Cardinality of Rationals

Cardinality of Rationals Mohammed Farhaan Abstract The Continuum Hypothesis (CH) is a profound and foundational statement in set theory concerning the possible sizes of infinite sets, specifically addressing whether any set exists with cardinality strictly between that of the integers (ℵ0) and the real numbers (c, the continuum). Its paramount importance stems not from its truth value, but from its independence from the Zermelo-Fraenkel set theory with the Axiom of Choice (ZFC), proven by Kurt Gödel and Paul Cohen. This independence implies that ZFC, the conventional foundation for nearly all of modern mathematics, is incomplete—it cannot resolve the size of the continuum. Therefore, the CH serves as a critical philosophical and mathematical benchmark, highlighting the limitations of our current axiomatic system and spurring the search for new, universally accepted axioms (such as large cardinal axioms) that might provide a definitive answer and solidify the structure of the mathematical universe. In this paper we invalidate any proofs of cardinality that stem from functions mapping a set to another non-distinct set. We try to build an equivalence of cardinalities purely using set intersections and complements to reason about wether two infinite sets can be considered equally sized. 1 Introduction Note the cardinality of Naturals we denote ℵ0. Given two infinite sets A0and B0we claim the following: 1. Construct A1and B1: A1=A0−A0∩B0 B1=B0−A0∩B0 Hypothesis: we claim that |A0|=|B0|iff |A1|=|B1| Motivation: the idea follows from the preference that if we have elements xs ∈A0∩B0we want bijections f:A→Bwith f(x)=x. All x∈A0∩B0must be fixed points under all bijective maps. This allows us to exclude maps that abuse and exploit infinity by exploiting a trick that seems obvious in the finite case. This Hypothesis should be easy to prove for examples where you consider A0=A,B0=A. And when one considers disjoint infinite sets, our Hypothesis claims they are all not unequal. Let A0∩B0=∅. Then constructing A1, B1: A1=A0−A0∩B0=A0−∅=A0 similarly. B1=B0 i.e. the question |A0|=|B0| is still undetermined by |A1|=|B1|. Now let’s use this new notion of equivalence between two sets A1, B via their cardinalities. Theorem Cardinality of whole numbers is not equal to cardinality of Natural numbers. 1 Proof W={0,1,2,3, ..} N={1,2,3,4, ..} Let fbe a bijection f:W−→ N W∩N=N Construct A0(let W=A0,N=B0), A1and B1: A1=A0−A0∩B0 A0−A0∩B0 =∅ Since |A1| =|B1|we conclude that |A0| =|B0|. Lemma 1 If A0, B0can be used to construct A1and B1, we can use A1, B1to construct A2, B2. A2=A1−A1∩B1 B2=B1−A1∩B1 Similarly we can An=An−1−An−1∩Bn−1 Bn=Bn−1−Bn−1∪Bn−1 So we claim that if An∩Bn=∅ then equality is indeterminable. Take An={x|xmod 2 ≤0, x ∈N} Bn={x|xmod 3 ≤0, x ∈N} An∩Bn={x|xmod 6 ≤0, x ∈N} Note: we are not constructing the set of multiples of 2 using A0={2× |x|∈N} to avoid constructing a set with the same cardinality as the naturals. A1={x|xmod 2 ≤0, x mod 6 ≤0, x ∈N} B1={x|xmod 3 ≤0, x mod 6 ≤0, x ∈N} A1∩B1=∅={x|xmod 2 ≤0, x mod 3 ≤0, x mod 6 ≤0, x ∈N} 1. A1and B1are disjoint. A1=A2···An B1=B2······Bn 2. We cannot determine if the cardinality of the set of even numbers is equal to the cardinality of the set of multiples of 3. Lemma 2: |A∪B1|= (A1+ 18) if |A∩B1|=∅ 3. We would like to build up to a proof that shows that |A|={x|xmod k≤0, x ∈N}=ℵ0 k 2 to prove this we would need to show that Xi={x|xmod k≡i, x ∈N} Claim k−1 [ i=0 Xi=N Then      k−1 [ i=0 Xi      =ℵ0 and k−1 X i=0 |Xi|=ℵ0 Since Xi∩Xj=∅for any i, j and |Xi|=|Xj|for any i, j if we make the above assumption we can show that k−1 X i=0 |Xi|=ℵ0=k|X0| |X0|=ℵ0 k On an equally daring note lets just denote the cardinality of pairs of natural numbers Let N2={(x, y)| x, y ∈N} |N2|=ℵ2 0 Let Q={(a, b)|a, b ∈Nand a bis an irreducible rational} using lemma 2 a=_|Ba|where Ba=b a b ais irreducible, a, b ∈N 1ℵ0 2ℵ0−ℵ0 2 3ℵ0−ℵ0 3 4ℵ0−ℵ0 2 5ℵ0−ℵ0 5 6ℵ0−ℵ0 2−ℵ0 3+ℵ0 6 . . . 30 ℵ0−ℵ0 2−ℵ0 3−ℵ0 5+ℵ0 10 +ℵ0 15 +ℵ0 6−ℵ0 30 . . . |Q|=X a∈N |Ba|or [ a∈N Ba=Q 3 so we need to compute Pa∈N|Ba| X a∈N |Ba|=ℵ0−ℵ0 2+ℵ0−ℵ0 3+ℵ0−ℵ0 5+ℵ0−ℵ0 2−ℵ0 3+ℵ0 6+··· =|B1|+|B2|+|B3|+··· =ℵ2 0−ℵ2 0 22−ℵ2 0 32· · · +ℵ2 0 62+ℵ2 0 102· · · − ℵ2 0 302· · · =ℵ2 0 ζ(2) =6ℵ2 0 π2 2 Further Interests Consider the set A0, A1and N A0={2x|x∈N} A1={x|xmod 2 ≡0, x ∈N} N=naturals. A0is basically a set that is created using a function f(x) f:N−→ A0 every element in Nhas an image in A0A1is created by constructing a subset of Nsuch that A1⊆N as we have shown in this paper |A1|<ℵ0 but |A0|=ℵ0?or we may say |A0|=ℵ0 we know |A0|=|A1|since A0∩A1=A0=A1 therefore only by transitivity we have a contradiction implying that 1. either transitivity of an equivalence relation is trickier with infinite sets 2. f:N→A0is not a cardinality preserving map. If our contradiction implies (2), We can assume that these contradictions will be restricted if we consider maps fsuch that f:M−→ Nand M∩N=∅ in this example , our sets are not disjoint hence A0∩N=A0 4 References [1] Gödel, K. (1940). The Consistency of the Continuum Hypothesis. Princeton University Press. [2] Cohen, P. J. (1966). Set Theory and the Continuum Hypothesis. W. A. Benjamin. [3] Jech, T. (2003). Set Theory, The Third Millennium Edition. Springer Monographs in Mathematics. [4] Woodin, W. H. (2001). 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