Recent content by krausebj0

This may help: It's a lost letter from Goedel. http://rjlipton.wordpress.com/

A will be the same regardless of how Ai is ordered. No element ri of RI is a subset of A unless h(ri) is an element of Ai whenever Ai is a finite sequence. Ai is a finite sequence unless A = D by necessity.

Here is another slightly easier brain teaser: For any finite set Ai, does the order of the elements of Ai matter when it comes to A? In other words, will A equal the same set regardless of how Ai is ordered, for each possible finite combination reals that could possibly comprise Ai? Answer...

You have now just blown off my work because you put words in my mouth. I said I agreed these were infinite sequences. Ai is not infinite until A = D though. That my friend, is one totally different statement. Your ignorance is none of my concern though, if you wish to leave this thread. By...

Kielejocain, I have tried to make RI more clear for you above. Thanks for ultimately improving my work by making me try to be more clear. Please quote one line of the op only now going forward, assuming you are comforable with RI.

Sorry, ri is in RI. There is only one set RI. I wish to define RI better for you. I consider it your quoted line of the proof (your one per post please).0.111... <---> { 0.1, 0.11, 0.111, ... }, where k(0.111...) = { 0.1, 0.11, 0.111, ... } 0.10101... <---> { 0.1, 0.101, 0.10101, ... }, where...

Again, you are clinging to the old textbooks, but I will acknowledge your argument by accepting that you think the cardinality of Bi is greater than that of aleph null. A finite set is a recognized term. It will have a cardinality less than aleph null. If I say that both Ai and Bi have finite...

This thread is going to turn into a nightmare if people do not start doing the one thing I have asked. Go in order. Pick the first statement you think is wrong. Ignore everything else. Quote the line of the op you think is wrong and nothing else. Discuss only it. This is really getting...

RI is a subset of the infinitely large elements of the powerset of dyadic rationals. I am not so sure your above notation checks out Kielejocain. I take that back. Let's just keep things simple, as I DO TRULY feel that the visual representation of RI is best for this. It is clearly obvious...

Kielejocain, Have you ever heard of the phrase, "My mind is open." You cling to the old textbooks above... Here is a real brain teaser: Let Ai = 0.10111..., 0.110111..., 0.1110111..., ... Is Ai an infinite sequence? Elementary mathematics of course teaches us yes. The answer...

I am very familiar with cardinalities, etc. I trust you are just trying to ensure we are both on the same page? I have addressed Cantor's Theorem in the proof and above, so I will disregard your putting words into my mouth by saying I cannot count the reals. Thanks for the notational...

Here is a real brain teaser: Let Ai = 0.10111..., 0.110111..., 0.1110111..., ... Is Ai an infinite sequence? Elementary mathematics of course teaches us yes. The answer is actually no. Why not? A does not equal D. Ai is only an infinitely large sequence if A = D. Proof: Consider...

The weight of these words broke your jaw gopher_p. Mine too though at first, so please do not think that I am degrading you. Until you have completely made sense of the following, I respectfully ask you to not post again and also thank you for your time thus far: "There is a reason that...

I agree that my conclusion must be false "according to Cantor's Theorem". That is the point of this thread. We agree (you just think we disagree). Ai is the tape. Step 3 defines the turing machine. You are very confused here. The purpose of this is to analyze different Ai. The...

Ai can be finite. This is like a turing machine, where Ai is the tape. If step 4 is complete, there exists an element of Ai such that iterating that element results with A = D... Hope the above helps...