uncool 212 Posted yesterday at 05:58 AM 59 minutes ago, wtf said: Secondly, suppose you do find a nonempty interval. Then the argument as I understand it says that there's a smaller element, ad infinitum, giving a countably infinite sequence of ever smaller reals. I don't see that in the proof being used. The proof seems to be based on the claim that a well-ordered set must be exhausted by the denumerable sequence of its minimum, the successor of the minimum, the successor of that, etc. This is distinct from the "halfway between" proof that discountbrains was using earlier, and is on surer footing notationally, but fails in that it cannot prove the above key claim (and in fact, the claim is false, as you have pointed out with omega + omega). 0 Share this post Link to post Share on other sites

discountbrains 32 Posted yesterday at 03:21 PM 17 hours ago, wtf said: I said IF <* happens to be a well order, your proof would fail. How do you know it isn't? If a set S is WO it has to have an element, m, such that S={x: m ≤*x} and this is not the same as S'= {x: a<*x, for some a}. I am saying there can always be an S'. This 'finite downward chain' was also part of my thinking, but I was really trying to show any z in T was also contained in my countable sequence of z's which might really be doable. The difference between S and S' should be sufficient though. 17 hours ago, wtf said: I said IF <* happens to be a well order, your proof would fail. How do you know it isn't? If a set S is WO it has to have an element, m, such that S={x: m ≤*x} and this is not the same as S'= {x: a<*x, for some a}. I am saying there can always be an S'. This 'finite downward chain' was also part of my thinking, but I was really trying to show any z in T was also contained in my countable sequence of z's which might really be doable. The difference between S and S' should be sufficient though. For some m in S' how do we know there is an x in S' such that a<*x<*m? We would just have to look at the particular order relation, <*, to find it. We would have to find what we think is m too. 0 Share this post Link to post Share on other sites

wtf 125 Posted yesterday at 05:31 PM (edited) 3 hours ago, discountbrains said: If a set S is WO it has to have an element, m, such that S={x: m ≤*x} and this is not the same as S'= {x: a<*x, for some a}. I am saying there can always be an S'. This 'finite downward chain' was also part of my thinking, but I was really trying to show any z in T was also contained in my countable sequence of z's which might really be doable. The difference between S and S' should be sufficient though. If a set S is WO it has to have an element, m, such that S={x: m ≤*x} and this is not the same as S'= {x: a<*x, for some a}. I am saying there can always be an S'. This 'finite downward chain' was also part of my thinking, but I was really trying to show any z in T was also contained in my countable sequence of z's which might really be doable. The difference between S and S' should be sufficient though. For some m in S' how do we know there is an x in S' such that a<*x<*m? We would just have to look at the particular order relation, <*, to find it. We would have to find what we think is m too. > If a set S is WO it has to have an element, m, such that S={x: m ≤*x} and this is not the same as S'= {x: a<*x, for some a}. I am saying there can always be an S'. Ok. Take the natural numbers. Then 0 is the smallest element. And S' might be all the numbers greater than or equal to 14. Ok. What does that prove? Of course there are always UPWARD chains, even infinite ones depending on the order. And can someone please point me to the definitive version of the proof? I seem to have missed the new proof. > For some m in S' how do we know there is an x in S' such that a<*x<*m? We would just have to look at the particular order relation, <*, to find it. We would have to find what we think is m too. Wait what? A moment ago m was the minimum of S, now you have some a that's smaller. What is the argument you are making? Edited yesterday at 06:28 PM by wtf 0 Share this post Link to post Share on other sites

discountbrains 32 Posted 22 hours ago 1 hour ago, wtf said: >Wait what? A moment ago m was the minimum of S, now you have some a that's smaller. What is the argument you are making? Just because I labeled a number, m, doesn't mean its a min of the set. I'm only saying we're picking a number and testing it to see if its the min of the set. But, we know and I agree there are some sets (natural numbers etc)that can easily be well ordered. And, we can come up with a lot of well orders for a lot of sets. BUT, we cannot come up with an order that well orders all sets. My little one line above should really be my proof. 0 Share this post Link to post Share on other sites

wtf 125 Posted 22 hours ago 24 minutes ago, discountbrains said: Just because I labeled a number, m, doesn't mean its a min of the set. I'm only saying we're picking a number and testing it to see if its the min of the set. But, we know and I agree there are some sets (natural numbers etc)that can easily be well ordered. And, we can come up with a lot of well orders for a lot of sets. BUT, we cannot come up with an order that well orders all sets. My little one line above should really be my proof. I didn't see an argument. Can you repeat it? All you said is that 0, say, is less than or equal to all natural numbers; and 14 is less than or equal to all of them except for some. But so what? There's no argument there. > BUT, we cannot come up with an order that well orders all sets. We can't "come up" with one, but we can easily prove that one exists in the presence of the axiom of choice. And even without choice, the reals might be well ordered even if some other set isn't. You haven't got a proof. 0 Share this post Link to post Share on other sites

discountbrains 32 Posted 21 hours ago 3 hours ago, wtf said: And even without choice, the reals might be well ordered even if some other set isn't. There is no ONE subset that has no min for all order relations. Its just, given any <*, there is always a set of the type {x: a <*x} that can be constructed. 0 Share this post Link to post Share on other sites

wtf 125 Posted 21 hours ago 13 minutes ago, discountbrains said: There is no ONE subset that has no min for all order relations. Its just, given any <*, there is always a set of the type {x: a <*x} that can be constructed. That's just not a coherent argument. I can't understand what you are trying to say. 0 Share this post Link to post Share on other sites

discountbrains 32 Posted 21 hours ago (edited) Allow me to ask for your take on my debunking of a portion of Einstein's relativity I posted on the physics forum a year ago. It involves one result of his theory and simple arithmetic. I believe uncool commented on my post there at the time: 1) Suppose someone is sitting at a table with a wire going across it perpendicular to him. He measures the length of the wire and finds it 3ft long. A current flows through the wire and he adds up the incremental distances between the near light speed moving charges. Because Einstein's length contraction formula says all these incremental lengths become much smaller the sum of all the incremental lengths becomes much smaller than the original length of the wire.The observer measures it one way and gets one answer; measures another way and gets much shorter. Which is it? Optical illusion? 2) Two rocket ships are traveling on the same path from one planet to another at the same speed of 0.9999c. One is just 100mi from the 2nd planet; the trailing ship is 100mi from the 1st planet. With relativity length contraction adding up the distances the planets, to an observer off to the side, are much, much closer than thought. If the ships are closer together the total distance becomes much larger. 28 minutes ago, wtf said: That's just not a coherent argument. I can't understand what you are trying to say. Wrote this hurriedly and edited it. I'll simply state: For any order relation, <*, on the reals (not natural #s etc) a set of the type {x: a<* x} for some a can be constructed. So we don't go around and around again on this we assume there are numbers in the set which are comparable by <*. Edited 20 hours ago by discountbrains add commas 0 Share this post Link to post Share on other sites

wtf 125 Posted 20 hours ago (edited) 50 minutes ago, discountbrains said: Allow me to ask for your take on my debunking of a portion of Einstein's relativity I posted on the physics forum a year ago. ... Wrote this hurriedly and edited it. I'll simply state: For any order relation, <*, on the reals (not natural #s etc) a set of the type {x: a<* x} for some a can be constructed. So we don't go around and around again on this we assume there are numbers in the set which are comparable by <*. I have no comments on matters of physics. > For any order relation, <*, on the reals (not natural #s etc) a set of the type {x: a<* x} for some a can be constructed. Ok. The usual order on the reals is a total order. We can form the set of all reals > 0 for example. What of it? Where is your argument? Premises, reasoning, conclusion. I see no argument. Edited 20 hours ago by wtf 0 Share this post Link to post Share on other sites

discountbrains 32 Posted 18 hours ago This is easy. Let S = all real numbers x such that x > 0. There is no least element in S. Of course this would be one of my sets. 0 Share this post Link to post Share on other sites

wtf 125 Posted 18 hours ago (edited) 13 minutes ago, discountbrains said: This is easy. Let S = all real numbers x such that x > 0. There is no least element in S. Of course this would be one of my sets. Correct. You just proved that the usual order on the reals isn't a well order. However you have NOT shown that NO linear order on the reals can be a well order. Your proof fails. I looked back a couple of pages but could not find the supposed new version of the argument. I see no argument at all being made. Of course it is true that you can find a countably infinite ascending sequence of reals in a given linear order. For example let [math]<^*[/math] be defined as: all natural numbers [math]<^*[/math] everything else Now you have a countably infinite well-ordered subset. But we could extend it up through all the countable ordinals and you still wouldn't have a proof. You have not presented a proof that there is NO possible well order on the reals. But our current problem is that you have not presented a complete argument in quite some time; and I do not understand the purpose of the things you're saying. You take the trouble to point out that the usual order on the reals isn't a well-order? Why? We all agree that it isn't. Why bring it up now? Please give a complete argument. Give me something to work with. Nothing you say makes sense. In YOUR mind you think you're filling in details to something you've already explained. From where I sit there is no official version of your argument that I can refer to. I have no idea what you are talking about. Please: a) Point me to the post in this thread that expresses the coherent argument you're assuming I know; or b) Give such a coherent argument here. Write a complete argument so I can understand what you're thinking so I can respond. What is the point of your one-liners? I have no idea WHY you are telling me these things. Edited 18 hours ago by wtf 0 Share this post Link to post Share on other sites