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discountbrains

Where to submit my proof that the set of real numbers can't be well ordered

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OK, of course its necessary for me to be precise. When I speak of (a+b)/2 I mean for a to be the min of an <* ordered set and b to be the very next number. So, in our above example 3 is not in our set with with min a=2. (a,b) is meant to be {x: a < x < b}  I hope I pretty much covered everything. I'm looking forward to seeing examples and corrections. Its fun to play with them. If I'm wrong, I'm wrong.

Edited by discountbrains

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1 hour ago, discountbrains said:

OK, of course its necessary for me to be precise. When I speak of (a+b)/2 I mean for a to be the min of an <* ordered set and b to be the very next number. So, in our above example 3 is not in our set with with min a=2. (a,b) is meant to be {x: a < x < b}  I hope I pretty much covered everything. I'm looking forward to seeing examples and corrections. Its fun to play with them. If I'm wrong, I'm wrong.

You haven't said anything meaningful here. You're very far from your original ideas. I simply do not have any idea what "our set" is. Is it like "our gang" from the 1930's films with those annoying rugrat kids? I want you to tell me WHAT SET you are talking about. Show me the set, tell me the order, and then make your point. You are constantly confusing the usual arithmetic with a nonstandard order. By refusing to notate your set with set brackets and then tell us what the order is, you are confusing yourself and giving me nothing to respond to. Your posts recently are incoherent. You were making much more sense (although you were still wrong) pages ago. You're not giving me anything to work with.

> OK, of course its necessary for me to be precise.

Yes. So do that. Here's the pattern I'm looking for.

* Identify some set by listing its elements in set brackets. {x, y, z, w, q}

* Tell me what the order is. <z, w, q, y, z>. 

* Then make your point about (a+b)/2, making sure you understand that this is the USUAL addition on the integers and does not necessarily apply to a nonstandard order.

I think that when you force yourself to be explicit and precise, you'll see that your argument has problems.

 

 I hope I pretty much covered everything.

You didn't cover anything. I don't know what "our set" is and you constantly change around orders and make vague claims that are neither true nor false since it's unclear what they refer to.

 

Edited by wtf

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I believe we are "not singing on the same page" (the original expression which people don't seem to get and is actually humorous to visualize). Maybe I should generalize my subset S from being (a,b) to any subset. Here is your example S = <1, 2, 3, 4, ....> in the usual order where 1 < 2 < 3 < 4 <... in this order I write (1 + 2)/2 = 3/2, (2 + 3)/2 = 5/2, ...None of these numbers are in S. You write S = < 1, 3, 2, 4, 5,...> which of course is well ordered. All my (a + b)/2 were calculated using the usual order and all come out  as fractions and therefore not in S so we can't talk about their order in S. Early on I stipulated (a + b)/2 must be a member of the set S. I'm taking S to be the very same set (a,b) as it was with the usual order. Every x in S is the very same x as it was originally. Order doesn't change that.

I'm just playing with your example here; I'm not trying to make any further point about what I previously stated.

I keep believing I can come up with yet another way to prove my original claim. I welcome seeing more counterexamples. 

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55 minutes ago, discountbrains said:

I believe we are "not singing on the same page"

I suspect you are mixing up two different idioms : "singing from the same hymn sheet /book" and "to be on the same page"

OK. Now back to the topic!

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On 11/10/2018 at 10:42 AM, discountbrains said:

I believe we are "not singing on the same page" (the original expression which people don't seem to get and is actually humorous to visualize). Maybe I should generalize my subset S from being (a,b) to any subset. Here is your example S = <1, 2, 3, 4, ....> in the usual order where 1 < 2 < 3 < 4 <... in this order I write (1 + 2)/2 = 3/2, (2 + 3)/2 = 5/2, ...None of these numbers are in S. You write S = < 1, 3, 2, 4, 5,...> which of course is well ordered. All my (a + b)/2 were calculated using the usual order and all come out  as fractions and therefore not in S so we can't talk about their order in S. Early on I stipulated (a + b)/2 must be a member of the set S. I'm taking S to be the very same set (a,b) as it was with the usual order. Every x in S is the very same x as it was originally. Order doesn't change that.

I'm just playing with your example here; I'm not trying to make any further point about what I previously stated.

I keep believing I can come up with yet another way to prove my original claim. I welcome seeing more counterexamples. 

Oh I see. By "not in the set," you mean not defined. 5/2 is undefined in the integers. Ok, now that I understand your terminology, your argument is somewhat less incoherent but is still unclear.

So now that I understand what you mean, can you clearly restate your argument? The fact that 5/2 is undefined in the integers surely doesn't show that the reals can't be well ordered. 

> I keep believing I can come up with yet another way to prove my original claim.

Another way? You haven't shown one way.

Edited by wtf

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ps -- Ok. <1,2,3,4>. And there's no integer between 2 and 3. What of it? Surely that's not surprising in a well-ordered set where addition is defined. Can you write down a complete, line-by-line argument? 

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I'm going to prove the reals cannot be well ordered another way: Let S be any subset of R such that for any a,b  S there is an x  a < x < b. Let abe the min of S since its claimed every subset has a minimum. Construct {a}\S. There is also a min for {ai}\S. Call it ai+1. There are x, y, z, ... such that  If these two as are minimums of S and  {a}\S,  x, y, z,... can't be between them; there are no x such that a<* x <* ai+1. Therefore, any such x, y, z, ... with  a< x < y <z,...<  ai+1  must be greater than ai+1 . But, what this does is reverse every element of any subset S or R. That is, now for any x, y, z,... the order is now z <* y <* x or I'll write       .... <* z <* y <* x. This says any time we try to WO we can't avoid merely reversing the order and the numbers are dense in R as before, yet well ordering requires them to be separated. And, we get another contradiction here. Its left to the reader to show any required reordering of the numbers from the usual order < results in a totally reverse order.

Try anytime you have a < x < b for any a and b moving x such that a <* b <* x and see what you get.

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I suspect you are mixing up two different idioms : "singing from the same hymn sheet /book" and "to be on the same page

Oh, no, no, No! Think about it, everyone in church has their hymn books out and are singing and you got someone singing louder than everyone else and in conflict with everyone because they're on the wrong page. I think this is hilarious.

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2 hours ago, discountbrains said:

I'm going to prove the reals cannot be well ordered another way: Let S be any subset of R such that for any a,b  S there is an x  a < x < b. Let abe the min of S since its claimed every subset has a minimum. Construct {a}\S. There is also a min for {ai}\S. Call it ai+1. There are x, y, z, ... such that  If these two as are minimums of S and  {a}\S,  x, y, z,... can't be between them; there are no x such that a<* x <* ai+1. Therefore, any such x, y, z, ... with  a< x < y <z,...<  ai+1  must be greater than ai+1 . But, what this does is reverse every element of any subset S or R. That is, now for any x, y, z,... the order is now z <* y <* x or I'll write       .... <* z <* y <* x. This says any time we try to WO we can't avoid merely reversing the order and the numbers are dense in R as before, yet well ordering requires them to be separated. And, we get another contradiction here. Its left to the reader to show any required reordering of the numbers from the usual order < results in a totally reverse order.

Is there some reason why you jam everything together into one run-on paragraph that makes your exposition difficult to read? Can't you please write one or two ideas per line and put in some whitespace? Like paragraphs, but for math.

Fortunately, there are fatal errors in the first couple of lines, so it's not necessary to read the rest of it. 

Let S be any subset of R such that for any a,b  S there is an x  a < x < b.

You have been meticulous -- for which I commend you -- in using < for the usual order, and <* for an alternate order. So I assume < is the usual order on the reals. In which case let S = the rationals ...

> Let abe the min of S since its claimed every subset has a minimum.

... and the rationals have no minimum. So your argument is dead in the water right here. It's over. I need read no further. 

* So, can I see if I can figure out if you mean something else? Perhaps by < you mean <* so that S is a subset of (R, <*). But then S can not be a dense order. Every subset of a well-ordered set is well-ordered. So if the reals are well-ordered by <*, no subset of the reals could be densely ordered by <*. So under this interpretation, where you meant to write <* instead of <, your proof is also dead in the water.

* Ok perhaps you mean that the reals are well-ordered by <*, and S is a subset that is densely ordered by the usual order. Let S be the rationals again. And let's say they're well-ordered by <* as \(q_1, q_2, q_3, \dots\).

But what can this have to do with anything? I didn't read the rest of it because there is no relationship between the usual order < and the well-order <*. I'm only mentioning this interpretation to ask if this is what you mean. That the reals are well-ordered by <* and S could be the rationals, dense in the usual order. Is this where you're going? And why? No argument you can make along these lines could work.

I skimmed the next sentence or two and as usual you are taking a well-ordered set, claiming you have a subset that's densely-ordered by the same well-order, which of course doesn't exist; then you try to argue a contradiction.

You keep making the same mistake over and over of confusing < and <*. If the reals are well-ordered by <*, so is every subset of the reals, under the same order. And any well-ordered set (or subset) can not be densely ordered by <*. So there is no S and your proof is nonsense after the first line or two.

ps -- I skimmed the rest, your idea of reversing a well-order is nonsense. You're just confused about < versus <*. Again, taking S to be the rationals, they're densely ordered and have no minimum element under <. And they can't be densely ordered under a well-order <*. Two different things. 

Edited by wtf

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ps -- If you would contemplate a well-order on the rationals, you will see why your reverse idea is wrong. There's a first rational and a second and a third and so forth. But the well-order doesn't need to reverse the usual order or whatever you're trying to say.

 Its left to the reader to show any required reordering of the numbers from the usual order < results in a totally reverse order.

Well-order the rationals by, say, the Cantor pairing function. You'll see that you're wrong.

https://en.wikipedia.org/wiki/Pairing_function

And think about it. If we reverse the order on the rationals, they're still dense. The reverse order on the rationals isn't a well order. In the usual order, 1/2 < 3/2. In the reverse order, 3/2 <* 1/2. SO WHAT????

Exercise: Show that the reverse of a dense order is also dense. 

Edited by wtf

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I discovered my Cantor type proof has some problems. I'll repeat my (a + b)/2 proof:

Let S =(0,1), R being well ordered means there's an a which is a min(S, <*). Consider {a}\S, There is a minimum, b, for {a}\S and there can't be any x in S with a <* x <* b. Now, (a + b)/2 ∈ (S, <) so a <*(a + b)/2 and it can't be less than b; therefore, it must be greater than b. We must have b <* (a + b)/2.

Now for some arithmetic: Suppose b = 0.4 and 2b = -2 and suppose b = 6 + 7 +5 where + = +*. In my case this doesn't matter because  b + b <* a + b is the same as b + (b - b) <* a.  b - b = 0 and b + 0 = b. Therefore, b <* a which is a contradiction. Finally this says assuming all subsets of the reals can be ordered so they have a least element leads to a contradiction.

I'll think about your rational number thing. I like my latest reordering proof. I need to clean it up and make sure its perfectly clear.

The problem with the Cantor type proof is I can show there is a countable set of minimums for subsets of S = (0,1) and another number exist in S that's not in this set of minimums, but there may need not be any reason why it should be in the set of minimums.

I

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Yeah, I know. 

10 hours ago, uncool said:

Please define your "+*".

Yeah, I know. I just threw that in. It seems to me though that whatever some sort of addition + means x + (-x) = 0 and 2x should be x + x.

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If you are trying to use certain properties for your addition, you will have to prove that you can define such an addition. 

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14 hours ago, discountbrains said:

 

I discovered my Cantor type proof has some problems. I'll repeat my (a + b)/2 proof:

Let S =(0,1), R being well ordered means there's an a which is a min(S, <*). Consider {a}\S, There is a minimum, b, for {a}\S and there can't be any x in S with a <* x <* b. Now, (a + b)/2 ∈ (S, <) so a <*(a + b)/2 and it can't be less than b; therefore, it must be greater than b. We must have b <* (a + b)/2.

Now for some arithmetic: Suppose b = 0.4 and 2b = -2 and suppose b = 6 + 7 +5 where + = +*. In my case this doesn't matter because  b + b <* a + b is the same as b + (b - b) <* a.  b - b = 0 and b + 0 = b. Therefore, b <* a which is a contradiction. Finally this says assuming all subsets of the reals can be ordered so they have a least element leads to a contradiction.

I'll think about your rational number thing. I like my latest reordering proof. I need to clean it up and make sure its perfectly clear.

The problem with the Cantor type proof is I can show there is a countable set of minimums for subsets of S = (0,1) and another number exist in S that's not in this set of minimums, but there may need not be any reason why it should be in the set of minimums.

I

I would like specific responses to the two points I made on your previous post.

* If <* well-orders the reals; and S is any subset of the reals; then:

1) <* also well-orders S; hence S can NOT be a dense order; and

2) If < is the usual order on the reals, then the rationals are a dense set (under <) with no least element, no matter what order you put on R.

Please clearly state your understanding or questions, agreement or disagreement, with these statements.

Edited by wtf

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I should have stuck to my original argument at the very start. You people got me side tracked; its all your fault. 

I started with: consider the subset of the reals S = ((0,1), <), < the usual order. If R is well ordered there must be a number a in S such that a ≤* x where x is any number in S and <* is the well ordering. Now if we delete a from S we have {a}\S = {x: x ∈ S and a <* x, a ≠ x}. BUT, we can plainly tell {a}\S contains no <* minimum number! We claimed <* well orders R, but clearly {a}\S ⊂ R and it has no least element which contradicts our assumption <* WOs R. NOTE: if <* is an order relation on R it applies to all elements in R; therefore, if <* is the order chosen for S it also applies to {a}\S. You can't pick a different order relation to give you what you want for every set you're considering. We defined {a}\S using <* and its claimed here that WO guarantees {a}\S has a minimum and the AC guarantees it. But, we can't find it!

I don't have to show S exists; its already the same numbers of the interval (0,1) with we started with; its just reordered by our supposed <* order relation.

I'll get to your latest questions, but the above is the most straightforward proof. PLEASE NOTE: All I need to do is show there is one set that no matter which order you choose no least element (minimum) can be found. wtf seems to think I need to prove the opposite (apparently) that all subsets have no minimum; hence, he keeps bringing up natural and rational numbers. 

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1 hour ago, discountbrains said:

I should have stuck to my original argument at the very start. You people got me side tracked; its all your fault. 

 

I read this and cracked up laughing. I hope you meant this as a lighthearted joke. But I for one will be very happy if you can stick to one topic so that we may attain clarity.

I'll get to the rest of this later. Just wanted to let you know you made me laugh, and hopefully that was your intention.

1 hour ago, discountbrains said:

wtf seems to think I need to prove the opposite (apparently) that all subsets have no minimum; hence, he keeps bringing up natural and rational numbers. 

Skimming ahead I found this line. On the contrary, I'm well aware that if the reals are well-ordered by <* AND you can find a single nonempty subset of the reals that is not well-ordered (ie that has no least element under <*) then you would have proved your claim.

But this can not happen. Every nonempty subset of a well-ordered set is well-ordered by the same order.

So your quest is futile. If the reals are well ordered by <* then so is every nonempty subset. 

I have pointed this out to you, I don't know, six or eight times. I can't keep this up much longer.

I ask you directly: Do you understand that every subset of a well-ordered set must be well-ordered under the same order? Yes or no?

And (again for the sixth or eighth time), I ask you to contemplate well-orders of the rationals because they are key to understanding the uncountable case. The rataionals are a set that is usually densely ordered, and you want to understand the nature of any well-order it admits. That is why I am telling you to understand the case of the rationals. If you understand the case of the rationals, you will see why you are wrong about the reals.

In your first post you asked for someone who understands this material. You found me. And you refuse to listen or even engage with the specific points I'm making. It's frustrating.

Do you understand that every nonempty subset of a well-ordered set is well-ordered by the same order?

Edited by wtf

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3 hours ago, discountbrains said:

consider the subset of the reals S = ((0,1), <), < the usual order. If R is well ordered there must be a number a in S such that a ≤* x where x is any number in S and <* is the well ordering. Now if we delete a from S we have {a}\S = {x: x ∈ S and a <* x, a ≠ x}. BUT, we can plainly tell {a}\S contains no <* minimum number!

Ok. I see the subtle error you're making. Let's walk through this. Let's say that <* well orders the reals. And let's write out the first few elements of the well-order <*. Let's say they are: 1/4, 1/5, 1/pi, 1/2. It's convenient that they're's all in (0,1), but they don't need to be, this is just for illustration.

I hope you can see that after the well-order <* is applied to (0,1), the elements of that set are still well-ordered. They must be, since any subset of the reals is also well ordered by <*.

However, under <*, (0,1) is no longer an interval. Its points are scattered to the winds, distributed all over the real number line. So the notation (0,1) is a little misleading. Under <, the notation (0,1) indicates a particular subset S of the real numbers. Under <*, S is still the same subset. But it's no longer an interval. It's simply a well-ordered set whose elements jump around the real number line randomly, when you re-order the real line by <*.

Ok, you are correct that there is some element \(x_1 \in (0,1)\) that is minimum with respect to <*. In fact we know what it is, it's 1/4, that's the well-order we're assuming for illustration.

So now let's consider \(S \setminus \{x_1\}\). By the way note that's how you notate set difference, the larger set is on the left of the backslash, and the set you're subtracting is on the right. Minor notational point.

So \(S \setminus \{x_1\}\) itself has a minimum under <*, which we'll call \(x_2\) and which has the value 1/5 in our example well-order.

You see we can keep doing this, because (0,1) is well-ordered by <*. But its elements are scattered all over the place, and your INTUITION is leading you ASTRAY. You say:

> BUT, we can plainly tell {a}\S contains no <* minimum number!

But it does!! It's 1/5. Or in general, \(x_2\). And after you remove \(x_1\) and \(x_2\), then \(x_3\) is the smallest element under <* of what's left, in this case 1/pi. 

You have made a CLAIM and not forced yourself to supply a PROOF, so you have made an ERROR. \(S \setminus \{a\}\) in your notation DOES have a smallest element. It's just the next element of the well-ordering of (0,1) under <*. 

Do you understand this? You made a claim that's false. I suspect it's because you haven't quite intuited that (0,1) does get perfectly well-ordered by <* and its elements are scattered all over the place, throwing off your intuition.

To sum up: If <* well orders the reals, it also well orders (0,1). But under <* you can no longer think of (0,1) as an interval. It's just a random set of points in the reordered real numbers. This is confusing your intuition. 

 

Edited by wtf

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I have long been aware the numbers could be scattered all over the place. That's why I wrote S\(a) = {x: x ∈ S and a <* x, a ≠ x} and not as an interval.

EDIT to my above post: I should check with my analysis book, but 'a' being the minimum of S makes it the greatest lower bound of  {x: x ∈ S and a <* x, a ≠ x} which means there is no l. b. for {x: x ∈ S and a <* x, a ≠ x} greater than a, thus {x: x ∈ S and a <* x, a ≠ x} has no minimum with respect to <*. This is what my argument hangs on.

OK, I SEE WHAT U ARE SAYING. You're saying if we assume <* to be a WO of reals than it can order (0,1) as you said. I can accept this and this leads to not my result. But, I am saying it also leads to my result which leads to a contradiction.

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2 hours ago, discountbrains said:

I have long been aware the numbers could be scattered all over the place. That's why I wrote S\(a) = {x: x ∈ S and a <* x, a ≠ x} and not as an interval.

EDIT to my above post: I should check with my analysis book, but 'a' being the minimum of S makes it the greatest lower bound of  {x: x ∈ S and a <* x, a ≠ x} which means there is no l. b. for {x: x ∈ S and a <* x, a ≠ x} greater than a, thus {x: x ∈ S and a <* x, a ≠ x} has no minimum with respect to <*. This is what my argument hangs on.

OK, I SEE WHAT U ARE SAYING. You're saying if we assume <* to be a WO of reals than it can order (0,1) as you said. I can accept this and this leads to not my result. But, I am saying it also leads to my result which leads to a contradiction.

You have to be careful to distinguish < from <*. Your argument certainly doesn't lead to your result, since S \{anything} always has a smallest element. It must since S is well ordered by <*. Of course S\{anything} may not have a least element under < but that doesn't prove anything.

but 'a' being the minimum of S makes it the greatest lower bound of  {x: x ∈ S and a <* x, a ≠ x} which means there is no l. b. for {x: x ∈ S and a <* x, a ≠ x} greater than a, thus {x: x ∈ S and a <* x, a ≠ x} has no minimum with respect to <*. This is what my argument hangs on.

Your argument is wrong. Please write it out line by line clearly. I already explain why you're wrong. S is well ordered by <*. 

but 'a' being the minimum of S makes it the greatest lower bound of  {x: x ∈ S and a <* x, a ≠ x} which means there is no l. b. for {x: x ∈ S and a <* x, a ≠ x} greater than a, 

Simply false. Try that argument with {1,2,3,4,5,}. 1 is the least element. 2 is the least element if you delete 1. And so forth.

You are confusing yourself by bringing in greatest lower bounds. We're talking about well-ordered sets, not topological completeness. 

Edited by wtf

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My friend @discountbrains, every nonempty subset of a well-ordered set is well-ordered by that same order. 

When you understand this you will be enlightened.

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"My friend @discountbrains, every nonempty subset of a well-ordered set is well-ordered by that same order. 

When you understand this you will be enlightened."

You are being a great help to me. You are helping me refine my thoughts and I'm gradually sneaking up on a proper statement of my theorem. I'll change my statement to: There exist some subsets of the reals  that  don't have a minimum for any order relation.  This will keep you from saying I declared R is well ordered by some <* which keeps you from claiming what follows in your argument.

Then I just proceed as I have stated many times above. I'll just say if S has a <*min the rest follows.

Are u stating that if a set is WO the order produces a set like <1, 2, 3, 4,...>? If so this makes the set into a countable set and if the set is  ((0,1), <) there are far more numbers that are not accounted for so we don't know whether they're less than the min or what. 

 Don't know if I ever said what motivated me to make my claim. I used to go into little spurts from time to time over the years to try to come up with a WO for the reals and concluded that for any order relation definition there would always be a set satisfying exactly the opposite definition.  

 

 

Edited by discountbrains

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You are unlikely to ever come up with an explicit well order for the reals. Most theorems that depend on the axiom of choice lead to the existence of something that is difficult to explicitly describe.

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1 hour ago, discountbrains said:

"My friend @discountbrains, every nonempty subset of a well-ordered set is well-ordered by that same order. 

When you understand this you will be enlightened."

You are being a great help to me. You are helping me refine my thoughts and I'm gradually sneaking up on a proper statement of my theorem. I'll change my statement to: There exist some subsets of the reals  that  don't have a minimum for any order relation.  This will keep you from saying I declared R is well ordered by some <* which keeps you from claiming what follows in your argument.

Then I just proceed as I have stated many times above. I'll just say if S has a <*min the rest follows.

Are u stating that if a set is WO the order produces a set like <1, 2, 3, 4,...>? If so this makes the set into a countable set and if the set is  ((0,1), <) there are far more numbers that are not accounted for so we don't know whether they're less than the min or what. 

 Don't know if I ever said what motivated me to make my claim. I used to go into little spurts from time to time over the years to try to come up with a WO for the reals and concluded that for any order relation definition there would always be a set satisfying exactly the opposite definition.  

 

 

> You are being a great help to me.

Thank you so much for the kind words.

> This will keep you from saying I declared R is well ordered by some <*

Shame on me for ever saying that. But that's the whole point. You claim it can't be done. But you haven't proved it can't be done. I'm perfectly well allowed to say the reals can be well-ordered, since it's a theorem of ZFC, unchallenged since it was put forth by Zermelo in the early 1900's.

> There exist some subsets of the reals  that  don't have a minimum for any order relation. 

Well this is falsified by any well-order on the reals, since any nonempty subset of the reals must also be well-ordered by the same order. And why can't I say that? You haven't proved no such well order can exist. You can't start your proof by assuming there's no well-order, that's assuming the thing you are trying to prove. 

The burden is on you to prove no well order on the reals exists, since it is a theorem that there DOES exist such a well order.

> Are u stating that if a set is WO the order produces a set like <1, 2, 3, 4,...>? If so this makes the set into a countable set and if the set is  ((0,1), <) there are far more numbers that are not accounted for so we don't know whether they're less than the min or what. 

Good point. No, I'm just illustrating my use of angle brackets. Well-orders in general can be quite wild. For example there's the even-odd order on the naturals: <0, 2, 4, 6, ..., 1, 3, 5, 7, ..>. And there's the mod-3 order: <0, 3, 6, 9, ..., 1, 4, 7, 10, ..., 2, 5, 8, 11, ...>. You can see how you could keep going and eventually get countably many copies of the usual order on the naturals. It can be defined explicitly. 

Of course those are all countable well-orders. There are uncountable well-orders, and that can be proved even without the axiom of choice.

>  I used to go into little spurts from time to time over the years to try to come up with a WO for the reals and concluded that for any order relation definition there would always be a set satisfying exactly the opposite definition.  

See this stackechange thread on the question of whether there's a know well-order on the reals.

https://math.stackexchange.com/questions/6501/is-there-a-known-well-ordering-of-the-reals

The technical fact is that there is some model of the reals in which there is a definable well-order of the reals. But in the usual model, there isn't. It's advanced set theory. In ‎Gödel's constructible universe there is a definable well-order of the reals.

https://en.wikipedia.org/wiki/Constructible_universe

But nobody thinks the constructible universe is the right model for set theory. Philosophers and set theorists have been at this for a century. Plenty of literature out there but it's highly advanced math.

For example see this. https://mathoverflow.net/questions/71337/definable-wellordering-of-the-reals

Now Mathoverflow is a question/answer site for professional mathematicians. It's not expected that any of us mere mortals can even read the page I linked. But at least you can see that it's a hard question that mathematicians wonder about.

Now, back to earth. YOU CLAIM there is no well-order on the reals. If that is true, then surely there is some subset of the reals that can't be well-ordered. I perfectly well agree. But you haven't proved your claim; and it contradicts known math, at least if you accept the axiom of choice. And even without choice, there is an uncountable ordinal. And if you accept the continuum hypothesis, then that uncountable ordinal must well-order the reals! So this is all a very subtle business that depends on the axioms you choose and even on the particular model of your axioms.

 

Edited by wtf

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"YOU CLAIM there is no well-order on the reals".

You say I begin by stating something as fact and then use the statement to prove itself; thus, making a circular argument.  I don't know how I;m doing that. I just went through a simple illustration to show that  there exists at least one set that doesn't have a min  with any order relation.

This is like its established that x is defined to have certain properties if its divisible by 3 and I show x is not divisible by 3.

FWIW, I would like u to look at a post I put on Physics Forum on this scienceforum board. I submitted a flaw in Einstein's special relativity. I got no responses contradicting me; all I got were personal attacks. Haven't looked at the board for months though. I used very simple straightforward arithmetic. I don't say the whole theory is wrong. Fatal flaw?

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42 minutes ago, discountbrains said:

"I would like u to look at a post I put on Physics Forum on this scienceforum board. I submitted a flaw in Einstein's special relativity. I got no responses contradicting me; all I got were personal attacks. 

To one answer that you got in that thread you replied: "I got some concrete answers this time pointing out my error." 

You were lying then, or you are lying now? 

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