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.

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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).

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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.

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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/

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: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.

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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.

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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.

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- 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'.

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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.

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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.

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