All of johnclark's Comments + Replies

5Cyan
We can do a constructive counter-example for this one too, if you don't like space-filling curves. Take any real number in [0, 1) and construct two real numbers each of which is also in [0,1) by concatenating the first, third, fifth, etc., digits to make one real number and the second, fourth, sixth, etc., digits to make the second real number. Treat those real numbers as specifying fractions of a revolution for the two clock hands, as in my previous comment. Now every clock face is associated with a unique number in a subset of the real number line and vice versa.
8Sniffnoy
...screw it, I'll reply to this one just to point out what you should be looking up. That is not Cantor's proof, Cantor used (invented) the diagonal argument[0]. Nor is that a correct proof; if it were, it would prove that Q would have a larger cardinality than Z. You may remember this surprise? Q has the same cardinality as Z but R has strictly larger cardinality? If this doesn't sound familiar to you, you need to relearn basic set theory. If this does sound familiar to you but you don't see why it applies, you need to better develop your ability to analyze arguments, and relearn basic set theory (going by your previous statements). EDIT: JoshuaZ points out a clearer counterexample to your argument in a brother comment. Here. Here are some Wikipedia links to get you started. http://en.wikipedia.org/wiki/Hume%27s_principle http://en.wikipedia.org/wiki/Galileo%27s_paradox http://en.wikipedia.org/wiki/Hilbert%27s_paradox_of_the_Grand_Hotel http://en.wikipedia.org/wiki/Equinumerosity http://en.wikipedia.org/wiki/Bijection http://en.wikipedia.org/wiki/Dedekind-infinite_set http://en.wikipedia.org/wiki/Cantor%27s_diagonal_argument http://en.wikipedia.org/wiki/Cantor%27s_theorem http://en.wikipedia.org/wiki/Cardinality http://en.wikipedia.org/wiki/Cardinal_number http://en.wikipedia.org/wiki/Injective_function http://en.wikipedia.org/wiki/Cantor%E2%80%93Bernstein%E2%80%93Schroeder_theorem That should do for a start, though a more organized textbook may be preferable. Now you have at least something to read and I will spend no more time addressing your arguments myself as the linked pages do so plenty well. [0]Yes, I know this was not his original proof. That is not the point.
JoshuaZ100

In the same way I can show you a mapping that associates every single real number with a unique clock face (all the clock faces a properly working clock can produce in this case) but I can also show you clock faces (an infinite number of them in fact) that are not involved in this mapping; I can show you clock faces not associated with a real number, thus the number of all possible clock faces must have a larger cardinality than the real numbers.

This is not sufficient to show that you have a larger cardinality. This is essentially claiming that if I hav... (read more)

johnclark-10

The set of all clock faces a working clock can produce - call this the set of all valid clock faces - has the same topology (and cardinality) as a circle.

Yes.

The set of all possible clock faces has the same topology (and cardinality) as a 2-dimensional torus.

Show me.

John K Clark

8[anonymous]
There are two hands, an hour hand and a minute hand. The set of all possible positions that the hour hand can take describes a circle. The same is true of the minute hand: its set of all possible positions describes a circle. Consequently, the set of all ordered pairs of possible positions (h,m), where h is the position of the hour hand and m is the position of the minute hand, is the Cartesian product of the two individual sets, and thus the Cartesian product of two circles. This is a two-dimensional torus.
1komponisto
And also, via a different association, to a face any clock (working or not) could display. You are the victim of a very common misunderstanding, which is to forget that mappings between sets are allowed to vary when we use them for the purpose of comparing cardinalities.
4Sniffnoy
This sort of reasoning only works with finite sets. I'm not going to bother to address the rest of your comment, because it's full of confusion and it's clear you really need to go back and relearn basic set theory. It would be a waste of all our time to continue this argument further.
0JoshuaZ
Sniffnoy's next remark was relevant. What he was saying was simply wrong was the idea that if one set A contains another set B then A must have higher cardinality than B. It seems that you have some confusion about how cardinality of infinite sets behaves. It might help to read the relevant Wikipedia entries starting with the basic one on cardinality or look at a standard textbook on set theory. Some of these issues will also be handled by a real analysis textbook.
4Cyan
Perhaps I can help explain why this is wrong by giving a constructive counter-example. Correct me if I'm wrong, but by "the set of all clock faces" you mean the set of all positions the two hands of a clock could take. You can specify the position of the clock hands by stating the angles they make relative to any fixed position -- say the 12 position for concreteness. Suppose the angles are expressed in radians and take values in the set [0, 2*pi). Multiply each angle by the conversion factor "1 rotation per 2*pi radians" to map the angles into the set [0, 1). Now you can express any specific "clock face" by a point in the unit square. Then you can return to my original point about space-filling curves showing that this has the same cardinality as a line segment.
6Sniffnoy
Evidently you are more confused than I realized. OK, last attempt at explaining this. The power set of R would be the set of all subsets of R, not just the set of all size-2 subsets of R. (I will ignore for now that you are talking about pairs and what you want here is ordered pairs.) The set of pairs of reals is in bijection with R. And any clock face, valid or not, can be described by an ordered pair of reals; there is no such clock face, valid or not, as {1, 3, 5} or {n in N | 2n+1 is prime}. Your conclusion that the set of clock faces has a higher cardinality than R does not follow, and in fact is false - as I pointed out in a cousin comment, R x R is in bijection with R, and as the set of pairs of reals injects into this, the result follows by Schroeder-Bernstein. You seem to be equivocating between "the set of all clock faces (valid or not)" and "the set of all ways of pairing up members of R" (which could mean any of several different things, but for now we'll leave it unspecified as the distinction is irrelevant - they'd have the same cardinality). The latter does indeed have cardinality greater than that of R, but this is an entirely different set than the former. You really need to be more precise with your language. "The set of all ways 2 real numbers can be paired together" would usually be interpreted to mean "the set of all 2-element subsets of R", not the much larger set above. Earlier you wrote: It's really not clear what sets you're referring to here. "All subsets of paired numbers between 0 and 12?" "All the ways a pair of 2 real numbers can be arranged?" I can guess at what you mean but I can't be certain I'm right - especially because you are using these as if they are self-evidently the same, while my best guesses for what you mean by each of them, if they were taken in isolation, would be very different sets! Please go back and learn the standard terminology so people have some idea what you're saying. I did not just say "the laws concerni
7[anonymous]
The set of all clock faces a working clock can produce - call this the set of all valid clock faces - has the same topology (and cardinality) as a circle. The set of all possible clock faces has the same topology (and cardinality) as a 2-dimensional torus. However, the cardinality of a 2-dimensional torus is the same as the cardinality of a square, which is the same as the cardinality of a line (as you yourself recognize), which is the same as the cardinality of a circle. Therefore the set of all valid clock faces has the same cardinality as the set of all possible clock faces. A power set indeed has a larger cardinality than the set it is a power set of. However, the set of all possible clock faces is not the power set of the set of all valid clock faces.

Some of the laws of physics could change from universe to universe, but there must be some laws that remain invariant across the entire multiverse because without rules it would behave chaotically and if the multiverse behaved that way so would all the universes in it, including ours. However there is order in our universe, but what is fundamental and what is not? I think we probably all agree that purely mathematical things like pi or e would remain constant in all universes, but consider some of the physical things that might change:

The Planck constant. ... (read more)

0[anonymous]
I will assume that by "universe" here you mean "Everett branch". I agree that they could (with low probability), just as they could slowly change over time or be different billions of light years away. This is a possibility because our reality has a property that physicists call locality. What I object to is the belief, which more than a handful of LW and SL4 participants hold, that there is something about many worlds (or about quantum mechanics for that matter) that increases the probability of stuff like that happening above what it would be if our reality had this locality property but no Everett branching. should increase the probability we assign to stuff like that happening. There is also the
6Sniffnoy
Firstly, thank you for stating what you meant by clock faces. You should really have stated that explicitly, though, as it's not a standard term. Also I had to read that twice to notice you were making a distinction between "clock faces" and "valid clock faces". But this is simply wrong: If S is strictly contained in T, and S is a finite set, then T necessarily has strictly larger cardinality than S. The same does not hold for infinite sets - this is just the old "Galileo's paradox"; Z has the same cardinality as N despite strictly containing it. EDIT: Sorry, I wrote something wrong here before due to misreading! Thanks to steven0461 for catching the real problem. You seem to be equivocating between C and the power set of C. C is in bijection with R, its power set is not. (And since C is in bijection with R, its introduction was really unnecessary - you could have just used the power set of R.) (You also seem to be using unordered pairs when you want ordered pairs, but that's a more minor issue.) In short this has a number of errors (fortunately they seem to be discrete, specifically locatable errors) and I suggest you go back and reread your basic set theory.
0[anonymous]
But what you're talking about at the end is not the set of all possible clock faces. It's the set of all possible ways you could divide that set into "valid" and "invalid" clock faces, that is, the set of all possible sets of clock faces.

Are atoms real? Whatever the answer to that question is imagine if it were exchanged, that is suppose that magically the reality of atoms became unreal or the reality of atoms became real, would the world be in any way different as a result? I think the clear answer is no, therefore regardless of what the status of atoms may ultimately be, the question "Are atoms real?" is not real because real things make a difference and unreal things do not.

John K Clark

0Ronny Fernandez
If I thought that atoms were unreal, I would not expect to be able to photograph them. I also wouldn't expect a single atom to be capable of casting a shadow. That's some ways (and there are many more) that I could be wrong about atoms being unreal mere pedagogical tools.
0jhuffman
Could you give me an example of something that is real?