Is there also a difference between "all" and "every"? Because you seem to be granting what you denied ... — Srap Tasmaner

It's a subtle point that is difficult to explain. Perhaps I should have used "any" instead of "every" in that second quote of mine.

Let me try to explain it another way.

You have a device that basically mimics the operation of f: N -> N0, f( n ) = n - 1. You can type in a natural number, starting at 1, to get a number from N0.

For one and the same input that you type in, you will get one and the same output from N0. Moreover, for every output from N0, there is exactly one number from N that can generate it.

That's what makes it a bijection in the weaker sense. And it's trivially true. We all agree about that.

But what that does not mean is that you can type in all of the numbers from N and produce all of the outputs from N0. That's a bijection in the stronger sense ( the one that matters. )

Suppose that you have all the time in the world. Suppose that means that the number of days at your disposal is larger than the number of natural numbers. Suppose that you decide to type in every number from N. On day 1, you type in 1. On day 2, you type in 2. And so on. How do you know that you will be able to produce all of the outputs? The fact that the function is bijective in the weaker sense does not tell you that.

Again, it is up to you to show any contradiction, not up to us to show there isn't one. — Banno

The onus of proof is always on the one making the claim. If you're making the claim that bijection between N and N0 exists, you have to show it, and that means, you have to show that such a bijection is not a contradiction in terms. That's what it means to show that something exists in mathematics.

You haven't done that. But I have done otherwise ( contrary to what you say. ) But you won't that accept because you're overly attached to your fallacious proof -- essentially, a circular argument -- that there exists a bijection between N and N0. "We can take any element from N and uniquely pair it with an element from N0, therefore, there's one-to-one correspondence between N and N0."

Again, if all you're going to do is spend all of your time justifying your chosen authorities, which is precisely what you're doing, then you want see the mistake they are making.

While it's good to see you using some formal notation, this isn't an example of a function. A function from A to B is a set of ordered pairs satisfying certain conditions. So writing f : A → B, f(n) = n − 1 is not merely symbolic stipulation; it is a claim that the rule maps every element of A into B. But in this case, as you point out, not every element of A maps to an element of B.

That's not a contradiction within a function, as you diagnose, but a failure to specify a function.

That is, <f : A → B, f(n) = n − 1, with A ={ 1, 2, 3, 4 } and B = { 0, 1, 2 }> does not construct a function. It's ill-formed. — Banno

You very clearly don't understand how language, definitions and oxymorons work.

Square-circles aren't squares either. They are also not circles. But they are also squares. And they are also circles. That's why they are oxymorons. They are two opposite things at the same time.

The same applies to the function that I mentioned. It is a function. But at the same time, it is not. The notation implies both. Not merely the latter. That's the mistake you're making. When you write, f: { 1, 2, 3, 4 } -> { 0, 1, 2 }, f( n ) = n - 1, that implies a function.

Symbols are capable of containing contradictions. It's not a new thing. It's the basis for the law of non-contradiction. It's not the case that symbols are necessarily X or not X. They can be both.

But either way, that's not really important, and it's nitpicking at best. The point is that the function does not exist. Which you agree with. And what that shows is that, just because you can define a function, it does not mean it exists. That was the entire point of that post.

In the argument we are considering, there is no such malformation. This is not just asserted, but demonstrated by the conclusion that the function is well-defined, injective and surjective. — Banno

And that's not true.

The only thing that you have shown is that you can take any element from N and uniquely pair it with an element from N0.

In other words, you have shown that, if we randomly pick an element from N, we can find its unique associate. Let's say we pick 1,345,219. Its unique associate would be 1,345,218. That holds true for every element from N. There are no exceptions.

But that does not mean we can put the two sets in one-to-one correspondence. That's a different thing. You haven't shown how many elements from N can be uniquely paired with N0. And your job is to show that you can take as many elements as there are in N0. Have you done that? Of course not.

All you have shown is that you can take an arbitrary subset of N that isn't larger than N0 and put it into a one-to-one correspondence with a subset of N0.

Do you understand the difference between the two?

The difference is that bijection means that you can take every element from N -- not merely any arbitrary subset of it -- and uniquely pair it with an element from N0.

You have to do it for all of the elements from N.

You don't know if that's possible. You just know that every element from N can be uniquely paired with an element from N0.

And remember, the onus of proof is always on the one making the claim.

You can't claim that bijection exists between N and N0 merely because you can take any element from N and uniquely pair it with an element from N0. That does not follow.

To argue that f(n)=n-1 does not map every member of N to a member of N0, you must show, as you did here, that there is an n for which f(n) is undefined or not a member of N0. — Srap Tasmaner

Not really. And that's a commonly made mistake.

Bijection does not mean that you can take ANY element from N and uniquely pair it with an element from N0. Of course you can do that with N and N0.

It means that you can take EVERY element from N and uniquely pair it with an element from N0. And that's what you can't do.

Do you see the subtle difference?

Let me illustrate my point.

Functions can be malformed. They can contain internal contradictions that effectively render them as non-existent.

Consider the following example.

Let A be { 1, 2, 3, 4 }.

Let B be { 0, 1, 2 }.

Consider the function f: A -> B, f( n ) = n - 1.

Since this is a function, and since functions are relations where every element from the domain is paired with exactly one element from the codomain, this is also a contradiction. Being a function means 1) it must obey the rules, and 2) every element form A must be paired with exactly one element from B. But both are violated. The rules say that the number 4 should be paired with the number 3, but no number 3 exists in B. It being a function means that the number 4 must be paired with exactly one element; otherwise, it is not a function. But it isn't paired with any element.

So even though this function is defined as a bijection, no such bijection exists.

What this means is that you have to show that f: N -> N0, f( n ) = n - 1 is not a contradiction in terms before you can conclude that it exists.

Has anyone done that?

Of course not.

Instead, the opposite has been shown.

Sorry, Magnus, but your "proof" merely begs the question. — Esse Quam Videri

You would have to prove that. I am, however, pretty sure you can't do it. But I can show, as I already did, that YOUR reasoning is circular. So what you're doing here would be sort of like a projection.

asserted impossibility without derivation — Esse Quam Videri

Again, you're doing the very thing you accuse me of. You're asserting something without proving it.

I presented a very clear process of derivation.

treated definitional existence as illegitimate by fiat — Esse Quam Videri

The onus of proof is on the one making the claim. You have to show that just because you can define a symbol as a bijection between N and N0 that it follows that such a bijection exists. I can show you why that does not follow. And I kind of already did.

accused others of fallacy and bad faith for not sharing your standards — Esse Quam Videri

Not for sharing my standards but for not being independent thinkers ( which is fine ) while pretending that they are ( which is not fine. ) Not everyone is an independent thinker -- and does not have to be.

refused to specify what would count as proof — Esse Quam Videri

Actually showing that bijection between N and N0 exists by employing definitional logic. So far, you've been merely asserting it and relying on something that is very much like a circular argument.

This is why the discussion keeps looping. — Esse Quam Videri

That's not the real reason. The real reason is that people do not know how to think outside of the box. People are missing the point all over the place. The main point of dispute is never addressed.

derive (not assert) an actual contradiction within the accepted mathematical framework — Esse Quam Videri

I already did that. But if your argument is a non-sequitur, it's not really necessary to to do so, isn't it? All I have to do is to show that it's a non-sequitur.

As it stands, Banno has already shown that combining your premise (1) with transitivity, antisymmetry and the existence of infinite partitions leads to contradictions. — Esse Quam Videri

He hasn't.

At this point there is nothing of substance left to discuss. — Esse Quam Videri

There is. Quite a bit. But all that is left is to think. No more room for quotations. But non-thinkers don't think.

Isn't it clear that this works, and that there can be no set of objects you could label starting at 0 that you couldn't also label starting at 1? — Srap Tasmaner

You can't do that. Logic prohibits it. There are more "labels" in N0 than there are in N.

And you're relying on a deceptive "proof". You think that, just because you can create a bijection between any proper subset of N with any equally sized subset of N0, that N and N0 are equal in size.

What is defined here is a function, not a symbol. This is a concrete mapping, not a mere linguistic construct, and it suffices to show that a bijection exists. — Banno

The symbol we're talking about is this:

f: N -> N0, f( n ) = n - 1

The definition of a symbol specifies what that symbol can be used to represent -- here, a specific bijection between N and N0.

However, that does not tell us anything about whether or not such a bijection is a contradiction in terms.

If it's a contradiction in terms, then bijection between N and N0 cannot possibly exist.

The definition you suggest cannot be used effectively with infinite sets. — Banno

Another lie.

Well, it's not just me... — Banno

It's not only you. I am aware of that. Your confidence is entirely grounded in what someone else said.

If your finitism is such that you cannot see that — Banno

Only a completely blind person can see any trace of finitism in what I'm saying.

This is false, since that definition applies only to finite sets. — Banno

That's a lie you've been shamelessly pushing forward.

The definition does not apply only to finite sets. It applies to all sets. The only reason you think it does not apply to infinite sets is because it leads to contradictions when you use it in combination with your mistaken premise, "If we can define a bijection, then bijection exists." That premise is your mistake. That premise is the main point of dispute. And so far, you've been conveniently ignoring it.

The other thing you conveniently ignore, because you clearly don't understand how definitions work, is that, you cannot blame the definition in this case. When you blame the definition, what you end up doing is changing the set of relations you're talking about, while calling the new relations the same names and insisting that they are the same relations as before.

It's like saying that, for red cars, "same size as" means something different than it does for cars that are not red. For example, that for red cars, "same size as" means "same color as". Therefore, all red cars are of the same size. And whoever says there are red cars that differ in size, you will claim they are wrong, simply because you got yourself in this bunker of confusion. In reality, all that you're doing is changing the relation that you're talking about ( from "same size as" to "same color as" ) while calling it the same name ( "same size as" ) and pretending that it's the same one ( "same size as". ) Instead of talking about the equality of sizes, you're now talking about the equality of colors, while calling it the same name ( "equality of sizes" ) and pretending that you're actually talking about the equality of sizes.

It's an ugly trick. But easy to see through for people who can actually think.

There's no need to list all of the elements. All this talk about constructivism, intuitionism and finitism misses the point ( I do not subscribe to any of these -isms nor do I have to in order to be internally consistent. )

PROOF

1) To say that S is larger than S' means that S' is a proper subset of S.
( A definition that applies to all sets, regardless of their size. )

2) N is a proper subset of N0.

3) Therefore, N0 is bigger than N.

This is an indisputable proof. As indisputable as 2 + 2 = 4.

However, if you're convinced by a fallacious proof, you will normally deny the validity of this one, like a cancer attacking healthy cells.

FALLACIOUS PROOF #1

The first fallacious proof they use to show that N and N0 are of the same size is the observation that, if you add 1 to infinity, you still get infinity. This is true but only in the sense that the result is also an infinite number ( i.e. larger than every integer. ) They make a mistake when they conclude that, just because "infinity" and "infinity + 1" are infinite numbers, it follows that they are equal. It's like saying that 4 equals 5 merely because 4 and 5 are integers.

FALLACIOUS PROOF #2

The second fallacious proof they use is grounded in the premise that, if you can come up with a symbol that is defined as bijection between N and N0, it follows that a bijection between N and N0 exists ( i.e. it's not a contradiction in terms. ) It's like saying that square circles exist merely because there exists a term called "square circle" that is defined as a square circle.

Here's the definition again — Banno

I can't quote the entire part, the LaTeX code gets messed up for some reason.

You are right that f( n ) = n − 1 by itself is not a complete definition of a function. A function definition requires a domain and a codomain. I conveniently left those out, assuming that they could be inferred from the context.

I was referring to the N -> N0 variant.

bijectivity would again depend on proof, not stipulation — Banno

When I say the function is bijective by definition, I do not mean that bijectivity is explicitly stated, but that it is an unavoidable consequence of the definition. The "proof" consists solely in unpacking what is already implied by the definition, not in adding any further stipulation.

f(n)=n−1 might be bijective, non-surjective, or non-injective depending on the domain and codomain. — Banno

That's correct. However, I was talking about f: N -> N0, f( n ) = n - 1. That specific variant does imply bijection.

The real problem is being ignored and that is that the fact that a definition implies bijection between N and N0 does not mean that such a bijection exists in the sense of being a logical possibility.

N0? — Banno

Not N0 but f(n) = n - 1. That function is a bijection by definition.

It is defined as f(n)=n−1 and then shown to be a bijection. That definition does not mention bijectivity at all. At this stage, the function could turn out to be injective, surjective, neither, or both. Nothing is being smuggled in. — Banno

Yes. It is not explicitly stated in the definition. However, the definition implies it. And because it implies it, it is a bijection by definition.

It's like defining the symbol "S" as "a closed figure with three straight sides". It does not explicitly state that it has 3 angles but it does imply it. So it is correct to say that S has 3 angles by definition.

"By definition" does not mean "explicitly stated by the definition". It means "fixed by the definition ( either explicitly or implicitly )".

In mathematics, functions are defined as sets of input-output pairs. f( n ) = n - 1, in this view, is a set of input-output pairs where every element from N is paired with exactly one element from N0 and vice versa. That makes it a bijection by definition. But that does not mean that the bijection between N and N0 exists, i.e. that it is a logical possibility. It merely means that's what the symbol can represent. It's similar to how simply saying that the term "square-circle" means "a shape that is both a square and a circle" does not mean that square-circles exist.

The seemingly devastating consequences of accepting that no bijection exists between N and N0 is that f( n ) = n - 1 does not exist either, i.e. that it is an oxymoron. The good news is that that's merely a consequence of an incorrect definition of functions. Functions aren't sets of input-output pairs. They are sets of rules.

It is defined as a bijection. The same way square-circles are defined as shapes that are both circles and squares. That does not mean they are logical possibilities, i.e. free from internal contradictions.

And there's a subtle difference between "You can pick any number from N and map it onto a unique number from N0" and "You can pick every number from N and map it onto a unique number from N0".

With finite set there's a contradiction.

With infinite set there isn't. — ssu

You're not responding to what's in the quote. You did not prove the following: that just because you can think of a function that is defined as bijection between N and N0, e.g. f(n) = n - 1, that it follows that a bijection exists between N and N0. You conveniently ignore that. And so does Banno. It's very convenient.

The problem is not with the definition of the word "same size", the problem is with the wrong conclusion that N is the same size as E or that N is the same size as Z. They aren't equal. And once you plug that in, the contradictions are gone. But neither you nor Banno nor mathematicians want to admit that you made a mistake that you've been stubbornly preserving for decades, so instead, you blame it on the definition, as if definitions can be logically proven to be wrong ( they can't, they are prior. ) The redefinition of the term "same size" merely covers up the fact that you screwed up.

OK, you really don't understand the Hilbert Hotel. — ssu

And you really don't have the ability to determine whether or not other people understand something.

And then when one gest, let's say G1, leaves, it's still full (meaning there's a bijection) because: — ssu

It's not full. There's no longer one-to-one correspondence between the two sets.

R1 R2 R3 ...
---- G2 G3 ...

R1 is unpaired. And it cannot be paired with any other guest because they are all already paired.

And if another guest comes, that G0, then the hotel fills up:

R1 R2 R3 ...
G0 G1 G2 .... — ssu

Nah, that's not what you get when G0 comes.

What you get is this:

R1 R2 R3 ...
G0 G2 G3 ...

G1 left. He's not in the hotel.

Their argument is that you can create an empty room for G0 by moving all the guests one place to the right. But of course, that's impossible. They employ sleight of hand to MAKE IT LOOK like it's possible. The trick is to hide the unpaired guest in the ellipsis and to never reveal it.

And understand that I went over this many times in the past.

they are not appeals to authority — Banno

Bullshit. You're literally copy-pasting textbook arguments. Zero thinking on your part.

I think ↪Banno has done a fine job of showing the inconsistencies that arise if we don't. — Esse Quam Videri

Not really. Banno's argument is flawed because it is based on the erroneous premise that I already covered:

"If we can think of a function that is defined as a bijection between A and B, it follows that there exists a bijection between A and B."

Because he can think of a function that is defined as a bijection between N and E, namely, f(n) = 2n, he concludes that N and E are the same in size. He never actually proves that such a function is not a contradiction in terms. And I can easily show that it is.

The inconsistencies that he speaks of do not come from "my" use of the term "same size" ( as if anyone else is using it any differently ) but from his own logical mistakes ( which aren't really his own, he merely copied them from a textbook. )

You’re treating “number of elements” as a notion whose inferential rules must be fixed by finite counting, and on that assumption the infinite case does look contradictory. Mathematics takes a different route — Esse Quam Videri

They don't. It's called ad hoc rationalization.

↪Magnus Anderson I think part of what’s driving the disagreement here is that two different notions of “same size as” are in play, and they come apart precisely in the infinite case. — Esse Quam Videri

I appreciate your response but I disagree with your conclusion, namely, that we're using two different definitions of the term "same size". I am quite confident that we're using the same definition.

When speaking of sets, the word "size" simply means "the number of elements". The word is defined the same way for both finite and infinite sets. That's how the word has been used for ages and it's the way it is used today.

With that in mind, the term "same size" simply means "equal number of elements". That's it.

The terms "bijection" and "one-to-one correspondence" refer to a relation between two sets A and B where every element from A is paired with exactly one element from B, and vice versa.

The observation is that, for every two sets A and B, if they are equal in size, they can be put into one-to-one correspondence with each other. And if they aren't, then they can't.

This means that, if we know that there's a bijection between A and B, it follows that A nd B are equal in size.

The problem they faced is that, with finite sets, one can determine whether or not they are equal in size simply by counting the elements and then comparing the resulting numbers; but with infinite sets, this isn't the case.

So they came up with the idea that, if we can show that infinite sets can be put in one-to-one correspondence with each other, we can conclude that they are equal in size.

So far so good. That is all true.

The problem lies in HOW they go about establishing whether or not any two sets can be put in one-to-one correspondence.

Their method is based on a hidden, and an erroneous, premise that, if we can think of a function that is defined as a bijection between A and B, it follows that there exists a bijection between A and B.

That's akin to saying that, if there exists a symbol that is defined as a shape that is both a square and a circle, then we can safely conclude that such shapes exist.

You've probably heard of Hilbert's Paradox. Hilbert's Paradox exposes a very serious contradiction in the way infinites are normally dealt with. Unfortunately, most pretend it's not a real contradiction, justifying themselves will sorts of silly rationalizations.

Suppose we have a hotel with a number of rooms equal to the number of natural numbers.

Suppose each room is occupied by a single guest.

That gives us a nice bijection between the set of guests and the set of hotel rooms.

R1 R2 R3 ...
G1 G2 G3 ...

Guest #1 ( G1 ) is in room #1 ( R1 ), guest #2 ( G2 ) in room #2 ( R2 ) and so on.

If there exists a bijection between N and N0, then it follows that we have a spare room for another guest. Let's call that guest G0.

---- R1 R2 R3 ...
G0 G1 G2 G3 ...

There is no longer a bijection between the two sets. G0 is not in any room. And if you try to add it to any room, you will either end up having two guests in a room ( not bijection ) or you will have to kick out one of the guests ( still not bijection. )

There's no way out of this conundrum . . . other than to pretend.

And that's what they do. They pretend.

A bijection does mean that sets can be put into a one-to-one correspondence. — ssu

You're missing the point. What has to be shown is that the fact that one can think of f(n) = n - 1 means that there exists one-to-one correspondence, or bijection, between N and N0. To do that, you have to show that f(n) = n - 1 is not a contradiction in terms.

I can think of the concept of square-circle but that does not mean square-circles exist. You have to show the concept is not a contradiction in terms. And in the case of the concept of square-circle, it very much is ( Taxicab geometry is not a valid counter-argument, it's merely a fashionable response, peddled by people who are not particularly good at logic. )

If you're a superficial thinker -- and most people are -- you will miss the subtleties.

I can very easily show that there is NO bijection between N and N0. But of course, it's not written in the books, so sycophantic ego-driven non-thinkers dismiss it.

No. There are injections and surjections, which aren't bijections (both injection and a surjection) and they are also called functions. — ssu

How exactly does that contradict anything I said?

:lol: — Banno

For a grownup man, that's a pretty childish response.

If the word "function" is defined the way mathematicians define it, namely, as a relation between two sets where each element from the first set is paired with exactly one element from the second, then, if a bijective function that maps N onto N0 is a logical possibility, i.e. if it's not a real oxymoron, then it follows that there's a one-to-one correspondence between N and N0.

Notice the requirement that it must be a logical possibility?

f(n) = n - 1 is a bijective function that maps N onto N0. But, understood through the lens of the above definition, is it a logical possibility? How do you know that? What if it's a real oxymoron that is useful?

You don't know that. And that's why you can't use it as a premise. You can't say, "There's a one-to-one correspondence between N and N0 because there's this function f(n) = n - 1 that maps N onto NO in a bijective way." You don't know if that function is a logical possibility. You have no proof of it.

And it kind of stinks of circular reasoning, doesn't it? "There's a one-to-one correspondence between N and N0 because there is f(n) = n - 1, a one-to-one correspondence between N and N0!"

Nor is your making shit up. — Banno

"Making shit up" is what people confuse with thinking when they know nothing other than to read books and / or be sycophants.

Reading a maths book isn’t just passive; it’s fuel for precise thinking, especially when you’re debating infinite sets. It shows how folk have thought about these issues in the past, and the solutions they came up with that work. — Banno

In your case, it's obviously passive. That you can't see it is your problem. It's pretty clear that you don't know how to think.

Your responses are now a bit too sad to bother with. Thanks for the chat. — Banno

You get what you ask for. But I'm sure you're innocent in your mind.

Well, it's one infinity amongst a few others... — Banno

That's a pretty bad excuse. The dispute was over the definition of the word "infinity". You were supposed to provide a definition that is different from mine. Instead, you provided a definition of a related term that I have no problem with. So what was the point? To show us that you read books? Are you really that pathetic, Banno? Obsessed over how you look in other people's eyes? Even though it's pretty clear that your thinking skills are . . . lacking, to say the least.

Your "definition" of infinity is not a definition of infinity. It's not false, it's just an intuitive approximation. — Banno

You surprise me with the amount of stupidity that you can spew. "It's not a false definition of infinity, it's just not a definition of infinity."

Yep. So I went the step further, presenting one of the standard definitions. — Banno

Of a different term, you imbecile. That I don't even dispute. And that does not go against my definition. How motherfucking stupid do you have to be?

It seems then that you haven't understood Cantor, either. — Banno

I am excused . . . I am not a fanatical book reader. But you're not allowed to make such mistakes. Cantor spoke of infinities that are not equal in size to the number of natural numbers. From that, one can conclude that he didn't believe your idiocy, "There is no last element, therefore nothing is left out."

A bijection exists between N and A — Banno

That a bijective function exists, cretin, does not mean that the two sets can be put into a one-to-one correspondence.

Can you please stop quoting books?

This is a philosophy forum, for fuck's sake, not a reading group.

You really should take ↪jgill' advice and read a maths book. — Banno

And you really should take my advice and use your brain for once. Reading isn't thinking.

Matching one to one from the left, the one left out is the 100. :meh: — Banno

Bravo!

With your
A = { 1/2, 1/3, 1/4, ... }
and
N = { 1, 2, 3, .. . }

There isn't last element. Nothing is left out. — Banno

Yikes. That goes against what Cantor said.

And I am pretty sure you won't be able to prove it ( asserting it isn't a proof. )

Yep, it only has every second number, so it must be half the size... Thanks for the giggle! — Banno

You're very clearly a non-thinker, Banno. Just a regular consumer of philosophy with ego issues.

...is not the definition of infinity. “Larger than every integer” is a heuristic, useful for intuition, but the mathematical definitions depend on limits or cardinality. Something like:

S is countably infinite ⟺∃f:N→S that is bijective (one-to-one and onto). — Banno

What you provided is the definition of the countable infinity. That's not the same as infinity. Furthermore, the provided definition does not contradict anything I said.

If you want to prove that my definition is false, you have to either argue that infinity is not a quantity or that it isn't larger than every natural number. You haven't done any of that.

Simply asserting that my definition is a heuristic that is useful for intuition is not an argument. Simply because your favorite books don't define it that way is not an argument. And it's not true anyways.

Sure. Infinities are not integers. — Banno

You're the king of missing the point. Of course they are not. But they are both categories of numbers. The only sense in which "Infinity + 1 = Infinity" is true is the same sense in which "Integer + 1 = Integer" is true. Unfortunately, that does not imply that the resulting number is equal to the one you started with.

The sets {1,2,3,...} and {2,4,6,...} are in one to one correspondence, satisfying the acceptable mathematical notion of "same size". But what happened to the odd integers in the second sequence?

Read a math book or two. — jgill

If you're going to take pride in your book reading skills, even though we're on a forum that is supposedly about thinking and not reading, at least don't conflate sequences with sets.

They aren't the same size. The set of even numbers is two times smaller. It has all of the elements that N has -- except for a half of them, namely, 1, 3, 5, etc. Doesn't matter what Cantor and mathematical establishment say. They aren't reality.

Here's another way one can explain why "Which one is left out?" question is problematic.

Let A be a finite set that is { 1, 2, 3, ..., 100 }.
Let B be a finite set that is { 1, 2, 3, ..., 99 }.

Obviously, these two sets aren't of the same size.

But suppose that someone comes along and makes the claim that they ARE the same because they can be put into one-to-one correspondence.

He shows you this:

1 -> 1
2 -> 2
3 -> 3
...

It might look convincing at first, but on closer inspection, you realize that he hasn't listed all of the pairs. He has listed only a subset of them.

You inform him of this and add that B has one element less than A.

He asks you, "Which one was left out?"

But how can you tell? There are 97 possible answers. He probably hasn't even chosen which one to leave out.

But you answer anyway . . . you say, 4.

He tells you, "Ah, no! I didn't leave that one! That one is paired with 4!"

You gasp and then say . . . You left out 100.

He smiles and says, "Wrong again! 100 is paired 100!"

You keep doing this for a while, failing to prove him each time. After some time, you might get tired, give up and concede. Or you might push him till the very end -- at which point, you win.

But with infinite sets there is no point at which this process can be finished. The person can keep you playing this game for as long as they want.

So there's no point in playing this game.

The correct response is to say that you can't tell which one was left out because there is an infinite number of possibilities. Moreover, in all likelihood, the person didn't even choose which one to leave out, meaning the game is rigged from the very beginning.

Furthermore, it's useful to add that a subset is not the entire set, and that the same applies to functions. If he can't show you the entire set of pairs, he hasn't proved anything.

Adding four to infinity is still infinity. — Banno

And adding four to an integer is still an integer.

The resulting category is the same. If you add four to a number that is larger than every integer, you get a number that is larger than every integer.

But the resulting number isn't the same.

Not for infinite sets. For obvious reasons. — Banno

Not quite. Definitions are prior. Nothing can invalidate them. If "add" means "increase in size", nothing can make it change its meaning. And the word "add" means "increase in size" for all quantities -- not merely for finite ones. The rest is ad hoc rationalization.

That is a proof of equal cardinality. Nothing is “pretended”. — Banno

And that is precisely what's being disputed. Your "function proof" is no proof at all. It's smoke and mirrors. The fact that f(n) = n - 1 exists merely means that you can use it on any natural number. That's all. It does not mean there's a one-to-one correspondence between N = { 1, 2, 3, ... } and N0 = { 0, 1, 2, 3, ... }. If you were to sit down and use every natural number starting at 1 as the input of that function, you won't end up producing all of the numbers from N0.

I can't think instead of you, Banno. If you can't do it, that's fine. But don't make it look like it's the other person's problem.

By definition, to add an element X to an existing set of elements S means to increase the size of that set. If you take a set N = { 1, 2, 3, ... } and you add 0 to it, you get a larger set. They are not the same merely because someone can pretend that they can be put into one-to-one correspondence. You can't list all of the pairs, can you? You can't. You can only list a subset. So that's not a proof you can put the two sets in one-to-one correspondence. On the other hand, the definition of the word "add" is indisputable. Nothing about infinities can change that.

Your question "Which element is left out?" is silly because I can't answer it, not because no element was left out, but because I can't tell which one you left out.

It's answerable if we're working with finite sets, e.g. A = { 1, 2, 3, 4 } and B = { 1, 2, 3 }. If you claim that A and B can be put into one-to-one correspondence and do the following:

1 -> 1
2 -> 2
3 -> 3
...

I can easily tell that you left out 4.

But with infinite sets, there's an infinite number of candidates. So how can I answer which one you left out? In fact, you didn't even CHOOSE which one to leave out. Yet, you want me to tell you which one you left out.

And you call that philosophy?

Which element is missing? — Banno

Silly question. The point is that you can't put them into one-to-one correspondence. In other words, one element must be left unpaired. Which one? You can pick any one. The ellipsis allows you to hide that.

There are not enough items in your second set to map one-to-one to the first set. Hence the cardinality of the firs tis larger than that of the second. Looks pretty convincing to me. — Banno

And there are not enough elements in the set A = { 1/2, 1/3, 1/4, ... } to put it into one-to-one correspondence with the set of natural numbers N = { 1, 2, 3, .. . }. It lacks exactly one element. Looks pretty convincing to me.

But that does not stop people from tricking themselves into believing that it's possible to do so by using the following "proof".

f(n) = 1/n - 1
1/2 -> 1
1/3 -> 2
1/4 -> 3
...

The fact that they can't list all of the elements is what makes it easy for them to trick themselves.

We should take your word for this? — Banno

Why not? If you can take Cantor's, you can take mine.

I gave an argument - albeit briefly. Fractions can be placed in a sequence, and so are no more than countably infinite.

Were did I go wrong? — Banno

You didn't do that. You merely asserted that you did it.

I can do the same for finite sets. Consider A = { 0, 1, 2, 3, 4, 5 } and B = { 1, 2, 3, 4, 5 }.

Here's me using the kind of proof that you're using to prove that A and B have the same cardinality.

0 -> 1
1 -> 2
2 -> 3
...

The fact that you can't list all of the pairs when working with infinite sets is what makes it easy to fall for that trick.

↪an-salad You are right that there are infinite infinities, but even with all those fractions, there are still only the same number as there are integers - ℵ₀, the smallest infinity - countably many. You can list them in a sequence, 1/1,1/2, 1/3, 2/3, 1/4, and so on, and so you can count them - line them up one-to-one with the integers. — Banno

That's not true.

Hi @Tristan L,

If the discovery-process is deterministic, the concrete instance of the solution exists from the start, although it only becomes “seeable” at the time that it manifests in a direct shape. Therefore, this is only creation in the broad sense, not in the strict sense. For example, the concrete software solutions that my algorithm will find already exist now, although not in a recognizable shape, so they can’t yet be used right now. They only become usable once the algorithm actually finds them, and that is the moment at which they are created (in the not-strict way).

If the discovery-process is deterministic, the discovered solution will necessarily be a solution that existed in a number of spaces long before it was discovered. Two of those spaces are:

1) the space of all possible solutions to all possible problems that can be discovered by following an existing algorithm (this is the space you're referring to)

2) the space of all possible solutions to all possible problems (this is where even solutions that were discovered through a random process existed long before they were discovered)

However, the discovered solution does not necessarily exist in the space of all possible solutions to all possible problems hitherto actualized (by humans, other living beings or machines.) And it is this space that ultimately matters.

One can very easily write a computer program that outputs every possible 32-bit 1920x1080 bitmap. The moment someone does so is the moment the first space (the one you're referring to) becomes filled with EVERY possible painting. If creativity is measured in relation to that set, that would make every subsequent painter an uncreative painter (even if they came up with a painting that depicts something of value that wasn't previously visually depicted.)

That is also what I think if what you call “possible idea” is what I call “idea” and what you call “actual idea” is what I call “concrete mental instance of an idea”.

I'd say so.

There is the set of everything someone can think of (possible ideas) and the set of everything people thought of (actual ideas.)

I agree with you if you mean the following: A mental instance of an idea in Alice’s mind has been invented by Alice, unless it was first invented by Bob’s mind, in which case Alice’s mind only discovers that instance of the idea.

I wouldn't say that Alice discovered it. I would say she reinvented it.

@Pffhorest

Creativity seems to be popularly held to be some kind of non-deterministic, random process of some kind of magical, metaphysically free will, but I hold that that is not the case at all. — Pfhorrest

Creativity is simply the ability to discover previously undisocvered solutions to problems. How you're going to discover such solutions is completely irrelevant. In other words, you can use a deterministic process but you can also use a random process. It does not matter.

On the other hand, I do agree with you that most people discover such ideas by following a deterministic process. (Most are merely not aware that what happens under the hood is largely, if not entirely, deterministic.)

I hold that there really isn't a clear distinction between invention and discovery of ideas: there is a figurative space of all possible ideas, what in mathematics is called a configuration space or phase space, and any idea that anyone might "invent", any act of abstract "creation" (prior to the act of realizing the idea in some concrete medium), is really just the identification of some idea in that space of possibilities.

I disagree with the bolded.

I will repeat what @Luke said.

"Discovery" implies that the thing that is discovered existed before discovery whereas "invention" implies that the thing that is invented did not exist before.

If you are talking about the set of all possible ideas, these can't be invented, since they already exist; they can only be discovered.

But that's because we're talking about the set of all possible ideas. The set contains all ideas that are possible -- there is absolutely no room for new ideas. If we're talking about the set of all actual ideas, however, one can introduce new ideas to it so as long it does not contain all possible ideas. An actual idea, that one that either existed within someone's brain at some point in time or did not, can be invented, provided there was no brain within which it existed previously.