Sure, but that is not really a “Turing machine of that size” - it’s a substantially bigger one—since it has to encode the input, and another program expressing decoding it and writing it out.
This fact does not cripple the proof. No binary turing decider of size x could run longer on an input of binary length n than x2^nBB(x), which is still an upper bound which can be used to solve the halting problem. (This is a tighter upper bound than that given by anon19 below, since BB(x) grows at a superexponential rate.)
Sure, but that is not really a “Turing machine of that size” - it’s a substantially bigger one—since it has to encode the input, and another program expressing decoding it and writing it out.
This fact does not cripple the proof. No binary turing decider of size x could run longer on an input of binary length n than x2^nBB(x), which is still an upper bound which can be used to solve the halting problem. (This is a tighter upper bound than that given by anon19 below, since BB(x) grows at a superexponential rate.)