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As black_knight said, you don't need the axiom of choice for Cantor's diagonalization proof---that particular proof is completely constructive and can be writte
by vilhelm_s 10y ago
As black_knight said, you don't need the axiom of choice for Cantor's diagonalization proof---that particular proof is completely constructive and can be written as an ordinary non-oracular program.
- jwtadvice 10y agoHuh looks like I need to do a lot more reading to understand all of this! :) I'm still trying to understand what the Probablistic Method looks like as a program. Any thoughts on that?