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Let $C$ be the space of all Cauchy sequences of rational numbers. Let $X$ be the space whose elements are equivalence classes of elements in $C$. Define an equivalence relation on $C$ by

$$x\sim y \Longleftrightarrow \lim_{n\to\infty}|x_n-y_n|=0$$

Define, for elements $[x]\,,\, [y] \in X$,

$$[x]*[y]=[(x_n*y_n)]\,\,,$$ $$[x]+[y]=[(x_n+y_n)]\,\,,\, n \in\Bbb N$$

I proved the above operations are well-defined, and this showed commutativity and associativity. I need to show that distributivity of "*" over "+" holds.

Let $x, y, x', y'\in C$ , such that $x\sim x'\,\,,\, y\sim y'$.

In order to show distributivity, would I need to verify

$$\lim_{n\to\infty}|(x_n*y_n)(x_n'+y_n')-(x_n'*y_n')(x_n+y_n)|=0\,\,?$$ Because this is what I did, but I want to make sure it's the correct method.


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up vote 2 down vote accepted

No, in order to show distributivity, you need to show that $[x]([y]+[z])=[x][y]+[x][z]$, that is: If $(x_n)_{n\in\mathbb N}, (y_n)_{n\in\mathbb N}, (z_n)_{n\in\mathbb N}$ are Cauchy sequences, then the Cauchy sequences $(x_n(y_n+z_n))_{n\in\mathbb N}$ and $(x_ny_n+x_nz_n)_{n\in\mathbb N}$ are equivalent, i.e. their difference conveges to zero. But this is trivial, as we have $x_n(y_n+z_n)=x_ny_n+x_nz_n$ for each $n$. In general, any equation that uses only $+$ and $\cdot$ and that holds for all rationals will hold also for this weird space $X$, simply because the equation is valid for each element of the sequence. Of course, (almost) all field axioms are of this type, thus you find (almost) immediately that $X$ is a field.

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Oh okay, I see what you mena. So I would basically show, |(x_n)(y_n+z_n)-((x_ny_n')+(x_nz_n'))| converges to zero? – Alti Oct 17 '12 at 18:18
There is no need for $y_n'$ besides $y_n$ etc. as you have already passed the stage of proving well-definedness. By definition, $[x]([y]+[z])=[x_n([y]+[z])_n]=[x_n(y_n+z_n)]=[x_ny_n+x_nz_n]=[([x][y])_n+([x][z‌​])_n]=[x][y]+[x][z]$. – Hagen von Eitzen Oct 17 '12 at 19:31

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