Magnus Anderson wrote:In any case, you seem to think that numbers that can only be described as approaching a limit without ever attaining the limit are a third kind of number after finite and infinite. Have a name for these kinds of numbers?

I don't.

Then answer the question. Is \(0.\dot9\) a finite number, an infinite number, or something else. If it's something else, what do you call it?

Magnus Anderson wrote:Take \(B' = \{p_1, p_3, p_5, p_7, \dotso\}\). You can relabel its elements by renaming \(p_3\) to \(p_2\), \(p_5\) to \(p_3\), \(p_7\) to \(p_4\) and so. That's what you're doing, right? By doing so, you can "prove" that \(B'\) is equal in size to \(A\). But by relabelling them in a different way, you can "prove" that \(B'\) is actually bigger than \(A\). Rename \(p_1\) to \(p_0.5\), \(p_3\) to \(p_1\), \(p_5\) to \(p_1.5\), \(p_7\) to \(p_2\), and so on. Voila! You have a set \(B' = \{p_{0.5}, p_1, p_{1.5}, p_2, \dotso\}\) that is "clearly" bigger than \(A\)!

That would only prove it to you. You're still using your own logic.

And you're missing the point. The point is to pair up each element in the set with a natural number. If for every natural number, you can match it with an element in the set, then you've shown that the set is infinite.

Imagine pairing every odd point in B with every point in A

before the points are removed from B. You would draw the correspondences the same way, right? You would draw a line straight across from a1 to b1. You would then draw a line at an angle from a3 to b2. You would then draw a line at a an even wider angle from a5 to b3. And so on. In this scenario, we agree that there are no differences between the length of the lines, and that they are both infinite. Therefore, you don't end up running out of points in line B before running out of points in line A, and visa-versa. You end up traversing line A at a faster rate than line B, but because neither line has an end, you never run out of points. The angles drawn between the lines just keep getting wider and wider and wider, without ever becoming 90 degrees. <-- This is exactly how I envision the lines being draw between points in the scenario

after the points are removed from line B and the remaining points filling the gaps. Therefore, it doesn't prove anything about line B being shorter (not to me). I agree that after removing the points from line B and shifting the remaining points to fill the gaps, drawing the mapping between the points would have to be done with these angled lines. But why would I think this proves that line B must be shorter when I just envision this scenario being no different from the scenario I just described

before removing the points and shifting the remaining points? You're barking up the wrong tree. You need to focus on proving that what applies to finite sets also applies to infinite sets, not on the fact that certain points were removed (I

know they were removed).

Magnus Anderson wrote:Let's take a look at an example involving a line consisting of a finite number of football players.

No! No more finite lines.

I'm very well aware that certain points in B got remove.

I'm the one who said to remove them! I'm not saying they magically come back. Stop belaboring the point!

Magnus Anderson wrote:The reason why you can relabel is because you're using a method of counting whereby you pair natural numbers with elements in the set.

And that's not a valid method of counting because the set of natural numbers does not have a size on its own.

Let's map \(N = \{1, 2, 3, \dotso\}\) to \(B' = \{p_1, p_3, p_5, \dotso\}\).

You can use bijection to do so:

\(f(x) = p_{2x - 1}\)

\(1 \mapsto p_1\)

\(2 \mapsto p_3\)

\(3 \mapsto p_5\)

\(\cdots\)

This makes the two sets equal in size.

But you can also use any other kind of function e.g. injection:

\(f(x) = p_{4x - 1}\)

\(\hspace{0.83cm} p_1\)

\(1 \mapsto p_3\)

\(\hspace{0.83cm} p_5\)

\(2 \mapsto p_7\)

\(\hspace{0.83cm} p_9\)

\(3 \mapsto p_{11}\)

\(\hspace{0.83cm} p_{13}\)

\(\cdots\)

This makes \(A\) smaller than \(B'\).

With this kind of "logic", you can literally prove anything you want.

Read my words very carefully, Magnus. I did not say the rule is: map the natural numbers onto members of the set any which way you want... I said: if there is a way to map

all the naturals onto members of the set such that

all members get mapped, then you know there are just as many members in the set as there are natural numbers. There just has to be

a way (really, you should watch the vsauce video). Incidentally, the scenarios you depicted

also show there are an infinite amount of members in the set. Mapping the naturals onto every odd number, for example, will show that there are an infinite number of odd numbers. Imagine then extending that to include the even numbers as well... wouldn't that for sure show that there are an infinite number of natural numbers?

Look:

Suppose A = {p1, p2, p3, ... }

\(1 \mapsto p_1\)

\(2 \mapsto p_3\)

\(3 \mapsto p_5\)

\(\cdots\)

^ That's one way.

\(\hspace{0.83cm} p_1\)

\(1 \mapsto p_3\)

\(\hspace{0.83cm} p_5\)

\(2 \mapsto p_7\)

\(\hspace{0.83cm} p_9\)

\(3 \mapsto p_{11}\)

\(\hspace{0.83cm} p_{13}\)

\(\cdots\)

^ That's another way.

\(1 \mapsto p_1\)

\(2 \mapsto p_2\)

\(3 \mapsto p_3\)

\(\cdots\)

^ That's a third way.

Can you spot the one where all the naturals map onto all the points? That's right, it's behind door #3! The rule: so long as there is

a way. And there is

a way. There's also a lot of ways

not to do it. But we're not picking any arbitrary method. We're asking: is there

a way to map all the naturals onto all the member. And the answer in the above case is yes. Therefore, this shows there are just as many members in the set as there are natural numbers.

Magnus Anderson wrote:I understand very well how the rule works. I'm simply saying it's not a valid rule.

If people say that a rule is valid, does that mean it's valid?

No, you have to think it through and determine for yourself. For myself, the determination is trivially simple: there are an infinite number of natural numbers. If you can pair them up one-to-one with members of a set, with no natural numbers remaining and no members of the set remaining, then the set must also have an infinite number of members. Trivial! I mean, like,

really trivial! It's how counting works.

If you don't think the rule is valid, then show me how it fails in the case of line B after removing the points. If it fails, then you

must run out of points before you run out of natural numbers. What is the last number you use before running out of points.

Magnus Anderson wrote:I'm certainly not a machine that convinces people (:

I'll say.

If you don't want to prove to me that what applies to finite sets also applies to infinite sets, that's your call. But I am telling you

what I need in order to be convinced. You can come up with scenario after scenario after scenario of ways to show that removing members of a set means that members were removed from the set--switching out points in a line for people in a queue or football players on a team or carts in a train or whatever--but since you have been informed that not only do I get the point (and am growing nauseous about hearing about it), but it's not what I need to be convinced, this is just an exercise in futility for you. So you go ahead and keep repeating the same argument over and over and over again; it's gonna get you nowhere.