What's the cardinality of a single equivalence class of Cauchy sequences in ℚ?


To clarify, I'm not asking for the cardinality of the real numbers, but for the cardinality of the set of Cauchy Sequences that are equivalent to any single real number.



Cardinality of Equivalence Classes of Cauchy Sequences
It's the same as the size of the real numbers. Here's a rough sketch of the proof.


For each element of (0,1) (which has the same cardinality as the reals), I'm going to construct a distinct sequence of rationals that converges to 0.


Think of an element of (0,1) in binary, so as an infinite sequence of 0s and 1s. For definiteness, we assume the sequence is never eventually constant 1 (this just gives a well defined bijection between (0,1) and the sequences, since, e.g., .1000000... = .0111111... ). Given such a sequence of 0s 1s $a\_1, a\_2, ...$, create the rational sequence $(-1^{a\_1}(1), -1^{a\_2} (1/2), -1^{a\_3} (1/3), ..., -1^{a\_n} (1/n),...)$.


Convince yourself all these sequences of rational numbers are distinct.


Thus, the size of the reals is less than or equal to the number of rational sequences converging to 0. For a bound in the other direction, note that the collection of ALL sequences of rationals = $\prod\_{\mathbb{N}} \mathbb{Q}$ and $|\prod\_{\mathbb{N}} \mathbb{Q}| = |\mathbb{Q}|^{|\mathbb{N}|} = |\mathbb{N}|^{|\mathbb{N}|} \leq |2^{\mathbb{N}}|^{|\mathbb{N}|} = |2^{\mathbb{N}}| =$ the size of the reals.


By the Cantor-Schroeder-Bernstein theorem, the set off all rational sequences converging to 0 has the same cardinality as the reals.


