Topology question

Ironically, this all started because I agreed with you.
Originally posted by matt grime
I'm slightly curious, curiousbystander, as to why you've told me this.

I actually ran short of time and didn't reach my point before I had to leave. I was only laying out the basics before I proceeded. Without the rest of what I had planned, the quote at the top of my previous post makes it sound as if you don't understand some very basic concepts, and the tone becomes condescending. That was not my intention and you have my apologies.

My goal (in brief) was to show the interdependency of the idea of a topology with the idea of a power set, show the relation between functions and special subsets of Cartesian products, and then use the set of functions from A to the set {0,1} to build the power set P(A) out of operations that are admissable in the system described by Phoenix. I wanted to do this carefully and concisely, and then use the usual generalizaton of Kantor's argument to show the differences in cardinality between the two sets. So if you're going to give up the power set you'd have to give up not only topologies but also functions between sets, or at least functions from the Universal set to the set {0,1}.

You should be aware that Phoenixthoth is not using ordinary set theory.
This is the part where I disagree I think that implicitly he is, and when all the terms have been clarified and the logical inconsistencies straightened out, his theory will closely resemble set theory. I've not yet been convinced (though I hope you will Phoenix) that a 3 valued logic system removes any of the difficulties.
 

matt grime

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Thanks for clearing that up.

I neither agree nor disagree with phoenix's theory 'cos i've not read it enough detail. My gut reaction? Well, I agree with you on that one.
 
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My goal (in brief) was to show the interdependency of the idea of a topology with the idea of a power set, show the relation between functions and special subsets of Cartesian products, and then use the set of functions from A to the set {0,1} to build the power set P(A) out of operations that are admissable in the system described by Phoenix. I wanted to do this carefully and concisely, and then use the usual generalizaton of Kantor's argument to show the differences in cardinality between the two sets. So if you're going to give up the power set you'd have to give up not only topologies but also functions between sets, or at least functions from the Universal set to the set {0,1}.
i just can't seem to give myself a satisfying proof that a set can't be mapped onto its powerset. so if you can do this and prove that all sets not already in bijection with U can't be mapped onto their power sets, that would be most excellent.

i'm wondering if these balls could be turned into a basis for a topology:
B(x,d)={y in U : x+y<=d}, where x+y means symmetric difference and <= means "can be injected into." i had this in an earlier version of the paper but i removed it...

perhaps a better one would be B(x,d)={y in U : x+y<d}.
 
This is the gist of the argument.

I'm a bit pressed for time, and I won't be able to get back online for more then a few miutes until Saturday but I didn't want to leave you hanging.

Let A be a set
Identification of a subset of A with a function from A to {0,1}
Let's use something small, and concrete to get the idea. Let A = {a,b,c,d}. We identify the subset {b,c} with the function f:A->{0,1} where f(a)=0,f(b)=1,f(c)=1, and f(d)=0.
Notice the input of f is an elt of A and the output of f encodes the answer to the question 'Is the input in the subset?'-- 0 means no, 1 means yes.

It's a characteristic function if you're more familiar with that terminology.

Similarly the function f(a)=0,f(b)=0,f(c)=1 and f(d)=0 corresponds to the subset {c}.

Any subset has a corresponding function (If you know which function corresponds to the empty set, then you've got the idea). Similarly, every function defines a unique subset.

I will define the power set of A to be this set of functions from A to {0,1} and I'll call it P(A)

For a finite set with n elts there are 2^n such functions.

The contradiction
Now let A be any set and P(A) the power set as described above. Assume we have a function g:A->P(A) that is onto. This parts a bit weird..
remember that g(a) is itself a function from A to {0,1}? that means (g(a))(a) is either a 0 or a 1-- The g(a) in the parenthesis gives me the function and then the second a in the parenthesis is the input to that function.

Okay so I'm going to build a new function h:A->{0,1}... this function will (by definition) be the in power set of A. h(a)=1-g(a)(a)...
So h is an elt of the power set of A... but it can't be in the image of g... you see g(a) corresponds to a subset of A and either a is itself in that subset, or it is not. If a is in g(a) then g(a)(a) = 1, and then h(a)=0 which means that a is NOT in the subset defined by h.
On the other hand if a is NOT in g(a) then g(a)(a)=0 and then h(a)=1 and so a IS in the subset defined by h.

There's our contradiction. We started by assuming g was onto, and we constructed an elt that can't be in the image of g.
 
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if you define [tex]P\left( A\right) =\left\{ 0,1\right\} ^{A}[/tex], where the RHS means the set of functions from A to {0,1}, then the funny thing about that is that both
1. [tex]\left\{ 0,1\right\} ^{U}\subset U[/tex] and
2. [tex]\left\{ 0,1\right\} ^{U}\in U[/tex].

i think one can fairly easily show that x~y iff P(x)~P(y) (where ~ denotes is in bijection with) and it's fairly routine to show that if you define P(x) the standard way that [tex]P\left( A\right) \sim \left\{ 0,1\right\} ^{A}[/tex]; so your proof translates into proving that even a powerset defined normally is "larger" than the set.

i'm banging my head against the wall trying to figure out where this proof goes wrong in the case of U and P(U). any thoughts?

it may have to do with this line right here: either a is itself in that subset, or it is not. that isn't the case in this system; those aren't the only two options. however, for a crisp set like U, there aren't even that many choices: there is only one choice for all sets are subsets of U and so your h function which is kinda a subset, or at least is isomorphic to a subset, would produce, i think, a set not in U which is itself a contradiction. so the argument in the case of U, and U alone, would say that U is not universal but it is so h doesn't exist. this sounds like the formal version of what organic has been trying to do all along... this argument only fails to work, if i even have it right, when the set one is considering is the universal set and i don't think any other time.

a theorem whose proof i didn't like was this (and i still believe the theorem): if x maps onto P(x) then either x is fuzzy or x=U. and this: if x is crisp and x maps onto P(x) then x=U.
 
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