You don’t run out of points, that’s for sure. For every point on line (B) there is an odd point on line (A). That’s where we agree. Where we disagree is that this means that (B) is equal in length to (A) with odd points taken out. I insist that it does not.
Do you think that (A = {1, 2, 3, \dotso}) and (B = {1, 2, 3, \dots}) are giving us enough information to conclude that the two sets are equal in size?
You obviously do. Like Silhouette, you think the two descriptions represent two infinite sets that are equal in size.
But I don’t.
This is evident in the fact that you can specify any kind of relation between the two sets. You can specify bijection but you can also specify injection. It’s an arbitrary decision.
You can say the two lines are equal. Fine. But if you remove one element from (A), you can no longer say they are equal. Indeed, the size of these two sets is no longer an arbitrary decision. So the fact that you can still specify a bijective relation between the two sets proves nothing.
I can say that (x = 3) and (y = 2). These are arbitrary decisions. But the result of their addition, (x + y), is not an arbitrary decision. It’s something that must logically follow from previously accepted premises. If you accept that (x = 3) and (y = 2), and that the operation of addition means what it normally means, then the result of (x + y) cannot be anything other than (5). The fact that you can change your premises (e.g. change (x) to (5)) to get a different result (e.g. (7)) does not mean that that different result is the result to this particular operation. That’s the kind of mistake that you’re making.
Is there such a notion as, more infinite? when a set is already in the state of being infinite… can something that is already endless, be more-so? Where is the end, to add to it?
Oh, obviously. I just question whether this is a commonly known definition (at least among specialists) or you’re just going off on your own inventing your own customized definitions.
So for you, a finite number must be expressible in terms of a finite sum of rational numbers. So if it can’t be expressed as a finite sum or if it can’t be expressed with rational numbers, then it isn’t finite. If we’re gonna talk about this, we might as well give it a name. I’m just gonna throw one out there. How 'bout “semi-finite”?
^ By this definition, would (0.\dot3) be a semi-finite number? I wouldn’t say so since it can be expressed as 1/3 (a finite sum consisting of just one rational number). You could say the same of (0.\dot9) since it can be expressed as 1/3 x 3, but this gives you 1 which you dispute (which is a whole other can of worms we could debate). It seems pretty obvious that you would say an infinite sum of irrational numbers isn’t finite (although it wouldn’t necessarily be infinite either). What about a finite sum of irrational numbers? For most irrational numbers, if you sum them, you probably still get an irrational number, which I’m guessing you’d say is semi-finite. But what about (\pi) + (4 - (\pi))? 4 - (\pi) gives us 0.85840734641… I don’t have a mathematical proof that this is also an irrational number but my gut tells me it is. So you’d have a finite sum of two irrational numbers which equals 1, obviously a finite number. But obviously, 1 can also be expressed as a finite sum of rational numbers; 1 is a finite sum of rational numbers consisting of just one term (1). But what about (\pi) + 1. This equals 4.14159265358979… which is obviously also an irrational number. I don’t know of any way to express this as a sum of rational numbers, but I derived it with a sum of one irrational number and one rational number. ← Do you agree that it still counts as a semi-finite number?
Anyway, I always thought of finite numbers as just not infinite numbers. You seem to think of (0.\dot9) or (\pi) as not finite because, what, they don’t have clear boundaries? I mean, a number like 4 clearly starts at exactly 0 and ends at exactly 4. But if you think there exists an infinitesimal between (0.\dot9) and 1, but you can’t say “the 9s end exactly here”, then the boundaries of (0.\dot9) are not clear, at least at the end closer to 1. ← Is that why you don’t think of it as finite? But then what do you say about (0.\dot3) which can be expressed as 1/3? I would think you’d say it doesn’t have a clear boundary either, at least not at the side furthest from 0. Does this make it semi-finite? Or does the fact that it can be expressed as 1/3 make it finite?
This wouldn’t prove that B’ has more members than N. Once again, the rule is: if you can map all the members of N onto all the members of B, then you know there are just as many members in B as there are in N (infinite). You’re twisting it to say: if you can map all the members of N onto some members of B and find that there are some members of B left over, then you know there are more members in B than in N. ← That’s not the rule (though I understand why you think it can be restated as such). What your example proves is that there are just as many odd members in B as there are in N (infinite). But because the number of odd members is infinite, and because adding more members to an already infinite set is still infinite, then adding the even members of B to the mapping still gives you the same amount (infinite). So you can map all the natural numbers onto every odd member of B, and you can map all the natural numbers onto every member of B (odd and even), and yes, Magnus, I’m saying they will both be the same amount (infinite) (this is what I’ve been saying all along). You really, really, really should watch the vsauce video I posted. He goes exactly into this–how to map the naturals onto two infinite sets. Here it is again:
You don’t need to categorize it in order to understand what’s going on inside my mind. I already told you that (0.\dot9) is a number that is greater than every number of the form (\sum_{i=1}^{n} \frac{9}{10^i}, n \in N) but less than (1). Why is that not enough? Why do you need it to be categorized?
That sounds good.
No. That’s because (0.\dot3 \neq \frac{1}{3}).
Do you agree that (0.\dot9) represents the infinite sum (\sum_{i=1}^{\infty} \frac{9}{10^i})? If so, you have to accept that (0.\dot9) is less than (1). Everything else is irrelevant. There is no (n > 0) such that (\sum_{i=1}^{n} \frac{9}{10^n} = 1).
I understand very well that’s not the rule (and what the rule is.) I am not saying that’s the rule. I am not restating the rule. I am questioning the rule.
What you don’t understand is that merely stating that’s the right way of determining whether any two infinite sets are equal in size or not does not make it the right way.
You adopted a conclusion that someone else arrived at without understanding the steps they took to arrive at it. And you think it’s correct because it’s widely accepted. And that’s all perfectly fine. What’s problematic is that you don’t seem to be able to comprehend that conclusions can be questioned and that the fact that they are widely accepted does not mean they are correct.
How about this: you really, really, really, really, REALLY, REALLY, REALLY should read my posts instead of living in a delusion that I know absolutely nothing about set theory.
Signs of frustration are another thing that can ruin discussions. Sure, it’s a fashionable thing, but that does not make it right.
Interesting video Gib… still got a 1/3 or so of it to go.
Some infinities can be bigger than others? Does anybody know what infinity means…? I now see why most say that infinity is hard to imagine/fathom, because most obviously cannot.
Sure, the elements in the set can be numbered… that was never my issue, it’s that infinity means infinite, not more infinite.
Oh how easily some forget their past unnecessary demeanours, and so think that all is well again, when it will never be so… standards of a certain level of decorum expected, not being met.
Too funny