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 Question from Justin, a student: Hello, I was just wondering why the infinity from real numbers is smaller than Beth Two in the context of Cantor's cardinality set theory?

Justin,

The phrasing "infinity from real numbers" is potentially confusing. We are not (or should not be) talking about the "infinity" that we adjoin to the real numbers as an upper bound in calculus. In fact, we don't usually use this word to talk about numbers like Beth2: we call them "transfinites".

Before we go on, you need to know (perhaps you do already?) that there is a theorem ("Cantor's theorem") that says that every set, empty, finite or infinite, has strictly more subsets than it has elements, in the sense of cardinality. "Cardinality" is a concept that generalizes "number" or "size" of a set to include infinite cases; it is defined in terms of the existence of 1-1 pairings. The proof of the theorem shows that any supposed such paring must break down, by a sort of "Spanish Barber" paradox. Suppose we have such a pairing: consider the set of all numbers that are not in the set they are paired with. Is it paired with itself or isn't it? [Evil supercomputer freezes and smoke starts to come out...] So there cannot be such a pairing.

Beth1 is equal to the cardinality of the real numbers, at least in standard axiomatizations. This is because Beth0 is the cardinality of the natural numbers, and each [non-limit] transfinite in the Beth sequence is the number of subsets of the one below it. Real numbers correspond (more or less; there are some technicalities about place value and 0.111... = 1.000... but these can be worked around) to binary (or decimal, but binary is easier) expansions, and every subset of the natural numbers corresponds to an expansion in which exactly those digits equal 1.

Beth2 is the number of subsets of Beth1 - by Cantor's theorem it must be bigger than Beth1.

Good Hunting!
-RD

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