3 Nov 19 '22 at 16:19
  • Why do you think "we can decide if it's rational or not"? Consider, for example, the Euler-Mascheroni constant (https://en.wikipedia.org/wiki/Euler%27s_constant). – Andreas Blass Nov 19 '22 at 16:44
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    Because it's the characteristic function of a countable set, which is an $F_{\sigma}$ set, it's at most a Baire $2$ function (i.e. very low in the ${\omega}_1$-length hierarchy of Baire/Borel-measurable functions). And since it's everywhere discontinuous, it's not a Baire $1$ function. All this is makes it from a non-computable standpoint hardly different from the characteristic function of the rationals. Thus there's nothing special about it from the standpoint of classical mathematics. – Dave L. Renfro Nov 19 '22 at 17:01
  • @AndreasBlass That we don't know whether Euler's constant is rational sounds like it has more to do with the definition of $\gamma$ than anything about the irrational numbers. One can contrive a constant such that it's hard to determine whether it's non-zero, but that doesn't impart any complexity to the indicator function of zero. – Erick Wong Nov 19 '22 at 19:51
  • @ErickWong I said nothing about the indicator function of zero. I disputed the claim that the indicator function of the rationals is much more manageable than the indicator function of the computable reals because we can detect rationality. – Andreas Blass Nov 19 '22 at 21:46
  • @AndreasBlass It was an analogy, but while you're commenting, would you dispute the claim that we can detect zero? A vague statement like "we can detect rationality" is open to a wide range of interpretation. – Erick Wong Nov 20 '22 at 06:44
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    @ErickWong There are reasonable interpretations of "we can detect" that make zero undetectable. For example, associate to each Turing machine $M$ the real number $r_M=0.a_1a_2\dots$ where the $n$-th decimal digit $a_n$ is $1$ if $M$, started on an empty tape, halts within $n$ steps and $0$ otherwise. So $r_M=0$ iff $M$ never halts. Detecting $0$ would solve the halting problem. – Andreas Blass Nov 20 '22 at 14:15
  • @AndreasBlass I agree such interpretations exist, and already mentioned this above. I guess what I objected to is that because interpretations of this sort render nearly every set undetectable (à la Rice's theorem), giving Euler's constant as some sort of evidence seemed besides the point. I don't agree so much with multiple of 3's claim, but I think one can be too literal with inferring their intent. – Erick Wong Nov 20 '22 at 16:25

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In a precise sense, your function $f$ is no more complicated than the Dirichlet function $D$.

Cantor showed that any two countable dense linear orders without endpoints are isomorphic. Let $i$ be an isomorphism between the rationals and the computable reals; by density, $i$ extends uniquely to a homeomorphism (= continuous bijection with continuous inverse) $\hat{i}:\mathbb{R}\rightarrow\mathbb{R}$. This map has the property that it "swaps" $f$ and $D$: $$D=f\circ \hat{i}\quad\mbox{and}\quad f=D\circ \hat{i}^{-1}.$$ So $f$ and $D$ are "topologically the same."

Noah Schweber
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