4 ms·
No, I'm not assuming any particular encoding of an algorithm. I'm just assuming by "algorithm" you mean a computable function, and we know there are only counta
by tshaddox 5y ago
No, I'm not assuming any particular encoding of an algorithm. I'm just assuming by "algorithm" you mean a computable function, and we know there are only countably many computable functions. This is not a cultural notion.
And yes, which particular encoding you decide to use is arbitrary, but the point is that you can enumerate the set of all algorithms, and thus you can select one without needing the axiom of choice.
- shadowlight 5y agoYour own example used the word "alphabetical." So your example is false because it uses a "particular" encoding. Try to select an algorithm out of the set of all algorithms without using an encoding. If you must use an encoding, please ensure that it's not a "particular" encoding. You can't. The point is all encodings in the known universe are "particular." Additionally, to even use an encoding you have to *select* and encoding from the set of all encodings.
- tshaddox 5y agoYes, I chose one arbitrary method of enumeration. That’s not important to the point, which is that algorithms are enumerable and thus you don’t need the axiom of choice to select one out of the set of all algorithms.
- shadowlight 5y agoYes I know that's your point. I'm saying you can't enumerate algorithms without selecting an algorithm. One way of selecting an algorithm is to select a way to encode the algorithms. It's easy to see that this is true. You chose an arbitrary example above. Try to do the same without choosing anything arbitrary. You can't.