TITLE: Are all infinities equal?
QUESTION [82 upvotes]: A friend of mine was trying to explain to me how all infinities are equal. For example, they were saying that there are the same amount of numbers between $0$–$1$ as there are between $0$–$2$.
The way they explained it, you could prove that there are the same amount of numbers by getting a match to each number.
For example, for any number in the range from $0$–$2$, you can find a matching number in the range $0$–$1$ by dividing the number by 2. In every case, the ending number will end up between $0$–$1$. In addition, if you multiplied any number in the range of $0$–$1$ by 2, you will always end up with a number between $0$–$2$ Therefore, there should be the same infinite amount of numbers between  $0$–$1$ as there are between $0$–$2$.
Is this principle accurate?
The problem that I had with it was, if in-fact there are the same amount of numbers between 0 and 1 as there are between 0 and 2, why is 2 greater than 1?

REPLY [3 votes]: What your friend was trying to explain is the concept of "trans-infinite numbers". The short answer to the title question is that many infinities are equal, but some infinities are not.
The proof you give is valid; for any $0\leq x\leq 2$, if $y=x/2$ then $0<y<1$, and vice-versa. Therefore, while both ranges have an infinite number of numbers, there must be the same number of numbers in both ranges, because we have a 1:1 transformation possible between the two sets, by which every number in one set can be converted to a number in the other set and vice-versa, with no overlap (no two values from the source set may produce the same number of the resulting set), no gaps (all values in the resulting set can be produced with some value from the source set) and no ambiguity (no value from the source set may produce two values in the resulting set). If there were not the same number of numbers in both sets, then there would not be a transformation (called a bijection) meeting all the above conditions.
This number of numbers is "aleph-null" $\aleph_0$, and is the cardinality of the set of all rational numbers (because the same proof holds for any $0\leq x\leq 2n$ and $0\leq y\leq n$ as $n \to \infty$), and for the set of all natural numbers (because for any natural $0\leq x\leq 10^n$ there is a $y=x/10^n$ such that $0\leq y \leq 1$, thus there are the same number of rational numbers between 0 and 1 as between 0 and any power of 10) and the set of all integers (there is a similar bijection that can evenly project any natural $n \in \mathbb N \to i \in \mathbb Z$ and thus there are as many integers as natural numbers).
However, there are a greater number of real numbers (rationals and irrationals) than aleph-null. This theorem is based on the difference between a rational and irrational; the number of non-repeating decimals of any rational number, though it may be uncountably large, is finite by definition; we may say it has finite "precision". A real number has infinite precision; we could hook an RNG (something based on non-computational randomness) up to a printer and have it print random digits until the end of time, and after it prints each new digit, we have an entirely new real number. There are aleph-null possible real numbers that can be produced by concatenating any arbitrary natural number's digits to one existing real number. So, the cardinality of the infinite set of all irrational numbers is equal to the cardinality of the set of all sets of natural numbers. This cardinality is generally notated as "beth-one" ($\beth_1$), which is defined by stating that $\beth_0 = \aleph_0$ and that any $\beth_{n+1} = 2^{\beth_n}$ (the cardinality of the "power set" - the set of all subsets of a set). We don't use an aleph number for this because the aleph numbers are defined to be cardinal; there is a number $\aleph_1$ which is defined to be one more than $\aleph_0$, while the beth numbers are defined by an exponential series, not an incremental one.
As to why 2 is greater than 1, given that there are the same number of numbers between any two numbers, it's quite simple; the value of 2 is not dependent on how many rational or real numbers there are between it and any other reference point (such as 0). It is dependent on how many natural numbers there are between it and zero, and that set is finite and countable.