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May 23, 2010 at 17:36 comment added François G. Dorais Never mind, that was a silly typo I made when copying your argument.
May 23, 2010 at 16:47 comment added Joel David Hamkins I think $W_e=N$ is correct. The point is that once you know $W_{\rho(e)}=N$, then $N-W_{\rho(e)}$ is empty, so it had better be that $W_e=N$ or else $f[N-W_e]$ won't be empty. Or have I misunderstood?
May 23, 2010 at 15:30 comment added François G. Dorais I think there's a minor error in the 6th paragraph: "that $W_e=N$" should be "that $W_e = ran(f)$."
May 23, 2010 at 4:10 vote accept François G. Dorais
May 23, 2010 at 4:10 comment added François G. Dorais This doesn't answer the main question, but I'm accepting it anyway since the argument is so beautiful!
May 23, 2010 at 4:04 comment added François G. Dorais This is great! I had gotten the finite-to-one part using a completely different argument, but not the computable bound on $f^{-1}(k)$.
May 23, 2010 at 3:36 history edited Joel David Hamkins CC BY-SA 2.5
deleted 256 characters in body
May 23, 2010 at 3:01 history answered Joel David Hamkins CC BY-SA 2.5