# Proving well definedness of addition in real numbers constrructed from cauchy sequences.

While studying real analysis, I got confused on the following issue.

Suppose we construct real numbers as equivalence classes of cauchy sequences. Let $x = (a_n)$ and $y= (b_n)$ be two cauchy sequences, representing real numbers $x$ and $y$.

Addition operation $x+y$ is defined as $x+y = (a_n + b_n)$.

To check if this operation is well defined, we substitute $x = (a_n)$ with some real number $x' = (c_n)$ and verify that $x+y = x'+y$. We also repeat it for $y$. i.e. we verify that $x+y = x+y'$.

Question:

Instead of checking that $x+y = x+y'$ and $x'+y = x+y$ seperately, would it suffice to check that $x+y = x' + y'$ in a single operation in order to show that addition is well defined for real numbers. Would it hurt to checking well definedness? Can any one explain me the logic behind ?

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You could certainly do it that way, but it looks more complicated to me. – TonyK Mar 21 '14 at 11:46

You have to differentiate typographically between (a) sequences and (b) equivalence classes of sequences, i.e., real numbers.

Write $x$ for the Cauchy sequence $(x_n)_{n\geq1}$ and $[x]$ for the equivalence class represented by $x$.

Since addition of real numbers is described in terms of representants: $$[x]+[y]:=[x+y]\ ,$$ we have to check whether this actually defines a binary operation on ${\mathbb R}$. It is sufficient to prove that $$x\sim x'\qquad\Rightarrow\qquad x+y\quad \sim\quad x'+y\ ,$$ for then we can argue as follows: When $x'\sim x$ and $y'\sim y$ then using commutativity we have the following chain: $$x+y\ \sim x'+y\ =\ y+x'\ \sim \ y'+x'\ =x'+y'\ .$$

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Wow. I was trying to answer an hour ago and wanted to come up with a substantial argument, and totally missed the fact that they're identical by commutativity. Silly me. – Tony Mar 21 '14 at 10:55

You can certainly check that the operation is well-defined by checking $x+y=x'+y'$ where $x$ and $x'$ are equivalent and $y$ and $y'$ are equivalent.

Let $(x_n)$, $(x_n')$, $(y_n)$ and $(y'_n)$ where $x_n-x_n'\to 0$ and $y_n-y_n' \to 0$, i.e., are equivalents.

So in ordered to prove the claim we need to show that $(x_n+y_n) -(x_n'+y_n')\to 0$, i.e., $x_n+y_n$ is equivalent to $x_n'+y_n'$. Given $\varepsilon>0$, choose $N$ such that $d(x_n,x'_n)<\varepsilon/2$ and $d(y_n,y_n')<\varepsilon/2$ at the same time for all $n\ge N$. Thus

\begin{align}|x_n+y_n-(x_n'+y_n')|=|x_n-x_n'+y_n-y_n'|\\\le |x_n-x_n'|+|y_n-y_n'|\\< \varepsilon \end{align}

Therefore, $(x_n+y_n)$ and $(x_n'+y_n')$ are equivalent and the addition is independent of choice of representatives.

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