ZFC Can Prove Everything Provable? But how do you define "provable"here??
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From: H. J. Sander Bruggink on 14 Jan 2010 09:41 On 01/14/2010 02:17 PM, Andrew Usher wrote: >> The so-called standard model is supposed to include all the sets that >> "really" exist. > > Is this provable? If not, how is it not just blind faith? You have to start somewhere, yes. You have to make some basic assumptions about the "real" sets. These basic assumptions are called axioms. The whole point about using a formal system is that it is completely clear what the axioms are. If the axioms of ZFC are actually true of sets, then the model of "real" sets is indeed a model of ZFC. [snip] > I didn't invent it. Anyway, what I mean is that the 'real' N and R > can't be defined by first-order logic (an immediate consequence of > LS), so the informal proof (e.g. Cantor's argument) that N< R can't > be, either. What I mean is that seeing it formalised, or knowing it > can be formalised, in ZFC does not make one any more certain that R is > not countable; on the other hand, it is open to objections (such as > mine, above) while the informal argument is not. The informal argument needs axioms also. It just doesn't spell them out. So you can still ask: is it provable that the informal argument is about "real sets"? If not, how is it not just blind faith? groente -- Sander |