Every Weird Paradoxes in Set Theory

[@ful7481]

video starts at 0:00

[@gametalk3149]

🦅

[@rzeqdw]

3:42 So if C^2 = C, that must mean that C = 1. So, 2^(aleph_0) = 1, therefore, aleph_0 is log2(1). SOLVE YOUR MATH FOR YOU, SET THEORISTS

[@MichaelRothwell1]

Thanks for this very enjoyable and informative video. Unfortunately the stated bijection between ℝ² and ℝ doesn't quite work as stated, due to the non-uniqueness of the decimal representation of real numbers. Suppose we are to take the finite decimal representation where one exists. Then 1/11=0.(09) in ℝ maps to (0, 0.(9)) in ℝ², which means (0, 1), which maps back to 1 in ℝ.

[@GabriTell]

When you write "0.(09)" you mean "0.090909...09...", right? 👀

[@MichaelRothwell1]

@@GabriTell Exactly so.

[@SiqueScarface]

You can resolve this by showing 1) ℝ is surjective to ℝ² 2) ℝ is a subset of ℝ². In this case, you show that card(ℝ) ≥ card (ℝ²) and card(ℝ²) ≥ card (ℝ), hence card(ℝ) = card(ℝ²).

[@sayaks12]

@@SiqueScarface i'm not sure i'd fully agree this resolves the issue. while yes it shows that a bijection exists, you no longer have an actual bijection to refer to. fully resolving this issue should involve constructing a different bijection imo.

[@guillaumelagueyte1019]

Cool video. If you make a follow up, another case of something that seems counterintuitive but is true, is the existence of sets that dont have a cardinality at all. I was pretty surprised with that one.

[@ColbyFernandez]

This is great!!

[@methatis3013]

3:15 this doesn't quite work. Consider x=1 y=0 This would give a=10 However, if we set x=0.9999... and y=9.999... we get a=9.99...=10 Thus, we have 2 different pairs of x and y mapping to the same output, meaning our function is not an injection (and thus isn't a bijection)

[@rjkrkkj]

0.999999… is equal to 1 and 9.9999999… is equal to 10, try to keep up

[@methatis3013]

@@rjkrkkj yes, that's not what Im disputing. My point is that if we use x1=1, y1=0, we get a1=10 So f(1,0)=10 But if we use x2=1, y2=10, we can also get a2=10 So f(1,10)=10 Thus, our function is not really an injection (thus not a bijection) It's also not really a function since, in the way it was defined, it can return 2 different numbers with the same input

[@MuffinsAPlenty]

@@methatis3013 Correct, there needs to be a patch to the argument in the video because decimal notation is not unique. However, there are fairly simple ways to patch the hole in the argument, such as making the choice that if there are two decimal representations of the same number, always choosing the one with trailing 0's.

[@methatis3013]

@@MuffinsAPlenty that is true. You would still need to check that the function you end up with is a bijection though, which is not entirely trivial just from stating it, precisely because of the funkiness with decimal representations

[@methatis3013]

@@MuffinsAPlenty for example, which pair corresponds to number 10/11? The number is 0.909090... Presumably, the pair that corresponds to it is x=0.999... y=0.000... But if we restricted ourselves to only one of these representations, we run into a problem. This means that there is no pair (x,y) such that f(x,y)=10/11 and thus our function is not a surjection (and finally, not a bijection)

[@cube2fox]

Just because a set theoretic paradox can't be formalized in ZFC doesn't mean that the paradox gets resolved by ZFC.

[@josesalazar7896]

You're (kinda) right. In ZF we mostly "resolve" paradoxes by actively avoiding them. In ZF the only things that exist are sets, so if we try to form a collection of all sets that do not contain themselves and we state that that collection is a set, then we get a contradiction, which means that that collection cannot be a set. If we stick to the word "collection" (or a better word, class), then we have that the class of all sets that do not contain themselves is a collection (or a class) but cannot be a set (why? the self-reference is avoided). All sets are classes (since sets are collections of objects), but not all classes are sets (we have that the class of all sets that don't contain themselves is not a set). Classes that are not sets are called proper classes. Most paradoxes that come from self-references are resolved by making the distinction of what things are in the theory. Proper classes are not part of ZF. They get proper treatment (full ontological status) in NBG set theory. Mathematicians often use ZF since they can avoid paradoxes while doing mathematics, while mathematicians specialized in logic (logicians) and in foundations often use ZF to do math and NBG in order to tackle the paradoxes (since NBG does not actively avoid the paradoxes). It's a matter of pragmatics. Why would you use something that is equivalent in expressive power but can get you in trouble, instead of using virtually the same thing while avoiding the paradoxes?

[@cube2fox]

@@josesalazar7896 All the paradoxes occur in natural language. They can't be solved (or really "avoided") merely by looking at a restricted formal language which lacks the expressive power of natural language.

[@joecaves6235]

There are only 3 sets, Aleph Null, and Aleph Valid, and Aleph (all if not null or valid)

[@cymberciara]

2:00. While it has been proven that the cardinality of the continuum is larger than ℵ0, it hasn't been proven that the cardinality of continuum has the same cardinality as the power set of ℵ0. It is a popular theory but it definetly hasn't been proven, so the cardinal arithmetic section doesn't really work

[@MuffinsAPlenty]

"it hasn't been proven that the cardinality of continuum has the same cardinality as the power set of ℵ0." Yes, it has. There are many ways to prove this. For example, you can embed P(N) into R in the following manner: First, map a subset S of N to an element of 2^N (a binary sequence) by specifying that S gets mapped to the sequence defined by f(n) = 1 if n is in S, and f(n) = 0 otherwise. This is an injection from P(N) to 2^N (indeed, it is a _bijection_ between these two sets). Then one can map a binary sequence f to R by sending it to the infinite series ∑ f(n)/10^n. This is an injection from 2^N into R. Thus, the composition is an injection from P(N) into R. On the other hand, you can also embed R into P(N): By the Dedekind cut construction of R, we know every element of R is a set of rational numbers, and hence, is an element of P(Q). Hence, R injects in P(Q). Now, since there is a bijection between N and Q, there is a bijection between P(N) and P(Q). Indeed, if g is the bijection between N and Q, then map the subset S of N to the set T = {g(n) : n is in S}. Here, T is a subset of Q, thus an element of P(Q). Thus, P(Q) injects into P(N). Hence, R injects into P(Q), and therefore, into P(N). Since P(N) injects into R and R injects into P(N), the Cantor-Schroeder-Bernstein shows that there is a bijection between P(N) and R. You are probably confusing this with another fact: We don't know if the continuum hypothesis is true. That is, we don't know whether there is a cardinality strictly between that of N and P(N). We do, however, know that P(N) and R have the same cardinality.

[@AidanDaGreat]

These sure would be puzzling if i was smart enough to know what was going on!

[@ciroguerra-lara6747]

What about Godel´s incompleteness theorems? They might fit in this list?

[@sayaks12]

it's not really a paradox of set theory per se, it's more a paradox of logic. sure it applies to set theory but only because set theory is a logical theory.

[@IsaacDickinson-tf8sf]

The contiuum element c is the same as 1/0 or 2/0, but n/0 is not congruent to m/0 if n doesn’t equal m. or rather, once multiplying by 0 on both sides and seeing n doesn’t equal m based on the fact that the definition of n/0 is that when * by 0 it = n. Therefore, you could say that 1/0 is not congruent to 2/0 since 1 does not equal 2 even if they both equal the continuum c. They are equivalent in this sense: if you take points of length 0 from a line of length 1 unit, then you will have 1/0 of them. 1/0 points or numbers from 0-1= 2^aleph_null. You could also say that 0/0 is congruent to all the numbers, but only equal to n/0. Then you can say 1 does not equal 2 but 1 is congruent to 2 plus 0/0. This would imply a new rule where you can only multiply both sides by 0 or n/0 if congruence is the same on both sides. Now how to define x^(1/0)?… (also we could name 0/0 as zillion and x^(n/0) as bajillion) All of this may break some axioms like the a=b and b=c implies a=c but it can be worked around using this congruence. (-1)!=1/0= -1*-2*…=unsigned infinity used in perspective geometry. This is the best way to divide by 0.

[@GabrielMirandaLima-hv7oe]

What about Cantor's paradox and Burali-Forti paradox?

[@RSLT]

Wow, very cool 😎

[@mathewgriffiths1870]

Bertrand Russell was Welsh

[@Gus-s9v]

That’s still “British “ you fool

[@brainpower782]

and Tarski was not an American

[@oniondesu9633]

although he was born in Wales, he identified himself as English

[@mathewgriffiths1870]

@@oniondesu9633 I stand corrected

[@MisterrLi]

Since cardinality can't tell the difference between any two countable infinite sets, maybe we should use a more precise measure? It seems obvious the naturals is bigger than its squares, since the intersection (pairing up identical elements) is all the squares numbers, leaving nothing paired up with non-squares, or zero paired up with infinity, which is a failed bijection (Galileo's paradox).

[@nzqarc]

Here is the thing, there is only one countable infinity, Aleph-null, the infinity after, Aleph-1 is uncountable. Both the sets of natural numbers and squares are countable, therefore they have the same "size". In fact, natural numbers, integers, rationals and algebraics are all countable. Sure, you could say they are subsets of one another but remember, we are talking about transfinite quantity here, transfinite numbers don't behave exactly like finite numbers.

[@IngvarLind]

@@nzqarc There is not only 1 set theory, thankfully. One way to see that you actually can compare "countable sets" more precisely is by using a Venn diagram. The intersection takes care of the elements that are the same in both sets, so the rest of the natural numbers show an excess of numbers compared to zero for the square set. But if you want a less precise comparison, just use cardinal numbers where everything countable is the same size (for transfinite countable numbers), so there is no obvious contradiction, only more or less precision. The lowest precision you can have of infinite values, of course, is using the "oo" infinite concept, where all infinities are the same size = infinite.

[@nzqarc]

@@IngvarLind you would be surprised to know that in some cases, (most cases actually) ∞

[@IngvarLind]

@@nzqarc That really sounds like a real paradox! So, a way to make sense of "most oo < Aleph-null" would be to use a density function on the set of natural numbers, for example, the Odd Numbers has density 50% of N. This wouldn't make it a smaller cardinal number though, so the "

[@nzqarc]

@@IngvarLind it's not a paradox because here is the thing Aleph null is a number. ∞ isn't, it's a "gap" Specifically, it's an unreachable gap from both sides. You can never reach it from below and above. (300×20¹⁰⁰⁰²)×10¹⁰⁰ < ∞ (√(log(log(log(ω)))÷10¹⁰⁰ > ∞

[@josegers5989]

1:47 0 should be in the Natural number Set, I think. I also don't know what the difference is between the Whole number Set and the Integer Set. (I am a math teacher in Belgium, maybe it's is a language difference.

[@Irfirt]

zero is not a natural number

[@josegers5989]

@@Irfirt ok! I just read on nl.wikipedia.org/wiki/Natuurlijk_getal that there is no agreement on the issue. Flemish schoolbooks describe zero as a natural number.

[@minirop]

integer means whole (from Latin), so they are synonymous, but I read on wikipedia that in the 50s and 60s, the US teachers tought that "whole" means "positive integer" (so synonymous or not to "natural" depending on your definition).

[@josegers5989]

@@minirop Thank you for the information. We teach that zero is natural and both positive and negative. Mathematics is truely about conventions.

[@MuffinsAPlenty]

@@josegers5989 You are correct that there is no consensus. Whether 0 is or is not considered a natural number is a convention, which varies depending on branch of mathematics or sometimes language. Typically, branches of mathematics like set theory, mathematical logic, abstract algebra, and category theory consider 0 to be a natural number (because the natural numbers has better structural properties when 0 is considered to be a natural number than when 0 is not considered to be a natural number). On the other hand, branches of mathematics like analysis typically do not consider 0 to be a natural number (because they often want to take reciprocals of natural numbers, which cannot be done with 0). Primary and secondary education in the US and many other countries take a more historical view and teach rigid definitions of "natural numbers", "whole numbers", and "integers" which don't accurately reflect actual practice by working mathematicians.

[@nektariosorfanoudakis2270]

The paradoxa vanish instantly when you realise infinite sets don't exist; at least completed infinities aren't real, only potential infinities. Infinite sets are "incomplete lists", given by formulae, and different "cardinalities" are just fundamentally different rates of "growth", or rather "computability". For example Cantor's theorem on the uncountability of P(N) is similar to Turing's negative proof of the halting problem ("diagonal" arguments). Some infinite sequences are "computable" in a sense, and some others are "uncomputable". For example the "set" of all computable reals, isn't a computable set of computable reals.

[@Seagaltalk]

Set theory is on some shaky ground... oh well we'll just ignore that

[@ivaniliev929]

Tell me you don’t know anything about mathematics without telling you know nothing about mathematics

[@martimlopes8833]

@@ivaniliev929 to be fair this is the attitude that most high level math mathematicians have. They're not necessarily wrong

[@ivaniliev929]

@@martimlopes8833 Most mathematicians do not think that set theory is on shaky ground

[@CutleryChips]

@@ivaniliev929but you have just made a contradiction and therefore OP can’t do that

[@CutleryChips]

@@ivaniliev929Let’s assume he tells you that he doesn’t know anything about mathematics. Then it follows that he told you that he knows nothing about mathematics. QED

[@yetmwerk3093]

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[@silver6054]

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[@yetmwerk3093]

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[@icarus-wings]

If you read this reply the devil has now claimed your soul.

[@yetmwerk3093]

Repeat accept Jesus as your lord and savior by pray simply say Jesus I believe in you and I accept your free give of eternal life please forgive me of all my sins 🙏. Believe the gospel 🙏 😊❤

[@rydawg8934]

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