For me it can be difficult to tell whether something's technical or not if I'm unfamiliar with it. — fdrake

Ok then what's wrong with, "Can you please explain yourself," versus, "My drill sergeant says you should suck it up, buttercup." What is the point? I do hang out on the Craigslist forums, where "Fuck you moron" is regarded as just as intellectually appropriate as quoting Aquinas. If this place is devolving to the CL forums I can play too. Hey @tim wood go fuck yourself. Everyone happy now? As if I don't know how to be an Internet jerk. I come to this forum in the hopes that I DON'T have to be an Internet jerk just to hold my own in a conversation.

come to this forum in the hopes that I DON'T have to bean Internet jerk just to hold my own in a conversation. — fishfry

Unfortunately almost no one is cordial in argument all the time. It pays to be understanding when someone gets uppity. Though it's hard to remain understanding when someone gets uppity.

Hey tim wood go fuck yourself. — fishfry

I accept. I ought to because I've delivered a few and felt justified doing so, so I ought to take if I've earned. But I still have a problem with bijection in uncountable sets: how do you do it? If defined by equal cardinality, then my bad. But that doesn't answer my question as to how it's done.

What is the usual order on the rationals? You aren't sure? Come on, man. The usual order is the usual order. — fishfry

Do you mean the integers? I accept that there is a "usual ordering" of the integers, but what would "the usual order" of the rationals look like? Or do you mean that any well-ordering is "the usual"?

Yet, following the same principle - pairing one element to another of these two sets - we arrive at a situation where every even number is paired with an even number in the set of natural numbers with the odd numbers left unpaired. — TheMadFool

Try this site:
https://www.mathsisfun.com/sets/injective-surjective-bijective.html
Bijective
"A function f (from set A to B) is bijective if, for every y in B, there is exactly one x in A such that f(x) = y

Alternatively, f is bijective if it is a one-to-one correspondence between those sets, in other words both injective and surjective.

Example: The function f(x) = x^2 from the set of positive real numbers to positive real numbers is both injective and surjective. Thus it is also bijective.

But the same function from the set of all real numbers is not bijective because we could have, for example, both f(2)=4 and f(-2)=4"

In terms of your example of even onto the even integers, and evens onto all integers, the latter, because the odd integers aren't matched, isn't a bijection.

Which can be well-ordered: definitely using the axiom of choice; and possibly even in the absence of choice. — fishfry

Hmm. Being called out is a good incentive to do some research and learn something. I find this among many sites that mention the problem. This one is brief and readable.
https://math.berkeley.edu/~kpmann/Well-ordering.pdf
I extract this as relevant: " So what do we believe today? First of all, it has been shown that if you want to believe the well-ordering theorem, then it must be taken as an axiom." The author goes on to observe that it's a pretty good axiom, and most accept it. He also notes in a footnote that "Moreover, it has also been proven that it is impossible to write down an explicit well-ordering for the set of real numbers."

There is no contradiction. The definition of cardinal equivalence is that there exists at least one bijection between the two sets. I don't know why you have a psychological block against grokking that.

Guy robs a bank, gets caught. In the interrogation room the detective says, "Fred we know you're the bank robber." Fred says, "Oh you are wrong. Here is a list of all the banks in the state that I didn't rob. I even have a notarized statement to that effect from the manager of every single bank in the country that I did not rob."

Is Fred a bank robber? Yes of course. He robbed a bank! He robbed one single solitary bank and DIDN'T rob all the others. But he's a bank robber.

It's an existential quantification, "there exists," and not a universal one, "for all." Someone is a bank robber if they ever robbed a bank, even if there are many banks they didn't rob. Two sets are cardinally equivalent if there is a bijection between them, even if -- as must ALWAYS be the case -- there are maps between them that are not bijections.

Someone murders someone, they're a murderer. No use parading before the jury the seven billion human beings they DIDN"T murder. That lady cop in Dallas a few months ago who shot a guy sitting in his living room eating ice cream. She was convicted of murder. She's in prison as we speak, ten years if I recall. No use trying to point to all of her neighbors who she didn't kill. She killed one guy. That makes her a murderer in the eyes of the law.

Why is this simple point troubling you? If you're on the jury do you say, "Well, the prosecutor showed that she murdered someone. But she didn't murder EVERYONE." You find her not guilty on that basis? Of course not! Right?

Even Hitler didn't murder EVERYONE. You think he got a bad rap? LOL. — fishfry

:rofl: :rofl:

You just keep repeating yourself. Is this North Korea? :joke:

What you leave out, and what has apparently been left out, of all of this is that the sets have to first be well-ordered. Then the bijection is a two-way Hobson's choice: next rider, next horse. And you never run out of either riders or horses. The problem with irrationals, is that they cannot be put into a well-ordering. (I.e., whenever you put two net to each other, you can then always put one in between - actually, a whole infinity of numbers in between.) — tim wood

What do you mean well ordered? Kindly explain.

But I still have a problem with bijection in uncountable sets: how do you do it? — tim wood

I gave you a worked example.

In terms of your example of even onto the even integers, and evens onto all integers, the latter, because the odd integers aren't matched, isn't a bijection. — tim wood

A bijection between evens and odds.

The set of all even numbers is given by {2k} for k in {0,1,2,...,}
The set of all odd numbers is given by {2k+1} for k in {0,1,2,...}

The even numbers look like {0,2,4,6,...}
The odd numbers look like {1,3,5,7,...}

Define f from the odds to the evens by f(x) = x-1
If x is odd, then x-1 is even. So this works.

f is an injection: f(y) = f(x) => y-1 = x-1 => y=x
f is a surjection: any even number is of the form 2k for some k, then 2k+1 is odd, then f(2k+1)=2k

f is therefore a bijection.

2 sets have the same cardinality if they have a bijection between them. f is a bijection between the evens and the odds. Therefore the evens and the odds have the same cardinality.

(If you want to work through the case for the evens or odds to the naturals; f(k) = 2k is a bijection from the naturals to the evens, f(k) = 2k+1 is a bijection from the naturals to the odds.)

With regards to the order stuff, this is an order preserving function

assume x<y then f(x) = x-1 < y-1 = f(y), so x<y => f(x) < f(y) which was to be demonstrated. The order is a well order since the evens and odds are subsets of the naturals. The naturals are generated by the following function: f(x) = x+1, repeated application of f to 0 gives every natural. The order defined by x<y iff y = f^n (x) for some natural n>0 and x=y iff f(x) = f(y). This is a total order.

That order is also a well order. Take some nonempty subset of the naturals. Assume for reductio that it does not have a least element. Let some element x belong to the set, then this element cannot be 0 (all others are greater). Then this element cannot be 1 (all others are greater and it can't be 0). Then this element cannot be 2 (all others are greater and it can't be 0 or 1)... Then this element cannot be x (all others are greater and it cannot be in {0,1,2,...,x-1}). This holds for arbitrary x by induction. Then arbitrary x isn't in the subset. Then the subset is empty. Contradiction. Therefore the set is well ordered. Call an even (odd) less than another even (odd) when it is less than it in this order on the naturals. That is also a well order through the same argument and symbol substitution.

Edit: this well order does not require the well ordering principle to construct. Therefore there are sets that can be well ordered without the well ordering axiom. This says nothing about whether well orders are first order predicable/definable for arbitrary sets in general, but shows that there are first order predicable well orders (well, once you rewrite the f^n thing into a formula using x+n for some n to save quantifying over a function symbol by instead quantifying over the variable n).

What do you mean well ordered? Kindly explain. — TheMadFool

I mean by well-ordered a set (of numbers) ordered in such a way that given a starting value, say, on the left, one could move to the right step-by-step and enumerate/list/count all the elements without missing any. The natural numbers, ordered on ≤, would be such an ordering.

I find this online:
The natural numbers are a well order under the standard ordering ≤.
The set {1/n : n =1,2,3,...} has no least element and is therefore not a well order under standard ordering ≤.

For rationals, a well-known ordering is 1/1, 1/2, 2/1, 3/1, 2/2, 1/3, 1/4, 2/3, 3/2, 4/1, 5/1,.... It's not ordered on ≤, but it demonstrates that the rationals can be ordered and counted.

fdrake proves that uncountably infinite sets of equal cardinality are bijective. Which is news to me, nor do I challenge it. I thought to do any kind of in-, sur, or bijection, you had to exhibit a way to do it.

This I cited above and reproduce here seems relevant:

https://math.berkeley.edu/~kpmann/Well-ordering.pdf
"So what do we believe today? First of all, it has been shown that if you want to believe the well-ordering theorem, then it must be taken as an axiom." The author goes on to observe that it's a pretty good axiom, and most accept it. He also notes in a footnote that "Moreover, it has also been proven that it is impossible to write down an explicit well-ordering for the set of real numbers."

Now, this refers to well-ordering. And in my opinion it is a) strange, and b) gets stranger the more you look at it. It's apparently very useful to suppose that something can be done in all cases, but then you prove that it cannot be done at all in some cases.

I mean by well-ordered — tim wood

You know, when you repeatedly encounter what looks like a special term in a largely unfamiliar area, the smart thing to do is to look it up, instead of making up your own definition.


To those who wish to engage Tim in a mathematical discussion it may help to know that he is firmly convinced that "all mathematics is counting." This hangup may help to explain the difficulties that he is having with rationals and reals.

I mean by well-ordered a set (of numbers) ordered in such a way that given a starting value, say, on the left, one could move to the right step-by-step and enumerate/list/count all the elements without missing any. The natural numbers, ordered on ≤, would be such an ordering. — tim wood

There's really only one order like this.

You have a set like the naturals: {0,1,2,3,...} which are presented in their standard order with the successor function S (@SophistiCat "counting") always giving the "next" element in that order. The successor function for the naturals will be called S.

You define a bijection f from the naturals onto some countable set, giving {f(0),f(1), f(2), f(3), ...}. This set also needs a similar successor function T. But you additionally require that f(S(x) ) = T f( x ); the mapping of a successor is the successor of the mapping. (successor functions are how you move right step by step)

If you have that x<y in the naturals, then for some n, y=S^n ( x ), then f( y )=f(S^n (x) ) so f( y ) = T(f(S^(n-1) x) ) > f( x ), that is, f is an order preserving bijection. This is called an order isomorphism. IE - the only possible mathematical order you could be talking about (up to order isomorphism) is the standard order on the naturals.

It's not much of a definition of well ordered. You're artificially constraining math with a poor definition (if you think it's necessary that we accept your definition).

I just don't understand the insult culture around here. Over the past couple of years I've had to take extended breaks from this forum because someone started piling on personal insults at me over technical matters on which they happened to be flat out wrong. Not because I can't snap back; but because I'm perfectly capable of snapping back, and that's not what I'm here for. I'd suggest to members that whenever they throw an insult in lieu of a fact, perhaps they should consider whether they've got any facts. — fishfry

Fishfry, you appear to be very sensitive. From my experience, if someone points out to you, your misguided way, you take it as an insult. This is philosophy, so you ought to learn to take this as an attack on the ideology which has guided you, rather than an attack on your person.

However, let's do something different. We take the same sets N and E. We know that N has the even numbers. So we pair the members of E with the even numbers in N. We can do that perfectly and with each member of E in bijection with the even number members of N. What now of the odd numbers in N? They have no matching counterpart in E. — TheMadFool

The problem with your argument is that the described function is not bijective. Instead, you have constructed an injection from the set of even numbers to the set of natural numbers. From this, you are only permitted to conclude that the set of even numbers has at least as many elements as the set of natural numbers. If you combine this with an injection from the set of natural numbers to the set of even numbers, then you can conclude that a bijection exists. Because these sets are countable, this bijection is constructible (and is given by the standard map from the natural numbers to the even numbers).

What do you mean well ordered? Kindly explain. — TheMadFool

How about learning actual math? Set theory in this case. It's easy in our time. Just google it. I'll even give a link here: Well-order. Or here: Well ordered set or Well-ordering principle.

Or just watch the video short video:


Showing that you don't quite understand what others are talking about isn't a great argument.

The problem with your argument is that the described function is not bijective. Instead, you have constructed an injection from the set of even numbers to the set of natural numbers. From this, you are only permitted to conclude that the set of even numbers has at least as many elements as the set of natural numbers. If you combine this with an injection from the set of natural numbers to the set of even numbers, then you can conclude that a bijection exists. Because these sets are countable, this bijection is constructible (and is given by the standard map from the natural numbers to the even numbers). — quickly

Showing that you don't quite understand what others are talking about isn't a great argument. — ssu

Thanks for bearing with my stubbornness but have a look at what I say below:

A = {x, y, z} B = {r, s, t} and C = {l, m}

Cantor got it right if his claim is that quantity/number is simply an abstraction of what is common between sets - specifically the possibility of putting their members in a 1-to-1 correspondence in such a way that each element in one set is paired with one and only one member in the other set with no element in either set left unpaired. This is called bijection I believe.

[Assume the notation that n(A) means the cardinality or number of set A.]



n(A) = n(B) because of the bijection between them. One way for the bijection is as follows:

x ---- r, y ---- s and z ---- t

The argument that the set of natural numbers has the same cardinality as the set of even numbers, i.e. a bijection is possible, is exactly like the above with one exception - we're dealing with infinite sets.

So far so good.

Consider now the set A and set C. There is no bijection between them and the set A has one extra member that doesn't have a counterpart in the set C. We then say n(A) > n(C) i.e. set A is bigger than set c. The attempt to pair the members of sets A and C will look like:

x ---- l, y ---- m, z ---- ???

The set theoretic way of saying n(A) > n(C) is that set C can be put in a 1-to-1 correspondence with a proper subset of set A.

There's no issue with any of what I've said so far. To summarize we agree on the following:

Facts:
1. Two sets have the same cardinality if and only if there's a bijection between them

2. A set G has a cardinality greater than a set H if and only if the there's a bijection between set H and a proper subset of G

[Definition: a set K is a proper subset of a set L if and only if all members of set K are present in set L and the set L has at least one member which isn't a member of set K].



Since all of you are on Cantor's side it implies that you understand the argument that the cardinality of the set of natural numbers is equal to the cardinality of the set of even numbers. This result follows naturally from fact 1 above.

However, we still have to consider fact 2 by which we can determine inequality of cardinality of sets

The set of natural numbers N = {1, 2, 3, 4,...}

The set of even numbers E = {0, 2, 4, 6,...}

It's clear that N can be separated into two proper subsets viz. the set of even numbers V = {0, 2, 4, 6,...} and the set of odd numbers, D = {1, 3, 5, 7,...}

Notice how set E has a bijection with set V and set V is a proper subset of set N. If so, then in accordance with fact 2 above, n(E) < n(N) i.e. the set of even numbers is less than the set of natural numbers.

This presents a problem doesn't it? One of what I called facts leads to the infinity of natural numbers having the same cardinality as the infinity of even numbers while the other (fact 2) leads us to the conclusion that the set of natural numbers is greater, not equal, than the set of natural numbers.

Comments...

This is known as Galileo paradox:

https://en.wikipedia.org/wiki/Galileo%27s_paradox

I share Galileo's rather than Cantor's opinion, but I am in the minority.

I share Galileo's rather than Cantor's opinion — Devans99

That's right! Maths is nothing if not opinion. Btw, do you have an opinion as to the temperature that water boils?

That's right! Maths is nothing if not opinion. Btw, do you have an opinion as to the temperature that water boils? — tim wood

100 °C

What is your point exactly?

OK I get you:

https://www.compoundchem.com/2016/03/22/boiling-point/

What is your point exactly? — Devans99

That maths isn't about opinions. That's my point. Now what's yours: you repeatedly post nonsense - in itself not so bad; I do some of that myself - but you refuse correction and insist on your nonsense. Why is that?

:grimace: You are full of s**t. If you had any counter arguments to my points you would post them... but you don't so shut up.

Ok. Can you count the counting numbers? Of course you can. Can you count the even integers? Of course you can. Each even integer taken in turn can be matched with each counting number in turn. What might that look like? One way: (1,2), (2,4), (3,6), (4,8), (5,10), ..., and so on. Looks like for each of one, there's one of the other and vice versa. A fancy way of saying that is that the two sets have the same cardinality. Less technical but simpler is to say they're the same size. Still have a problem?

Oh yes. I answered your question, but you haven't answered mine. Do you think you might? Fair play and all that?

God be praised - a counter argument from @Tim Wood - a miracle!

In each finite segment of the naturals, the number of naturals is approximately twice that of the even naturals. Yet each natural can be doubled to give a one-to-one correspondence to the even naturals. This is Galileo's paradox - by one measure, the cardinality of the two sets is different, by another measure its the same.

What Cantor did was to ignore half the paradox and define 'size' in terms purely in terms of one-to-one correspondence. That is expressing an opinion on the nature of size that is incompatible with all finite subsets of the naturals - so yes MATHS IS OPINION - and in my opinion Cantor got it wrong.

Infinity is by definition unmeasurable so it has no size - how can you assign a measure to something that only exists in our minds and our minds says goes on forever?

If you disagree, tell me the size of the naturals. And no alpeh-zero is not the answer - that's a symbol without meaning.

"Same size" is a meaningful concept, yes? What is the usual method for determining if two things are the same size? Quantitative measurement, usually counting. You count until there's no more to count, the result being that one is more than the other, or they're both the same size.

What if the counting will go on forever - not a trivial problem. Answer: you define counting and establish some rules for it, then you count and if the count is even at every point, then the two things are same size.

And I thought Galileo's paradox had to do with squares, and how there seem to be many more integers than squares. Of course the right way to look at it is that every integer has a square. One-to-one, therefore, same size.

What if the counting will go on forever - not a trivial problem. Answer: you define counting and establish some rules for it, then you count and if the count is even at every point, then the two things are same size. — tim wood

Yes and the rules set theory defines are broken. If something goes on forever, you can't count it - even with an infinity of time it is not possible to measure something that goes on forever. This is what Galileo recognised and what Cantor ignored - and it leads to spurious results, such as the number of naturals is the same as the number of rationals - how can anyone swallow that? For each natural, there is an infinite number of rationals... One-to-one correspondence gives nonsense results.

One-to-one correspondence gives nonsense results. — Devans99

And what we're left with is that you're right and everyone else is wrong. Or might there be some other alternative? My guess is that you insist on trying to understand transfinite maths on finite terms, and you're the poster child for that not working.

You're in the position of the primitive who counted one, two, many, and simply denied the possibility or sense of three, four, and so on. What in fact he was denying was that he could learn. By which standard, incidentally, we demonstrate that the stupid (because he will not learn) primitive is alive and well and well-represented among us.

- Infinity does not exist - see the OP - so this whole conversation is about MARSH GAS

- Even if infinity existed, finity is a subset of infinity and what works for infinity needs to work for finity also. One-to-one correspondence does not work for both so it is flawed

- You are very closed minded. You think that maths has it 100% correct - that is an unbelievably naive assumption to make. At what point in the past was human knowledge 100% correct? At what point in the future will it be 100% correct? It was not, is not and will never be 100% correct. The blind (=you) swallow everything without question. I question things. Now I may be wrong but at least I keep an open mind rather than just dumbly reciting the received 'wisdom'.

Infinity does not exist — Devans99

Did we define "exist"? Seems like it might be a good idea.

- You are very closed minded. — Devans99

I am, about some things, like 2+2=4. But for you I'll reopen a very open-minded offer I made to you earlier. I give you one dollar bills, and for each one you give me a five dollar bill. Can't get much more open-minded than that, or is that too much for you?

Did we define "exist"? Seems like it might be a good idea. — tim wood

OK give me an example of an actually infinite set from nature.

I am, about some things, like 2+2=4. But for you I'll reopen a very open-minded offer I made to you earlier. I give you one dollar bills, and for each one you give me a five dollar bill. Can't get much more open-minded than that, or is that too much for you? — tim wood

Arithmetic is defined, actual infinity has no sound definition. So its fair game for the open-minded.

You seem to have a childish obsession with point scoring through cryptic remarks.

You wrote, "Infinity does not exist." I ask you, what do you mean by exist?

actual infinity has no sound definition. — Devans99

Not according to you. Apparently that's all that matters to you. That's why I refer to you as being in your playpen.

Does infinity have existence beyond our minds? All sorts of things like fairies, square circles, can exist in our minds but only a subset can take a concrete form in reality.

I believe that an actually infinite set or some substance that extends 'forever' cannot exist in reality. We can only imagine such things.

Does infinity have existence beyond our minds? All sorts of things like fairies, square circles, can exist in our minds — Devans99

Yes. Let's be clear. You said Infinity doesn't exist, now that it does. I agree that it does, as an idea. And I'm inclined to agree that it cannot exist in reality - although reality itself does present some challenges to that (e.g., a ruler, and by extension the rest of reality).

So just for the heck of it, let's call maths a virtual reality and ask if transfinite math works in it. And it does. That's all you get. It works. Completely indifferent to your mistaken attitudes, beliefs, and opinions about it. But a clear message to you to fix them, if you can.