What does "$\tilde{0}$ is the equivalence class of the constant Cauchy sequence $(0,0,\dotsc)$" mean?

Question

What does "$\tilde{0}$ is the equivalence class of the constant Cauchy sequence $(0,0,\dotsc)$" mean?

I already know what Cauchy sequences, equivalence relations and equivalence class are.

Answer 1

I suppose (you don't specify, so I must guess) that you're talking about a standard construction of the set of reals $\Bbb R$ from $\Bbb Q$ using Cauchy sequences.

We have the set $S$ (say) of all Cauchy sequences in $\Bbb Q$ (with its standard distance) and an equivalence relation $R$ on $S$ and we want to use $S{/}R$ (the set of all equivalence classes of elements of $S$ modulo $R$) as a "model" for $\Bbb R$. This has to be an ordered field like $\Bbb Q$ is, and this sentence tells us what the new $0$ of that set will be: $S$ contains the constant sequence $(0,0,0,\ldots)$ (all constant sequences are trivially Cauchy), and its class $\{(s_n) \in S: (s_n) R (0,0,0,\ldots)\}$, the set of all rational Cauchy sequences equivalent to $(0,0,\ldots,)$ is given a new "name" $\tilde{0}$ as an element of $S{/}R$, or "$\Bbb R$".

Presumably the proof will then go on to show that $\tilde{0}$ has the required properties to be called a zero-element, after we have defined order/operations etc. on the set of equivalence classes. It's a leap in abstraction: from "simple" rational numbers to sets of equivalent sequences of rational numbers. Maybe that is what's bothering you.

Answer 2

Let $X$ be a set and $\mathscr P$ a partition of $X$. Recall the theory

$\quad$Partitions and equivalence relations

If $x \in X$ it belongs to a (unique) block $B$ of the partition $\mathscr P$. If $\sim$ is the corresponding equivalence relation for $\mathscr P$ we use the notation $\stackrel{\sim}{x}$ to represent $B$.

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For the OP's problem, $X \subset {\mathbb Q}^{\mathbb N}$ is equal to the set of all Cauchy sequences. The constant sequence $(0)_{n \ge 0}$ is a Cauchy sequence and so we have the corresponding block of all equivalent rational sequences (the ones converging to $0$), denoted by

$\tag 1 \widetilde {(0)_{n \ge 0}}$

As indicated in Henno Brandsma's comment, we have a natural inclusion

$\quad {\displaystyle \iota :\Bbb Q \hookrightarrow {\mathbb Q}^{\mathbb N}}$

Also, each such constant sequence is a Cauchy sequence. So it is OK, when no confusion is possible, to let $\stackrel{\sim}{0}$ represent the expression in $\text{(1)}$.

One of the first thing you learn when studying this topic is the following

$\quad \text{IF } p, q \in \Bbb Q \text{ and } p \ne q \text{ THEN } \stackrel{\sim}{p} \ne \stackrel{\sim}{q}$