Does infinity exist?

You mean to include ignite between 1 and 0 - yes but thats the same as why naturals and rationals are the same class.
what I try to figure out is how you can show one set (irrationals) is infinite by mapping a greater set (reals) onto it.

No, completely different proof and idea. Well related, but not really the same.

I did. The set of all positive irrationals can be bijected onto one if its proper subsets, namely the irrationals strictly between 0 and 1.

Are you confused about the reals versus the irrationals? The reals include both the irrationals and the rationals. You can use the same proof idea for the reals or the irrationals. If you only care about the irrationals, you need to exclude the rationals.

Please tell me which part of this isn’t clear. It’s clear in my mind so perhaps I’m not understanding your question.

We show the reals are infinite by mapping them onto a proper subset.

We show the irrationals are infinite by mapping them onto a proper subset.

You could map the reals onto the irrationals bijectively, but it’s a bit tricky and not worth the trouble.

Remember to show a set is infinite, I only have to biject it to SOME proper subset of itself. I don’t have to biject it to any particular proper subset.

I do in fact know how to biject the reals to the irrationals, but it’s a tricky construction and not worth going into detail about unless you want me to.

“I do in fact know how to biject the reals to the irrationals, but it’s a tricky construction and not worth going into detail about unless you want me to.”

This bijecting the reals to the irrationals is indeed what I was inquiring about, since what you said earlier about the real line hinges on it. The rest was clear to me before.

No that is not true, and it’s a point you seem unclear on. Please take a moment to engage with this point, it’s important.

A set is infinite if it can be bijected to at least one of its proper subsets. So the naturals are infinite because they can be bijected to the even naturals, or the odd naturals, or the primes, or (as Galileo noted in 1638) the perfect squares.

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

Likewise the reals are infinite because (0,1) is a proper subset and the tan/arctan functions biject the reals to (0,1). Or if you haven’t taken trigonometry, you can biject (0,1) to the set of positive reals via f(x) = 1/x.

Please I request that you spend some time to understand this point.

Now, bijecting the reals to the irrationals is a curiosity. I don’t need it to show the reals are infinite, the (0,1) examples already do that. But bijecting the reals to the irrationals is an interesting exercise, and shows how in general to get rid of a countable set within an uncountable one without altering the cardinality of the uncountable set.

So, here’s a function that maps the reals to the irrationals.

First, the rationals are countable so they may be placed into an order like this: (q_1, q_2, q_3, \dots)

Now we need to choose any countable sequence of irrationals. It doesn’t matter which one we choose, but for definiteness let’s pick the sequence (\pi, 2 \pi, 3 \pi, 4 pi, \dots)

We define our function (f(x)) as follows. If (x) is rational, it’s one of the (q_n)'s.
We map each rational (q_n) to (2 n \pi). That is, we map (q_1) to (2 \pi), (q_2) to (4 \pi), and so forth.

If (x) is irrational and one of the (n \pi)'s, we map it to ((2 n - 1) \pi). For example (\pi) goes to (\pi), (2 \pi) goes to (3 \pi), etc.

Finally, if (x) is anything else – that is, if it’s irrational and not one of the (n \pi)'s – we map it to itself.

If you think this through (and I don’t claim that’s easy, this takes some work), you will see that we have a bijection between the reals and the irrationals.

This proof is due to Cantor. He used the irrational sequence (\frac{\sqrt 2}{2^n}) in order to show a bijection between the unit interval of reals and the irrationals in the unit interval.

See this thread for several variants. The example I showed is based on the answer by MJD (fifth and final answer on the page). math.stackexchange.com/question … -irrationa

Again, please note that showing the reals are an infinite set does not depend on this somewhat complicated example. We know the reals are infinite because we may biject them to (0,1), a proper subset of the reals. But if you ever need to show that there’s a bijection between an uncountable set and that same set minus some countable set, this is the construction to use.

Ah, this is the sort of reply I was hoping for. Yes, I will take some time. Thanks wtf.

You’re very welcome. It’s such a great construction and I enjoyed reviewing it myself.

I would feel better if I understood why you think it’s important, because of the reasons I already mentioned … that it’s the (0,1) example that shows the reals (or the irrationals) are infinite, and the real → irrational bijection is just a curiosity, although a nice one. But if you’re happy I’m happy, and I’ll stand by for questions. I’ll be off the air the rest of the day but I’ll be back tomorrow.

I was moved to make this little sketch. I never understood this construction so clearly before. Thanks for asking about it.

I really like it when abstract things are worked out to the concrete detail. It always, always always turns out to be valuable. Because it takes real effort from real minds.

I was pushing this topic a bit because I had been waiting for a good reason to get somewhat deeper into math. That worked very well, I now also have a better perspective on Russell, because I looked down on him pretty dramatically but I never knew of his type theory. I think that may be a bit underestimated, so far. I think it may become important in the future.

And yes I understand the proof now!
I wish I was good enough at math to compare this to type theory.
Still, I can begin to work at it now.

Yes I agree. And it’s cool that Cantor himself came up with this proof. It also serves as the standard technique for getting rid of countably many pesky problems. For example in Cantor’s diagonal argument, we are listing the decimal representations of real numbers, but some real numbers have two distinct representations. For example .5 = .49999… So when we form the antidiagonal, how do we know that even if it’s not on the list, its OTHER representation is?

The way this is usually handled is that we vigorously wave our hands (that part is essential) and say, “There are only countably many such pesky dual-representations so it doesn’t matter.” A thoughtful person might respond, “How do you know it doesn’t matter?” Our example shows how to prove that the dual representations don’t matter.

I don’t think any of us are in a position to do that! He was a great man in multiple endeavors. I don’t know much about him, but why do you dislike him? Is it the pipe?

How does that change your estimation of Russell?

There is no question that type theory is making a comeback as a potential foundation for math. The drive in this direction is from automated checking of proofs. There’s a big movement called homotopy type theory (HOTT) that’s the buzzphrase and Wiki page to know.

One thing I know from type-theoretic approaches to foundations is that they are associated with denial of the law of the excluded middle. I know very little about this entire area. It makes sense that this idea would gain currency in our age of computers; since there are many sets of natural numbers such that neither they nor their complement are computable. So neither is “true” in a world where truth is whatever can be determined by an algorithm.

The mathematical philosophy of all this goes under the name of intuitionism (if it’s classic 1930’s Brouwer type) or neo-intuitionism (if it’s more recent).

h

Yay! Me too! I had a vague idea about it before and now it’s beautifully simple.

What leads you to your interest in type theory? Just curious. Well if HOTT eventually gains majority mindshare, I will think of that as Brouwer’s revenge. There is another alternative
foundation that’s made huge inroads into much of modern research-level math, Category theory. Attitudes toward foundations are a matter of historical contingency.

I agree. Infinity can’t exist.

So you concede then that the infinite is boundless? How else could inch cubes and feet cubes correspond? (Sorry, I meant to say cubic feet and not square feet)

What I want to know is if there is any way sets can biject without being boundless.

If true, then it means you and god are the same, so when you look at god (or think about god), you’re looking at yourself and see the infinite regression.

If you are not god, then god is bounded by you and he is not infinite because where you exist is a place that god does not.

It suddenly occurred to me that definition is backwards. It should be: a set can be bijected to at least one of its proper subsets if and only if the set is infinite.

So instead of: A set is infinite if it can be bijected

It should be: A set can be bijected if infinite.

This is because the bijection isn’t possible without the infinite, so bijection is conditional to the infinite and not the infinite being conditional to the bijection.

Infinity comes first, then one performs the bijection. Obviously we can perform bijection before the set exists.

How do you show infinite f(x) without having to first prove there are infinite x?

I responded to your own hypothetical about infinite space. Your response is disingenuous. You asked, IF space is infinite, is the number of cubic inches equal to the number of cubic feet. You posed a hypothetical. Jeez man.

Yes, the unit interval [0,1] bijects to the interval [0,2]. Both intervals are bounded.

Galileo noted in 1638 that the set of counting numbers can be bijected to one of its proper subsets, namely the perfect squares.

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

This was 240 years before the advent of set theory. Bijection is a perfectly sensible notion even without a theory of infinite sets. The fingers on your hand are in bijection with the fingers on your glove.

I don’t understand the question. Do you deny that the unit interval (0,1) is in bijection with the positive reals via the map f(x) = 1/x? This is a fact familiar to high school students.

The chain of logic is as follows.

  • I note that (0,1) is a proper subset of the positive reals.

  • I note that (0,1) is in bijection with the positive reals via the map 1/x. If you doubt that you need to review your high school math.

  • Since the set of positive reals are in bijection with one of its proper subsets. the positive reals are infinite by definition.

  • I can do the same thing for the entire set of reals using the tangent/arctangent, but then you needed to have taken high school trig. The 1/x example has a more modest mathematical prerequisite so it’s preferred for this conversation.

Jeez man, answer the question: How else could inch cubes and feet cubes correspond?

Then what is the bound?

Show me the bound where the bijection ends.

If you cannot show me the bound that terminates the bijection, then you simply must concede that the set is unbounded.

There is no escaping this one: either show me the bound or admit it doesn’t exist.

This is irrelevant history and demonstration of finite bijection which still proves my point that the fingers must exist before the bijection.

The infinite set must exist as an infinite set BEFORE bijection with a subset of itself can happen, so bijection CANNOT be a definition for infinity.

The a posteriori cannot be the cause of the a priori.

Consider the real number line. If we measure it in feet, it’s infinitely many feet long. If we measure it in inches, the same. The cardinality’s the same. How many copies of [0,1] are there in the real line? Countably many. How many copies of [0,2]? Countably many.

The unit interval [0,1] is bounded by 0 below and 1 above. If this is unclear to you, please tell me what’s unclear about it so I can attempt to explain.

In this particular case, the positive reals are unbounded above. (Though bounded below, by 0). Of course SOME infinite sets are unbounded. You’re having quantifier problems.

I’ve shown you this same elementary example, obvious to a high school student, half a dozen times already.

I’ve shown it over and over and over. You just keep coming back with the same question, as if you either didn’t study any math after eighth grade, or just want to argue for the sake of arguing. I can’t go back and forth with you anymore. You’re denying high school math.

No, it’s a very easy question: how do you show there are infinite f(x) without already having infinite x?

You must have, in your possession, infinite x in order to plug all infinite x into f(x) in order to show there are infinite f(x). So, first you must prove there are in existence infinite x and THEN you can move to the next step, which is proving there are infinite f(x). So your example (0,1) is moving the goal posts, which I pointed out on page 1 and have been trying to get you to respond to ever since.

You must prove there are infinite x before you can perform any function of x and that includes bijection with one of its subsets. First prove the set is infinite, then show the bijection, then show the f(x) is infinite.

  • The bijection with a subset cannot exist unless the set is infinite, so prove the set is infinite BEFORE showing the bijection.
  • The infinite set of x must exist BEFORE infinite f(x) can be shown, so prove there are infinite x before proving there are infinite f(x).