In an infinitely repeating pattern of three integers, said integers being 1, 2, and 2, could one say that there are more twos than ones?
Another manner of phrasing this question: "is a sum count of even numbers equal in size to a sum count of all whole numbers because all infinities are equally infinite?"
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So in a countable subset there are more twos but, in an infinite set, measurements cannot be made. Did I read that right?
Correct-ish. There are as many 1's as 2's if the set is infinite.
Thanks, I intuited that answer but I really don't trust myself when it comes to math on that scale.
That's in terms of cardinality, but in terms of measure, it depends on which measure you use (and although I've never had a formal course about it I'm sure it can give infinitely many different answers).
Pull the ones out of the pattern and stack them in their own.
You are left with:
>An infinite string of 2s
>An infinite string of 1s
If the original pattern was infinite, then you have infinite of each, and you can't tell if there is more of one or the other.
>There are as many 1's as 2's if the set is infinite.
>If the original pattern was infinite, then you have infinite of each, and you can't tell if there is more of one or the other.
Brainlets pls leave.
what exactly does the fact that real numbers have a greater cardinality than integers, have to do with this case, brainlet?
shut the fuck up, reading wikipedia does not make you understand shit, especially when you're completely fucking wrong
retard, the anons obviously know cantor's classic argument if they're able to build bijections between countable sets
>what does the fact one infinite set can have more elements than another infinite set have to do with the question "can one infinite set have more elements than another infinite set?"
Seriously, brainlet?
fuck off, retard. the question is about subsets of a sequence.
More in what sense?
Same measure, same cardinality, same density. It makes more sense to say that there's an equal amount.
instead of calling other people brainlets and posting random wiki links the best you can do to further your case is to provide actual solid arguments (hint: there are none)
>Cantor's_diagonal_argument
I guess you want to tell them, that they should use the term "countably infinite" instead of infinite. But it still has nothing to do with the diagonal argument, asshole.
Set 1: {1,1,1,1,1,1,...}
Set 2: {2,2,2,2,2,2,2,2,2,2,2,2,...}
both countably infinite. both equal in amount. aleph null!
[math]\aleph_0 [/math]
cardinality of all the natural numbers, including even and odd, i.e. 1 and 2. Therefore amount of recurring 1s is the same as the amount of recurring 2s.
the cardinality of the set
A={1, 2, 3, 4, 5, ...}
is equal to that of
B={1, 4, 9, 16, 25,...} where they are the square of A.
OP here, new to this, sorry, so cardinality, here, refers to the number of unique identities in an infinite set of whole numbers. the set {1, 2, 2, 1, 2, 2, ...} has two unique identities so would that be Aleph 2?
If that is correct then, when separated into two distinct sets that of twos and ones would be equal in cardinality, correct?
What the fuck is a "unique identity"?
Two sets have the same cardinality if you can make a bijection between them. Having the same cardinality is an equivalence relation. So, we define the "cardinality" of a set to be a special set which it can biject to. Aleph null is the size of the natural numbers.
Ok, so {1, 1, 1, ...} and {2, 2, 2, 2, 2, 2, ...} are sets of equal size as they would both be equal to the total of natural numbers?
(Sorry I'm such a dumbass at this)
Well, {1, 1, 1, ... } is actually the same as {1} as a set, sets don't have repeated elements. The sequence (1, 1, 1, ... ) usually represents the set {(1,1), (2,1), (3,1), (4,1), ... }.
Excluding technicalities, yes, all infinite sets which you can write is a list have the same size, that of the natural numbers. The proof is easy: there's a bijection between them and the natural numbers, built by assigning the first element to 1, the next to 2, and so on.
Alright, so in the sets {1} and {2} both 1 and 2. respectively, would biject to the natural "one" as they only contain one, non repeating element, thereby both sets are of equivalent size, correct?
Yeah, exactly. They both biject to the natural number one, which is usually the set { {} }. That's the idea.
Alright, thank you for being so patient.
so then what would their measure be in comparison to their cardinality? what is measure?
measure is another way of gauging how "big" something is. consider the (riemann, the usual) integral in the set of real numbers R. from what we have talked, you might see that the sets [0,1] and [0,2] have the same cardinality: you can easily create a bijection between them. But if you integrate the constant function 1 in each of them you'll get 1 and 2 respectively, showing that one is "bigger" in that sense. the integral is a measure (it can be extended into what it's called the lebesgue measure) but there are many other measures depending on what "size" means.
But two sets are not infinite sets. How are you going to make a diagonal on two infinite sets?