taeto

No problem at all. I am being vague on purpose in this thread. If caught, it is fine too. Somehow maybe I will escape...

For \(x = \omega\) you have \(y = \omega +1\), so no escape clause needed. One is generally enough. Excellent. So you have now seen how to prove that between two real numbers there is a rational number. Now that you have seen how that works, why do you insist that there is anything wrong with it?

So all the proofs are just wrong, but you are not going to tell us where they go wrong?

A well-ordering is pretty much precisely the example of an ordering < such that if x is any given element, there is one particular element y such that x < y, and there is no z for which x < z < y. The usual ordering < of the real numbers does not have this property, since if x and y are any real numbers with x < y, then z = (x+y)/2 does lie between x and y and is different from both.

By speaking of the "next highest" irrational number, you appear to assume a well-ordering of irrational numbers. However, the usual and familiar linear ordering of the reals is not a well-ordering. But I think that in the context of the OP, we are supposed to deal with just the usual ordering. Would you like to see me write out a proof that for any two real numbers x and y with x < y there are infinitely many rational numbers q such that x < q < y, or would you prefer to work out the easy proof for yourself?

Then you would have to agree that between any two distinct reals there is also a `full set' of rationals, i.e. as many rationals as in \(\mathbb{Q}\) altogether?

There is indeed. But the point was that even though the set of rational numbers has a smaller size than the set of real numbers, it is also true for any two distinct real numbers that there are infinitely many rational numbers between them.

That is a very reasonable comment. It brings us to a more familiar scenario. Suppose we started with the set of all real numbers, and then we took away all the rational numbers from it. Then we have \(\mathbb{R}\setminus \mathbb{Q}\). Removing a countable set \(\mathbb{Q}\) from the uncountable set \(\mathbb{R}\) produces another uncountable set, which has all its pairs of distinct points separated. Are you comfortable with that?

The wiki article also has Norton's "argument" for uncountability of the endpoints. Do not trust everything that you read there. The left over points are those real numbers in [0,1] that have representations in base 3 with 0 and 2 as the only digits. That makes it particularly convenient to prove that the set of left over points is uncountable; changing every digit 2 in the base 3 representation to a 1 defines a bijection with the binary representation of the whole interval [0,1]. E.g. 1/4 represented in ternary by .020202... maps to .010101... which is 1/3 represented in binary. The endpoints are precisely those points that are represented in base 3 with only finitely many 2's and no 1's.

Good point. I mean the set of natural numbers (without zero): \(\{1,2,3,\ldots\}.\)

It is not supposed to be rigorous; it is sufficient that it explains what goes on, because your understanding is slightly off. You like the example of the power sets of {1}, {1,2}, {1,2,3},... We can stick to that, since it is simpler. Each set {1,...,n} has n elements and its power set has 2^n elements. The limit of the sequence of sets {1},{1,2},{1,2,3}... is equal to \(\mathbb{N}\). Why? Because every element n of \(\mathbb{N}\) belongs to one of these sets, and to all the following sets as well. The latter is important, if n drops out again after a while, then n would not be in the limit of the sequence. In fact even supposing that n re-enters again, but drops out again infinitely often, then n will not be in the limit. That is how limits of sequences of sets work, do you agree? The power set of a set {1,2,...,n} in the sequence contains the 2^n subsets of this set. Consider the infinite sequence of these power sets. Which elements belong to some of the sets in this sequence? They have to be subsets of at least on set {1,2,...,n}. In order to stay in the limit of the sequence they have to be subsets as well of all of the sets {1,2,...,n+1},{1,2,...,n+2}... after that. In our case that condition is obviously satisfied, so it is enough that a subset of \(\mathbb{N}\) is a subset of just one set {1,2,...,n} to guarantee that it will be in the limit. Hence this condition becomes necessary and sufficient for a subset to be an element of the limit of the sequence of power sets. If you still think that there are infinite sets in the limit of the power sets of {1},{1,2},{1,2,3},... please point to any one in particular, then we can consider whether it is in the limit or not.

I thought that I did just that. Or you do not believe that the \(2^{n-1}\) new endpoints that are created at each stage can be numbered with integers from \(2^{n-1}+1\) to \(2^n\)? You have to think more carefully about the "limit" of those power sets. Their limit has to be the same as just their union. The subsets of \(\mathbb{N}\) that appear in the union of these power sets are precisely those subsets that appear in at least one of the sets \(\{1,2,\ldots,n\}\). Such sets are finite subsets of \(\mathbb{N}\). When you take the limit, you will never produce an infinite subset of \(\mathbb{N}\). And the set of finite subsets of \(\mathbb{N}\) is in fact countable.

It is tempting to think so. But then you would be assuming a kind of `continuity property' that is not true for cardinalities: it is not true that \( \lim_{n\to \infty} 2^n = 2^\infty,\) where you can replace \(\infty\) by \(\aleph_0\) if you feel like it. It is not a coincidence that the \(\infty\) symbol is used in such limits and not \(\aleph_0\). Indeed, you can clearly count all the endpoints as you proceed. An endpoint never gets destroyed once it has been created. At the stage when it gets created, you assign a unique number between \(2^{n-1}+1\) and \(2^n\) to it, so that all endpoints that are created at that stage get their own number, which they keep forever after. Every endpoint gets created at some stage, hence each gets a number, and all get different numbers.

The Cantor set can be described as the set of points left over in the closed interval [0,1] after successively removing the middle third open subset of every remaining interval. Then each removed set has the form (m/3^n,(m+1)/3^n). It leaves endpoints m/3^n and (m+1)/3^n. These are rational. Since \(\mathbb{Q}\) is countable, so is the set of endpoints.

What does it mean to "solve 2 different similar triangles"?

Very good point indeed. Though it is more like the other way around. The arithmetical statement that Gödel constructs is really an encoding of the statement "I have no proof from the axioms of arithmetic", which is what creates the sort of dilemma that if it does have a proof then it is false, whereas if it has no proof then it is true, but it is lacking of proof. This does not work for an axiomatic system based on an infinite amount of axioms, because you cannot build a finite statement like that which encodes all the axioms at the same time. It turns out that TA is safe from that point of view, from attacks by Gödel type statements. Moreover it is provably consistent and complete. At the cost of a huge amount of axioms.

You are not entirely wrong. But the textbook example of a massive star losing mass due to gravitational collapse only predicts a loss of around 75% of its initial mass. If it starts with a mass 16 times the solar mass, it should get to 4 solar masses left, which is enough that a complete collapse into a black hole will happen. That is a pretty massive star to begin with. But a star like \(\rho\) Cassiopeiae which is visible with the naked eye from the earth is 40 times more massive than the Sun.

Okay, nice pics. A Briggsian logarithm has base 10, but the base doesn't really matter here. Why do you want to transform the slope between peaks by a logarithm, does that have any meaning at all? If two different peaks have the same height, why would it make sense to equate that to a disaster, since the logarithm of zero is undefined. And when peaks get very close in height, the logarithm of their difference in height decreases to minus infinity. Your logarithmic transformation does not seem to make sense.

What does it mean to "log transform a variable" the way you use the expression? Reference, if possible. If negativity is a problem, then why can you not use that -90 degrees is the same as +270 degrees? Why can no angle be smaller than -90, i.e. \( -\pi/4\). What would be wrong with an angle of -135 degrees?

That tells me that it does not make sense to introduce infinity. Now answer this: who introduced infinity into this thread, you or I? That is a stupid and dishonest statement. There are lots of different uses of \(\infty\) in mathematics, and I had to ask which one you mean to address, which I did. That is not evasion. You on the other hand have evaded this question entirely. You explained \(dx\) as something having to do with \(\infty\), but you did not explain what you mean by \(\infty\). The sources do not mention the "duality", because they do not know about it. How would they have learned about it until your posts here? You are completely dishonest. You are the one who told us all that \(dx\) is not zero. You just applied a limit argument right there. I can verify that. It means that you are making up your own stuff, because you seem convinced that \(1/\infty\) is not zero. Which problem? There is no problem in calculus. The problem is to somehow explain things to you so you may grasp it.

Not really. Uncool gave a nice example in group theory. The statement "G is not abelian" can be formulated in group theory. If G is not abelian, then there are elements \(x,y\) in G for which \(xy \neq yx.\) It is not a theorem in group theory that G is abelian, nor is the negation a theorem. It can happen for a group G that no such \(x,y\) exist, but there is nothing in the rules or axioms which suffices to deduce that this is the case. Peano Arithmetic (PA) is much similar. To say that a given Diophantine equation of a particular shape has a solution can be stated in the language of PA, but there is nothing in the rules or axioms to say whether it is true or false. If you argue that if there is nothing in the rules or axioms to prevent it, then it must be true, then you are wrong. If there is something to prevent it, then whatever it is, it is not in the rules or axioms. True Arithmetic (TA) is different, see https://en.wikipedia.org/wiki/True_arithmetic. Any correctly formulated statement has a proof or a disproof from the rules and axioms. However, TA is mostly avoided for being too cumbersome. In particular it is impossible to formulate TA with finitely many axioms; you need an infinite collection.

Why would you need a "method" of choosing a least element? There is a least element or there is not. If there is one, there is no "choice", since it is already there and it is unique.

Any nonempty subset. If you leave out AC it is undecidable if a WO of the reals exists.

Your scenario is very different. The question that you ask has to be a valid statement within a particular theory. The variables that you use are constrained to be objects of that theory. If you are working in Peano arithmetic, they have to represent integers. In rational arithmetic they represent rational values, etc. It does not make sense to answer a question about the squareroot of two in rational arithmetic by referring to the theory of the real ordered field. Exactly in the same way that it makes no sense to answer a question about dividing 1 by 2 within the ring of integers by invoking rationals. In fact, you would be given the information that no solution exists, since there is no rational number the square of which is equal to 2. This is correct, and whichever counterassertion you make will be flawed. The scenario that I am referring to is one in which you have a clear cut question about an equation involving integer constants and variables. You are restricted to attempt a solution solely within the realm of the integers. You are given two pieces of information: nothing prevents the existence of a solution, and nothing prevents that there is no solution. In particular, there is no proof of the existence of one, nor a disproof. Whether there is a "solution" outside the given restrictions of having to assign integer values to the variables has no relevance.

You are given a Diophantine equation. A humperdink is defined as a solution to the equation. You are told that nothing in the rules or axioms prevents that a humperdink exists. You are told that nothing in the rules or axioms prevents that a humperdink does not exist. Both are correct. At this point you cannot assert that a humperdink exists, nor can you assert that none does, based on rules and axioms alone.