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 April 7th, 2017, 05:35 AM #11 Senior Member   Joined: Mar 2015 From: New Jersey Posts: 1,214 Thanks: 91 I can't construct anything from a set of points because a point has no size. $\displaystyle \infty$x0 doesn't exist. Points don't exist. Lines exist because I can draw them, and I can call locations on a line points. I can't create a line (or anything else) from points. Points are locations in space. No location, no point. I can't rotate locations, only the objects which occupy the locations, and an object doesn't consist of points. And you didn't answer my question. Does infinity occur any where in the proof? Actually, the proof is academic, since you are starting with something that just doesn't exist.
April 7th, 2017, 08:40 AM   #12
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 Originally Posted by zylo I can't construct anything from a set of points because a point has no size. $\displaystyle \infty$x0 doesn't exist. Points don't exist. Lines exist because I can draw them, and I can call locations on a line points. I can't create a line (or anything else) from points. Points are locations in space. No location, no point. I can't rotate locations, only the objects which occupy the locations, and an object doesn't consist of points. And you didn't answer my question. Does infinity occur any where in the proof? Actually, the proof is academic, since you are starting with something that just doesn't exist.
Have you glanced at the proof on Wiki? There are certainly infinitely many points in three space, and there are infinitely many rigid motions generated by the two main rotations about the x and z axes. Any axiom of choice proof always involves making a choice from an infinite collection of sets.

I'm confused. I don't recall you making these objections when we discussed limits. Limits are points, right?

Points in 1-space are real numbers. In 3-space they're triples of real numbers. I know you know this, so I don't understand what you're trying to say here. If points don't exist why do you study real analysis?

 April 10th, 2017, 07:31 AM #13 Senior Member   Joined: Mar 2015 From: New Jersey Posts: 1,214 Thanks: 91 You can't create geometry from points. Points are locations in space. For example, consider the line [0,1]. Color the rational points red and the irrational points green. Now rigidly move the red dots away. Do you have two lines? No 1) Both "lines" are missing points. 2) You can't rigidly move points because points are locations. You can only move the geometry on which they reside. You can rigidly move geometry (line, area, volume..) by specifying a new location for selected points, two points for a line, for example. 3) You can remove points from geometry but you can't create geometry from points. For example, remove .5 from [0,1] to get [0,.5) and (.5,1]. As a very, very rough visualization, you can remove atoms from the ocean with a spoon but you can't put atoms together to form an ocean, or car with a file. The same holds true by extension to any geometry: you can't create a square by lining up lines next to each other, but you can identify individual lines in a square. You can't create a cube by lining up squares but you can identify individual squares in the cube. If you know what you are looking for, you can spot this fallacy in the academic proofs you find by googling Banach Tarski Paradox. There is also a utube video* with over eleven million viewers that illustrates the same thing, but you have to ignore the blah blah and see the essential difficulty: you can't create geometry from points. * After you skip ad, go back to beginning. Ref: Point Set Geometry post#21.
 April 10th, 2017, 07:49 AM #14 Senior Member   Joined: Mar 2015 From: New Jersey Posts: 1,214 Thanks: 91 Ref https://en.wikipedia.org/wiki/Banach...Tarski_paradox In step 3 of the proof, A sphere is partitioned into orbits (shells). You can't do that anymore than you can partition a line into points or a square into lines. See previous post.
April 10th, 2017, 12:39 PM   #15
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 Originally Posted by zylo Ref https://en.wikipedia.org/wiki/Banach...Tarski_paradox In step 3 of the proof, A sphere is partitioned into orbits (shells). You can't do that anymore than you can partition a line into points or a square into lines. See previous post.
They're not shells, they're orbits under a group action. When you say shells I assume you are thinking of concentric spheres. If it were only so simple! It's not. The orbits are messy but perfectly well defined.

Ah, I see your confusion. These are not "orbits" as in astronomy. The orbit of a point is the entire set of points that the original point can be taken to by some isometry in the isometry group under consideration.

For example the "orbit of 1" in the integers under the action of "adding or subtracting 2" is 1, 3, 5, 7, ... and their negatives. It's all the results you can get by applying the actions repeatedly.

In any event, I don't understand your line of thought in this thread. When we discuss real analysis and someone says, "Consider the unit interval," you don't immediately object that no such thing can exist because it's made of points. You've never raised such an objection. Why do you raise this objection in three-space but not in one-space?

Last edited by Maschke; April 10th, 2017 at 12:42 PM.

 April 10th, 2017, 12:47 PM #16 Senior Member   Joined: Mar 2015 From: New Jersey Posts: 1,214 Thanks: 91 You can't create a unit interval from points. Place a point at 0. Now place another point next to it. Keep going. Report back when you get to 1. Same as trying to cover a square with lines or a sphere with orbits.
April 10th, 2017, 01:51 PM   #17
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 Originally Posted by zylo You can't create a unit interval from points. Place a point at 0. Now place another point next to it. Keep going. Report back when you get to 1.
Ok, fine. But that's not my question. My question is, why do you post questions about real analysis, which is subject to the exact same objections? Why do you ask questions about Rudin, rather than object on the basis that the real numbers can't exist because they're made up of points?

Quote:
 Originally Posted by zylo Same as trying to cover a square with lines or a sphere with orbits.
The sphere isn't being "covered" with orbits. If you believe the sphere exists, the orbits are just particular subsets of it. Of course if you are arguing that mathematical spheres don't exist, that's one thing. But if you grant that a sphere exists, an orbit is just one of its particular subsets. And orbits of group actions are constructively defined, they don't involve any axiom of choice magic.

Last edited by Maschke; April 10th, 2017 at 02:02 PM.

 April 10th, 2017, 05:20 PM #18 Senior Member   Joined: Aug 2012 Posts: 1,679 Thanks: 436 ps -- I'm going to hijack the thread to talk about orbits. Orbits are very cool and also very strange. Before tackling the orbits used in the Wiki proof of B-T, let's first think about the orbit of a simple rotation of a circle in the plane. Going back to basic trigonometry, a rotation of an object in the plane through an angle of $\varphi$ radians, denoted $r_\varphi$, can be thought of as a particular linear transformation of the plane. Any linear transformation of the plane to itself is completely characterized by its action on the standard unit basis vectors $e_1= (1,0)$ and $e_2= (0,1)$. A moment's thought (or a few minutes, or hours, or years ...) will convince you that $r_\varphi(e_1) = (\cos \varphi, ~\sin \varphi)$ and $r_\varphi(e_2) = (-\sin \varphi, ~\cos \varphi)$. You can work this out using high school trigonometry or possibly even Euclidean geometry. I'm not sure what a pure geometric proof would be, it's easier if you start by believing in $\sin$ and $\cos$. In the notation of complex exponentiation, $r_\varphi = e^{i\varphi} = \cos \varphi + i \sin \varphi$. Composition of rotations satisfies $r_\varphi r_\psi = r_{\varphi + \psi}$. If you write this out in exponential form it's just the usual exponent rules, which turn multiplication into addition. The other thing we know that turns multiplication into addition is logarithms; and indeed, the entire business of composing rotations is just the complex exp and log functions in disguise. Euler was the first to really see all this. Now suppose $\varphi = \frac{n}{m} 2\pi$, in other words $\varphi$ is a rational multiple of $2\pi$. Then if we compose $r_\varphi$ with itself $m$ times, it must come back to where it started. Because $(e^{\frac{n}{m} 2\pi})^m = e^{2n\pi} = 1$. The collection of all rotations of the plane forms a group; and in this case we say that $r_\varphi$ has finite order in this group. Any rational multiple of $2\pi$ eventually iterates itself back to the identity rotation. But what if the angle is not a rational multiple of $2\pi$? Then the orbit of the rotation -- the set of points hit by the successive compositions -- is dense in the unit circle. If you pick any point whatsoever on the circle, call it $x$; and choose any $\epsilon > 0$; then there is some $n \in \mathbb N$ such that $d(f^{(n)}(e_1), ~x)< \epsilon$ where $d$ is the usual Euclidean distance and $f^{(n)}$ is the $n$-th iterate of $f$. In other words the iterates of the rotations through an irrational multiple of $2\pi$ take $(1,0)$ arbitrarily close to every point on the circle. Or to say it yet another way: If you pick any point on the circle, then the sequence of iterates applied to $(1,0)$ contains a subsequence that converges to that point. Now that is really amazing. And it's the example they'd like you to have in mind when the Wiki page talks about the orbits of $H$ I think they call it. Just as we talked about the iterates of a single rotation in the plane; in the Banach-Tarski theorem, we are considering all the iterated combinations of an irrational rotation around the $x$-axis, and an irrational rotation around the $z$-axis. So that is the mental picture that is an essential part of the their use of the word "orbit." There's quite a bit of meaning packed into that one word. Not just mathematical symbology, but conceptual context. And if you followed the bit about $F_2$, the paradoxical free group on two letters; once we have a pair of independent rotations of 3-space, then 3-space contains a copy of $F_2$ and then 3-space itself becomes paradoxical under isometries. At heart it's a very simple proof. There are a lot of moving parts, but each part can be understood. Last edited by Maschke; April 10th, 2017 at 06:14 PM.
April 10th, 2017, 06:15 PM   #19
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 Originally Posted by zylo You can't create geometry from points. Points are locations in space.
That seems to be a very close-minded way of looking at things. Quite the contrary, I think; geometry is about relationships between points and sets of points.

April 10th, 2017, 06:24 PM   #20
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 Originally Posted by greg1313 That seems to be a very close-minded way of looking at things. Quite the contrary, I think; geometry is about relationships between points and sets of points.
You know ... from a philosophical standpoint ... I am not sure I agree with this. It's true that modern geometry is about sets of points. But geometry itself? That is less clear to me.

I was recently exposed to a philosopher named Charles Sanders Peirce, that's the correct spelling. He talked about the continuum. I'm probably mangling his idea but this is the sense I get. He said that if there's anything at all that we know about the continuum, it's that every piece or part of the continuum is exactly like the continuum itself.

In the mathematical model of the continuum given by the real numbers, this is true about open intervals; but its false about singletons. Mathematically we may write $\mathbb R = \cup_{x \in \mathbb R} \{x\}$. Since a single point is patently not a continuum; therefore the mathematical real numbers can not be the right model of the continuum.

As I say I'm really ignorant of this subject, I just ran across Peirce in passing. But it makes a lot of sense to me. We have no evidence that Euclid was thinking of set theory. He probably wasn't.

The foundation of geometry on set theory and the real number model of the continuum is a philosophical assumption. It's not God-given. It's not necessarily the right thing to be doing.

Of course this has no bearing on math itself. But on the philosophy of math, it's very much relevant.

The tl;dr is: When we identify a geometric line with the modern set $\mathbb R$ with all its set-theoretic baggage; that identification itself may legitimately be questioned on philosophical grounds.

This is how I understand Zylo's concerns.

I will take this one step further. As I understand it, in the field of algebraic geometry the concept of "point" has changed. A point is characterized as the zero-set of some class of polynomials. This is all the Grothendieck business people might have heard of. In advanced geometry they don't think about set theory any more. I wish I knew more about this but I think I have the broad outline right.

Last edited by Maschke; April 10th, 2017 at 06:37 PM.

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