Let $S$ be the set of all cauchy sequences.Does the set countable?

Question

Let $S$ be the set of all Cauchy sequences.Does the set countable?

Can we define a mapping from $\mathbb N$ to $S$ such that $f(n)$$=$$a_{m_n}$ for all $n\in \mathbb N$ where $a_{m_1}$,$a_{m_2}$, . . are Cauchy sequences and from that can we say that the set $S$ is countable ?

Do you mean real sequences, being Cauchy in the Euclidean metric? — Dimitris, May 12 '14 at 07:10
"Does the set countable?" Of course not. That would mean that $\mathbb R$ is countable. — TonyK, May 12 '14 at 07:10

André Nicolas · Accepted Answer · 2014-05-12T07:16:51.727

2

Consider the set of Cauchy sequences of shape $(a_n)$ where $a_n=\frac{\delta_n}{2^n}$, and each $\delta_n$ is $0$ or $1$, or if you prefer, $-1$ or $1$. This set already has the cardinality of the continuum, so is in particular uncountable. It follows that the set of Cauchy sequences of rationals is uncountable.

edited May 12 '14 at 07:16

answered May 12 '14 at 07:11

André Nicolas

507,029

score 2 · Answer 2 · answered May 12 '14 at 07:16

2

The irrational numbers are limits of sequences of rational numbers. These sequences must be Cauchy since they converge in $\mathbb{R}$, but the irrational numbers are uncountable therefore the set of all Cauchy sequences must be uncountable.

answered May 12 '14 at 07:16

EgoKilla

2,538

Let $S$ be the set of all cauchy sequences.Does the set countable?

2 Answers2