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Am I missing something or is the author of this pdf's proof equally flawed. They assume each function is idempotent and injective ... meaning that it is the ide
by Russell91 12y ago
Am I missing something or is the author of this pdf's proof equally flawed. They assume each function is idempotent and injective ... meaning that it is the identity function. And the proof doesn't follow. A much more natural way to have "fixed" the original proof would be to require that V: X -> R be injective.
- 4bpp 12y agoAssuming by function you refer to the "agents'" \Phi, where do they make the assumption that it is injective? I see nothing to the effect in the text, and the functions in the counterexample on page 6 are not injective.
- Russell91 12y agoOh, oops. I was reading their statement: This interpretation is incorrect without the additional hypothesis that V (x) is a one-to-one function. as - all of the phi functions must be one-to-one. Looks like they made the same assumption that I would have. This was bothering me though so thanks for your response.