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The halting problem states that there is no Turing machine that can determine whether an arbitrary Turing machine halts on $\epsilon$. But I try to ask something different, can we find a specific Turing machine $M$ such that $L_M=\{ 0^n : M$ halts on $\epsilon \}$ is undecidable? Or at least can we prove there exists such $M$?
The language itself isn't the issue, it is either $L=\{0^n : n \in \mathbb{N} \}$ or $L=\phi$ and the way to distinguish the cases is only by the right side of the definition of $L$.
In other words, can we define a Turing machine such that we (or a Turing machine) can't determine whether it halts or not on $\epsilon$?
You might want to take a look at How can it be decidable whether π has some sequence of digits?
– Aaron Rotenberg – 2020-01-29T03:25:17.530Hang on a minute, what does $0^n$ mean in the definition of $L_M$? – Rick Decker – 2020-01-30T16:46:50.637
A string of zeros of length $n$ – oren1 – 2020-01-31T07:31:04.373