As if you are a mathematician — AgentTangarine

In another thread in which you were posting, I already wrote that I am not an expert.

Why I write inf^1 is to highlighten the concept of cardinality. — AgentTangarine

It doesn't highlight any concept. It only highlights that you don't know anything about this subject.

FZ lived 120 years ago. — AgentTangarine

Who is FZ? And why does it matter that he lived 120 years ago?

The axiom of choice is based on finite sets. — AgentTangarine

The axiom of choice is not needed for finite sets. Every finite set has a choice function, irrespective of the axiom of choice. In the context of our exchanges, one of the important points about the axiom of choice is that it implies that every infinite set has a cardinality and that every infinite cardinal is an aleph.

aleph1.6 — AgentTangarine

Delusional and disconnected from reality. Check for fever.

Who is FZ? And why does it matter that he lived 120 years ago? — TonesInDeepFreeze

Fraenkel and Zermelo. Old-fashioned. They, like Cantor, overlook one infinity.

Repeat for page 2:

That would imply that the covered piece has the same cardinality as the whole square. — AgentTangarine

This is getting painful to watch. A simple example shows that the "number" of points in the interior of a cube {p=(x,y,z):0<x<1,0<y<1,0<z<1} , is exactly the "number" of points on the line {r:0<r<1}:

1:1 correspondence demonstrated by r=.3917249105... <-> p=(.3795..., .921..., .140...)

Extending these ideas shows the cardinality of R^3 is the same as that of R. — jgill

By the way, I did look at that Quora page you suggested in the thread that has since been deleted. The first post there is a proof that card(R) = card(RxR), which is exactly what you deny!

Fraenkel and Zermelo. — AgentTangarine

Fraenkel and Zermelo. Aren't they that old vaudeville comedy team that played the Borscht Belt years ago?

But I thought maybe you meant Frank Zappa. He's kinda old school too by this time.

I jumped up a dimension to make a point. R^n has the same cardinality as R.

You guys need to get a hotel room.

Algorithm for Eliza:

Step 1. Open browser.

Step 2. In the search field, type:

continuum hypothesis

Step 3. Click on the first link that appears to be an encyclopedia article.

Step 4. Read the part of article that states the continuum hypothesis.

Step 5. Click on bookmark for The Philosophy Forum.

Step 6. Click on thread 'Infinities outside of math?'

Step 7. In the posting box, type:

Now I see, TonesInDeepFreeze. You are right. Thank you.

Step 8. Click on 'Post Comment'.

Step 9. Stop.

Delusional and disconnected from reality. Check for fever. — TonesInDeepFreeze

That's what is said about geniuses in general. Like being a crackpot. Untill now I haven't seen one bit of math, only parrot references to the net. The cardinality of RxR being the same as R. I gave you a link to a so-called proof of a bijection between R and RxR. A wrong one. I asked you why it's wrong. No reply. Cantor didn't take the infinity of the real line into account in determining how many powers of an infinity are needed. You only replied that you can't raise infinity to a power. You just need inf^3 times to enumerate all points on a line. Saying that the aleph of a 2d infinite plane is the same as that of an infinite line is the same as saying R is the same as N.

That's what is said about geniuses in general. — AgentTangarine

So what? In general geniuses drink water. I drink water. That doesn't make me a genius.

In an any case, you show no evidence of genius. Very much to the contrary.

Untill now I haven't seen one bit of math — AgentTangarine

I gave you the primary formulas that you need to start with. You ignore them then complain that I haven't given you any math.

only parrot references to the net. — AgentTangarine

There is no fault in my recommending that you look up the continuum hypothesis.

I gave you a link to a so-called proof of a bijection between R and RxR. A wrong one. I asked you why it's wrong. — AgentTangarine

I don't need to defend someone else's proof that I hadn't referenced. I already know a more general proof, as can be found in a textbook on the subject.

You only replied that you can't raise infinity to a power. — AgentTangarine

No, I said a lot more.

You guys need to get a hotel room. — jgill

Disgusting.

I don't know whether the following post by me appeared in the thread that was deleted today:

You cannot project the naturals to R one to one. — AgentTangarine

Take away the word 'project' (a projection function is a certain kind of function and it is not needed to mention regarding whether there is a bijection from N onto R). So use 'map' instead'.

Also, trivially we can map N into R one to one. So in this context instead of 'to' we must say 'onto'.

Then your claim becomes:

We cannot map N onto R one-to-one.

And that directly contradicts your claim now that we can map N onto R one-to-one.

You are very very confused.

And that directly contradicts your claim now that we can map N onto R one-to-one. — TonesInDeepFreeze

We can! An infinite times infinite times actually. One time onto 0.1-0.9999999. This can be done an infinite times for [0-1]. And an infinite times for the whole real line. Hence aleph1.4, and not aleph1.

We can! — AgentTangarine

"She certainly can-can." - Cole Porter

She certainly can-can." - Cole Porter — TonesInDeepFreeze

And anything goes!

And anything goes! — AgentTangarine

I'll give you a point for that one.

I gave you the primary formulas that you need to start with. You ignore them then complain that I haven't given you any math — TonesInDeepFreeze

Which functions? Of repeated fractions?

In an any case, you show no evidence of genius. Very much to the contrary. — TonesInDeepFreeze

I have no intention to be one. But that's what they say. Like being crackpots.

Which functions? Of repeated fractions? — AgentTangarine

I didn't say I gave you functions. And I have never said anything about repeated fractions. You seem to have confused me with another poster.

Like being crackpots. — AgentTangarine

You haven't proven any math. But you have proven yourself to be a crank.

This is getting painful to watch. A simple example shows that the "number" of points in the interior of a cube {p=(x,y,z):0<x<1,0<y<1,0<z<1} , is exactly the "number" of points on the line {r:0<r<1}: — jgill

N can be mapped infinite times onto [0-1]. It can be mapped infinite times on a vertical linepiece. So infinite times infinite times on the square.

You haven't proven any math. But you have proven yourself to be a crank. — TonesInDeepFreeze

Then prove me wrong!

I already explained to you that proving that there is a bijection from N onto R requires stating your axioms, definitions, and rules of inference and using only those axioms, definitions, and rules of inference to show that there is a function whose domain is N, whose range is R, and is 1-1. All three clauses: domain, range, 1-1. Such a proof, if it were in ZFC, would contradict the theorem that there does not exist a bijection from N onto R, thus proving that set theory is inconsistent, and would make you among the very most famous people in the entire history of mathematics.

On other matters such as alephs, you're proven wrong by simply referring to the definitions and by fhe fact that the assertion that card(R) = aleph_1 is famously independent of ZFC.

1:1 correspondence demonstrated by r=.3917249105... <-> p=(.3795..., .921..., .140...) — jgill

This is the correspondence between infinite and infxinfxinf. And if r goes to 0.04566..., or 0.005667... or 0.000556654... (you get my point, I guess...) even infinite times are included. So the map becomes one between inf^2 and inf^4. Aleph1 and aleph2.

inf^2 to inf^4. Aleph1 and aleph2 — AgentTangarine

Incorrigible.

Show the correct answer then. I already gave you a link to a supposed bijection between R and RxR. How many times can N be mapped on the real line? Just offer a tasty recipe.

Okay, N^2, as you wish. Same as inf^2.

I'm embarrassed. "Heat Wave" is by Irving Berlin not Cole Porter.

CORRECTION to the post below. This answers proving that card(R) = card(RxR), which is not what was asked of me. Instead, what was asked of me is "how many times N can be mapped on the real line?" If that is taken in the sense of what is the cardinality of {f | f is a function from N onto R}, then of course the answer is 0.

I'll leave my remarks about card(R) = card(RxR) anyway:

I have been referencing the more general theorem that for an infinite set S and natural number n>0, we have card(S) = card(S^n). S = R is a special case of that. I'm not sure, but it seems perhaps the particular proofs mentioned in threads here lately for R don't use the axiom of choice (?). I have not studied those proofs to verify them for myself though I get the gist of them and they seem okay to me to that extent.

The proof I have studied of "for an infinite set S and natural number n>0, we have card(S) = card(S^n)" is in Enderton's 'Elements Of Set Theory'. It is pretty involved, two pages, requiring a number of previous lemmas, a proof that the axiom of choice implies Zorn's lemma, closure under unions of chains, and more (and even an illustration to aid intuition). I would not spend my time and labor composing it all for you in the confines of a post, and it would do you no good anyway since you are utterly unfamiliar with even the basics of set theory that are prerequisite let alone the mathematics of Zorn's lemma, chains, et. al. And I admit that I am rusty myself on some of the details now, though I have previously studied it in every detail to verify for myself that it is perfectly correct.

The best I can do for you is to recommend that you get a textbook and study it from page 1. Enderton's 'Elements Of Set Theory' in particular is widely used, highly regarded, beautifully written, and pedagogically exemplary. Though, I would actually first recommend at least gaining a basic understanding of symbolic logic.

I just asked you the question how many times you can map N on R.

I have been referencing the more general theorem that for an infinite set S and natural number n>0, we have card(S) = card(S^n). — TonesInDeepFreeze

That's for countable sets. For R this doesn't hold.