All the repeating sequences have finite Kolmogorov complexity. So, if you want, you can say that the sequences with infinite Kolmogorov complexity have even lower repetitiousness than the ones we can produce algorithmically. And this will go on and on through the countable ordinals.
Isn't this just true?
The set of algorithms is countable, and the set of sequences of real numbers is uncountable.
You're kind of "begging the question" here, where you're assuming that non-computable numbers exist and then using that to show that some numbers are non-computable. You can definitely show that these things exist, but that relies on "believing" the set of axioms that you used to prove it.
My understanding is that we can reason about these sorts of non-constructible sequences but we can't really know if they "really exist" because it's not clear what that even means if their existence has no implications for physical reality (which is possible but not known). So now we're into the territory of whether there is such a thing as abstract truth independent of reality.
I wrote a blog post in the past about my reading on this topic, and I can't claim that's it's accurate but at least I tried to talk about this subject: https://blog.kevmod.com/2022/04/09/do-the-real-numbers-exist...
That is assuming finite algorithms. I’m curious if infinite algorithms exist and/or considered in maths.
The internet says that an algorithm is finite by definition, but it’s hard to search for “infinite algorithm” because everyone talks about termination instead.
Edit: let’s not forget the rich theory involved in examining if sets/sequences contain arithmetic (or geometric) progressions! The well known Green-Tao theorem about primes containing arbitrarily long arithmetic progressions is one such result. In fact, the real Green-Tao theorem says that any set which has nonzero upper density with respect to the set of primes contains arbitrarily long arithmetic progressions.
Could you also be more precise about what you mean by true randomness?