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UBA - CienciaS :: Ver Tema - Duda sobre axioma de completitud y Q
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Duda sobre axioma de completitud y Q
http://ubacs.com.ar/ubacs/viewtopic.php?f=57&t=2143
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Autor:  FJL [ 03 Sep 2010, 14:18 ]
Asunto:  Duda sobre axioma de completitud y Q

Hola!

Estamos definiendo axiomáticamente los reales (bajón, yo quería algo tipo Dedekind o Cantor :(). Hace unas clases, vimos un ejemplo de como no satisface la completitud, en particular era . Acá, "mostró" que, con solo los axiomas de cuerpo ordenado, no existe un supremo para este conjunto. O sea, dice , pero podría decir ordenado.

Ahora, luego, introdujo el axioma de completitud. Oh sorpresa, ahora se puede probar que si cumple ese axioma, entonces ese conjunto si tiene supremo. Mi pregunta es, esto no implica un error en la demostración anterior?

O sea, ahora nos dieron más información - cumple una propiedad más. Pero si agarramos la demostración que usamos para cuando no tenía esta propiedad, debería seguir valiendo, si no habíamos mentido antes! O sea, si me dicen "Asumí que tenés 6 manzanas. Probame que podés comer 3.", yo lo pruebo, y luego me dicen "Ahá! Pero ahora, China es comunista. Tu prueba anterior dejó de valer."
Yo en ningún lado usé la no-comunistidad (...comunión?) de China en mi demostración de las manzanas, porque no sabía si China era o no comunista, y ahora al agregar ese axioma (que, a priori, no es contradictorio con los anteriores), mi demostración deja de valer?

Gracias por cualquier aclaración :)

Autor:  Quimey [ 03 Sep 2010, 23:44 ]
Asunto:  Re: Duda sobre axioma de completitud y Q

Fijate como es la demostracion de que el conjunto que pusiste no tiene supremo, casi seguro que tenes que usar que no existe la raiz cuadrada de 2. Pero cuando agregas completitud eso
deja de ser cierto...

Autor:  FJL [ 04 Sep 2010, 14:22 ]
Asunto:  Re: Duda sobre axioma de completitud y Q

Si, probablemente la demostración de eso solo valía para .
Gracias :)

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