Equivalence relations and binary operations

Question

Let S be the set of all sequences of real numbers. Define a relation $\sim$ on S by $\{x_n\} \sim \{y_n\}$ if $x_n - y_n \rightarrow 0$.

(i) Prove that $\sim$ is an equivalence relation.

(ii) Let $[\{x_n\}]$ denote the equivalence class containing $\{x_n\}$ and let $\mathcal{C}$ be the set of all equivalence classes in S with respect to $\sim$. Define $F$ : $\mathcal{C}$ x $\mathcal{C} \rightarrow \mathcal{C}$ by $F([\{x_n\}], [\{y_n\}]) = [\{x_n + y_n\}]$. Show that $F$ is a well-defined function; in other words, $F$ is a binary operation on $\mathcal{C}$.

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I have proved part (i), but I am stuck on (ii), can anyone help?

Answer 1

Each equivalence class has many members. Thus, if $C_1=[\{x_n\}],C_2=[\{y_n\}]\in\mathscr{C}$, it’s conceivable that $F(C_1,C_2)$ might depend on which members of $C_1$ and $C_2$ you chose. If that were the case, saying that $$(C_1,C_2)=[\{x_n+y_n\}]$$ would be meaningless, since you might have $\{u_n\}\in C_1$ and $\{v_n\}\in C_2$ such that

$$F(C_1,C_2)=[\{u_n+v_n\}]\ne[\{x_n+y_n\}]=F(C_1,C_2)\;.$$

To show that $F$ is well-defined, you must show that this cannot happen: no matter which $\{u_n\}\in C_1$ and $\{v_n\}\in C_2$ you choose, $[\{u_n,v_n\}]$ is always the same member of $\mathscr{C}$, i.e., the same equivalence class.

To do this, show that if $\{x_n\}\sim\{u_n\}$ and $\{y_n\}\sim\{v_n\}$, then $\{x_n+y_n\}\sim\{u_n+v_n\}$.

Answer 2

A well defined function, particularly, a well defined binary operation $\phi: X\times X\to X$ has the following property:

$$ \forall(x_1,y_1),(x_2,y_2)\in X\times X:(x_1,y_1)=(x_2,y_2), \phi(x_1,y_1)=\phi(x_2,y_2)$$

Here, showing that $F$ is well-defined on $C\times C$ is as simple as showing that $$[\{x_n\}],[\{y_n\}]=[\{\alpha_n\},\{\beta_n\}] \implies F\big([\{x_n\}],[\{y_n\}]\big)=F\big([\{\alpha_n\},\{\beta_n\}]\big)$$

Can you take it from here?