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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.

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5 Answers 5

up vote 7 down vote accepted

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.

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How about just taking subsequences of the sequence $a_n=1/n$?

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The answer is that the cardinality is equal to that of the continuum: i.e., that of the real numbers, which is (independent of the continuum hypothesis!) also equal to the the cardinality of the power set of the natural numbers.

Here's a sketch of a construction; you should try to fill in the details.

(i) Essentially by the above comments, the cardinality of the set F of all functions f: Z^+ -> {1,-1} is that of the continuum.

(ii) Suppose {x_n} is a Cauchy sequence, and let f be an element of F. Consider the collection of all sequences {x_n + 2^{-n} f(n)} as f ranges over all elements of F.

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It is independent, I just sketched a proof using it because I'm lazy. –  Harry Gindi Dec 8 '09 at 15:45

Here is an easier proof that the cardinality is continuum. (Perhaps this is what Gowers was suggesting.)

First, any Cauchy sequence is equivalent to any of its subsequences. Second, if the original sequence has no repetitions, then there are continuum many such subsequences, since every subsequence is determined by an infinite subset of the indices, and there are continuum many such subsets. Finally, every Cauchy sequence is easily seen to be equivalent to a Cauchy sequence with no repetitions. Voilà!

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Violin, Cello, Bass?! –  alekzander Dec 8 '09 at 23:50
Oops. Please pardon my French... :-) –  Joel David Hamkins Dec 9 '09 at 0:44

The cardinality is strictly greater than ℵ0.


We can construct distinct but equivalent Cauchy Sequences from the set of all irrational numbers between 0 and 1 by removing all elements from some Cauchy Sequence that correspond to zeros in the number's binary decimal form, since all infinite subsets of a Cauchy Sequence form equivalent Cauchy Sequences (I assume).

However, I can't prove whether it's equal to ℵ1.

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