Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 ?

share|improve this question
    
You could certainly do it that way, but it looks more complicated to me. –  TonyK Mar 21 at 11:46

2 Answers 2

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: $$x+y\ \sim x'+y\ =\ y+x'\ \sim \ y'+x'\ =x'+y'\ .$$

share|improve this answer
    
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 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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.