I don't mean that litteraly. — AgentTangarine

Then what do you mean? Why won't you now look up 'continuum hypothesis'?

The only riddle you gave as an actual answer is that you can map N zero times on R. — AgentTangarine

So what? This is not a game with a scoreboard for how many questions I've answered.

Anyway, I've addressed TONS of your claims.

I have done that years ago already. I don't agree though. I'm not a parrot like you. — AgentTangarine

So years ago you looked it up, but still don't understand it now.

You don't agree with "it"? The continuum hypothesis? You have been claiming the continuum hypothesis for a zillion posts yet you say you don't agree with it! You are one really confused crank.

You are parroting your own world now as this is the second time you've said I am parroting, And even though I amply refuted that a while ago.

I really would rather not know about the bedtimes of you and your wife.

Now you take refugee behind empty verbiage. — AgentTangarine

Not at all. It's telling that you could figure out for yourself, but you won't, that your claim is false for the simple reason that I have not just given references but have posted formulations for you (as I already mentioned I have done that to your PREVIOUS claim that I only give references) and explanations too.

Eliza is now all tangled up in confusion. My answer was to the question about mappings from N onto [0 1] not about why your claim that I have done nothing but give references is false.

I'm not surprised that your attention span wouldn't provide recalling my question: Why won't you look up 'continuum hypothesis' on the Internet?

But why false? — AgentTangarine

Eliza is getting tired and reduced to token replies.

That you don't understand the answer is not my problem. I did answer it. Now your turn to answer my question,.

Then what is false? — AgentTangarine

The claim you made, as I quoted it, silly.

You haven't answered one! — AgentTangarine

Now you have flat out lied.

how many times you can map N on R
— AgentTangarine

Do you mean N onto R?

The answer is 0. — TonesInDeepFreeze

Again now, your turn to answer my question.

You only make references. — AgentTangarine

Not the zillionth time, but it is the second time you have made that false claim.

conformal to the relation between N and R. — AgentTangarine

No one ever told you that mathematics is not throwing around words like 'conformal' while not knowing what they mean.

I answered one of your questions. Your turn to answer one of mine.

You have to read what I write. — AgentTangarine

I do. You don't conversely.

There are infinite bijections between between N and [0-1] — AgentTangarine

Read not just what I write, but what is written anywhere you would look: There is no bijection between N and [0 1].

What you wrote is contradictory:

You write that there is a bijection between N and [0 1] and you write that [0 1] is uncountable,

You don't know what you're doing.

Like the interval [0-1]. Uncountable. — AgentTangarine

You contradict yourself. You claim there is a bijection from N onto R, but above you admit that the interval [0 1] is uncountable. And we don't write '[0-1]' to denote and interval.

You are totally mixed up.

And it is beyond me why you think that the stuff you make up in your own head - without rigorous principles and without any understanding of the mathematics you allude to and the terminology you abuse - is somehow true and correct while the mathematics that has been rigorously formulated, studied, scrutinized, critiqued, and checked and thousands of times over rechecked by professionals and students all over the word is all wrong. In other words, what makes a crank tick?

The relation between R and RxR is the same as the relation between R and N. — AgentTangarine

That is purely arbitrary unfounded assertion.

On the other hand, in set theory, from rigorously stated axioms, definitions, and rules of inference we prove things.

And I answered a question for you, but you still have not answered my question for you. So the next question for you is: Why do you think one would feel need to answer your questions when you don't answer questions yourself? Wait, I have the answer: Because you are a crank.

Every cardinal is either countable or uncountable. Every countable cardinal is either finite or denumerable. There are denumerably many countable cardinals. There are denumerably many finite cardinals. There is only one denumerable cardinal.

That is easy. The hard part I described is PROVING that for an infinite set S and natural number n>0, we have card(S) = card(S^n).

After proving that, cardinal arithmetic is indeed beautifully simple:

If K and L are cardinals, and the larger of them is infinite and the smaller of them is not zero, then

K + L = K * L = max(K L)

That's for countable sets. — AgentTangarine

Wrong. For any infinite set S.

how many times you can map N on R — AgentTangarine

Do you mean N onto R?

The answer is 0.

My many previous remarks amply imply that.

Now that I've answered your question, howzabout you answering a question I've been asking you: Why won't look up 'continuum hypothesis' on the Internet?

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

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

Incorrigible.

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.

Like being crackpots. — AgentTangarine

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

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.

And anything goes! — AgentTangarine

I'll give you a point for that one.

We can! — AgentTangarine

"She certainly can-can." - Cole Porter

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.

You guys need to get a hotel room. — jgill

Disgusting.

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.

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.

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.

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!

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.

Then you have a different notion of aleph one — AgentTangarine

Indeed I have a different notion from yours! My notion is the usual one in mathematics.

You, on the other hand, are unfamiliar with the ordinary mathematical definitions.

The ordinal in Aleph one is just related to how many times the infinity is present. — AgentTangarine

Aleph_1 is the least cardinal greater than Aleph_0. That is the ordinary mathematical definition.

inf^1 — AgentTangarine

Please stop writing 'inf' that way. It doesn't have any apparent meaning.

line inf^2, so aleph1 — AgentTangarine

No, bad Eliza, bad.

Anyway, you said you were off, Eliza. Too bad you didn't mean it.

I just mean inf x inf. — AgentTangarine

Which has no apparent meaning.

I guess what you mean is SxS where S is infinite.

But then there are more than card(NxN) real numbers between 0 and 1.

card({x | x is a real number between 0 and 1:) = 2^N, which is greater than card(NxN),

You are abysmally confused.

"inf^2" is not a recognizable notion. Probably what you mean is 2^N.

And you ignorantly, wantonly persist about aleph_1

The claim that aleph_1 = 2^N is the continuum hypothesis.

It's boring — AgentTangarine

It's not so much boring as it is unfortunate that you persist to post misinformation while you won't even bother to look it up on the Internet.

I can't find the recent thread about the Russell set.

And when I click the link to 'Feedback' it does nothing.

Are you seriously implying that the cardinality of the 2-d continuous plane is the same as that of the continuous line? — AgentTangarine

I have not mentioned continuousness. I have merely pointed out the utterly well known fact that it is a theorem that card(R) = card(RxR),.

There could be cardinalities between 1 and 2. — AgentTangarine

That is risibly wrong.

There are no cardinalities between 1 and 2. And there are no cardinalities between aleph_1 and aleph_2. However, without the continuum hypothesis, there could be cardinalities between card(N) and card(R), as, without the continuum hypothesis, card (R) could be aleph_x for some x>1, as I pointed out to you over and over in the other thread.

You don't know what you're talking about,.

It is a theorem that card(R) = card(R^n) for any natural number n>0. This is known by anyone who has read a basic textbook in set theory. Just read the proof for yourself.

Moreover, you keep claiming that card(R) = aleph_1, thus precluding that card(R) might be greater than aleph_1 and thus precluding that there might be cardinalities between card(N) and card(R) . That is only the continuum hypothesis, not settled mathematics.