uncool

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).

I don't believe either of us said that any order relation has a minimum. You are the one who made a positive statement; it's your job to support it.

Why? Please write out the proof of this statement.

What I mean is that he could basically say "OK, maybe some intervals are empty", and continue with his proof. The existence of empty intervals doesn't affect the rest of his proof.

Why?

Sorry, my typo. I meant to say "A set being well-ordered isn't based on that set having a minimal element. It's based on every nonempty subset of that set having a minimal element." I did not change what you said. Check your post. However, I took your meaning correctly and my post remains accurate even if the word "every" is removed. While that is true, the method you have used proves that they would have to have the same order. Your proof says there are z0 < z1 < z2 < ..., and that every element of T must be listed as some z.

It producing the same order is an immediate implication of the way you tried to prove it. If z_i <* z_j, then i <* j, by the definition of the z_i. And this conclusion can't be true, as shown by the existence of countable well-ordered sets that aren't isomorphic to N. The problem is in this line: "since for any z you choose in T it is the min for every subset of T, {x: z ≤* x, z and x in T}, so its also in this set of minimums". There is no reason to think that the list z0, z1, z2, ... will include the minimum of every subset of T. Side note: this reformatting of text to get larger and larger is distracting. It doesn't. A set being well-ordered isn't based on that set having a minimal element. It's based on every nonempty subset of that set having a minimal set. The empty set has no nonempty subsets, so trivially, every nonempty subset of the empty set has a minimum.

I take the condition the set is nonempty in the established definition of WO to mean we just don't consider empty sets. We "don't consider" empty subsets when determining whether a set is well-ordered. That's not the same as saying that every subset of a well-ordered set is nonempty, which your reasoning seems to imply. Nor does it mean that a well-ordered set can't itself be empty. I sincerely doubt that there is a workaround. I'm surprised you aren't aware of the objection, as it's essentially the same objection I've been making for a while now.

I do think that the error you've identified is a relatively small error that can be worked around. The point he's aiming for is that the set T must be denumerable; if he instead uses countable, then should the sequence ever stop because a subset is empty, then T must be finite. The larger error is in his claim that the "mins" must eventually exhaust T.

That is incorrect. The empty set has a unique order, and that order is (trivially) a well-ordering. I am saying that the definition of a well-ordering is that a set S and order <* is a well-ordering if every nonempty subset T of S has a minimum. It says nothing if T is empty. Going back to your proof: your proof, between the lines "Now if we assume <* well orders T" and "Hence z is an element of this string of minimums above" can be condensed to the claim that any infinite well-ordered set must be order-isomorphic to the positive integers. And that's easily shown to be false. Your proof cannot work on those lines.

I can't quite read this line: "Note T can be nonempty since we can choose any [???] and b elementof S"; however, you have already chosen a and b at this point. Your proof would be better if you chose a and b appropriately to begin with. "To be sure, suppose z is an arbitrary element of T. z is the min of some subset of T since there exists a subset of T of all elements greater than or equal to z. Hence z is an element of this string [...]" This doesn't follow.

Why? Well-ordering says that any nonempty set has a minimum. If you are using the definition of well-order to say that T has a minimum, you must show T is nonempty. What is it, at this point? Please post it in full - with no elisions, no "This is obvious", no "This should be easy to see".

It is perfectly natural. This doesn't follow. Naturally defined sets can be empty.

As you said: many small things can add up to something bigger. But those small things, too, must exist. My point is that gravity must arise out of something that exists, even on the microscopic scale.

But its properties are based on the properties of those molecules.

If gravity "doesn't exist in the microscopic world", then how can it affect anything in the macroscopic world? The microscopic and macroscopic worlds aren't entirely separate; the macroscopic world has to "grow out of" the microscopic world.

Why? At most, all you've done is restate the basic premise: more particles exist than antiparticles. Just saying "More particles became real" doesn't explain the broken symmetry.

Newtonian physics and classical (non-quantum) physics are the standard examples. Of course, when saying that, The Relativity of Wrong is required reading.

As a note, the term "lower bound" refers to a specific order; something that is a lower bound with respect to one order may not be for another order. I have been assuming that in each case, "lower bound" referred to the <* order.

I truly don't. I don't even see a reason why 0 should be a lower bound, let alone the greatest lower bound. It's an order, yes. That doesn't mean you can directly copy statements that are true for other orders. Arbitrary orders are defined as relations: given two numbers x and y, either x < y, or it isn't, with the requirements that either x = y, x < y, or y < x, and that if x < y and y < z, then x < z. That's it. wtf's list clearly defines an order on the reciprocals of positive integers, which is even a well-ordering.

Why?

That's not how arbitrary orders are defined. Why must there be such a q? Your earlier attempt ("S = {m,n,o,...}") was better; at least it dealt with the actual definition and question at hand, and the argument could at least be repurposed to something somewhat meaningful. This "argument" is an attempt to ignore definition.

I'm sorry, I couldn't make heads nor tails of what you were trying to say there. What is your proof that for any order relation <*, the set S = {0 <* x <* 1} has no minimum? I am not asking you to show that S is a set. It is a set. I stipulate that. But your proof depends on your claim that S has no minimum with respect to <*, and you have yet to prove that claim.

Why not? Then prove it.

Why should this exhaust S? You keep trying to wave away key parts of your proof. Something you should learn when attempting a proof: the things you most want to handwave are often the most important parts of the proof. Your attempted proof seems to come down to the claim that any infinite well-ordering is isomorphic to the usual ordering on the positive integers. That is, if a well-ordering is infinite, then it looks like m<n<o<..., and every element must be represented in that sequence. Correct?