# Constructing $\mathbb{R}$ from $\mathbb{Q}$ and showing $\mathbb{Q}$ is dense in $\mathbb{R}$

This is a very long, multi-part problem that we were told to figure out by any means possible. There are no limits on getting help or finding answers online. I haven't had much luck at all solving this for myself, can anyone help me with some of this?

Thanks!

EDIT: I have made it to number (vi), which is where I need some help. I don't really understand that I am expected to do. Can someone assist from there? Thanks!

• Each step is comparatively easy! Just start and work closely to the definitions. If you have specific questions regarding any of these steps, then ask again. Apr 22, 2014 at 19:06
• For instance, can you do part (i), namely, define the addition and multiplication by $\mathbb{Q}$ on $\mathcal{C}(\mathbb{Q})$? If you can't get any of it, you should look it up in a book, since it's quite a lot for someone to write out for you. Apr 22, 2014 at 19:08
• This very basic approach becomes surprisingly easy once one knows some ring theory... Apr 22, 2014 at 19:12
• The order in (iv) is not defined correctly; consider, for example, the sequence $x_n = -1/n$. A correct definition could be for every $\epsilon>0$, there exists $N$ so that for $n>N$, $x_n>-\epsilon$. Apr 23, 2014 at 1:00
• @cQQkie I made an edit with where I now need help. Any suggestions? Apr 25, 2014 at 2:53

When you're thinking about things, understand that decimal expansions are good examples of Cauchy sequences. This justifies writing $$\sqrt{2} = 1.414213562373....$$ This is really a sequence of numbers whose square gets closer and closer to $2$. That is, in your context, $$\sqrt{2} = [\{ 1, 1.4, 1.41, 1.414, 1.4142, 1.41421, \cdots\}]$$ The square of this sequence is another Cauchy sequence that is equivalent to the constant sequence $[\{2,2,2,2,2,\cdots\}]$ because the following difference sequence tends to $0$: $$\{2-1^{1},2-(1.4)^{2},2-(1.41)^{2},2-(1.414)^{2},\cdots\}$$ Using these definitions, there are multiple ways to express the number $1$: $$1 = [\{1,1,1,\cdots\}]=[\{0.9,0.99,0.999,0.9999,\cdots\}] = 0.999999\cdots$$ To remove the ambiguity, you're lumping together all Cauchy sequences whose differences have an actual limit of $0$. That's what the equivalence relation is all about. An equivalence class $[\{ x_{n}\}]$ consists of all Cauchy sequences $\{ y_{n}\}$ such that $\{ x_{n}-y_{n}\}$ has a limit of $0$.

The first thing you have to do is show that you can add these equivalence classes of Cauchy sequences. This is done by taking one Cauchy sequence from $x=[\{ x_{n}\}]$ and another from $y=[\{ y_{n}\}]$ and showing that (a) $\{ x_{n}+y_{n}\}$ is Cauchy sequence, and (b) no matter what one you choose from $x$ and what you one you choose from $y$, you always get a Cauchy sequence which is equivalent to the first one $\{ x_{n}+y_{n}\}$. In other words, $\{ x_{n} \} \sim \{ x_{n}'\}$ and $\{ y_{n}\} \sim \{ y_{n}'\}$ implies $\{ x_{n}+y_{n} \} \sim \{ x_{n}'+y_{n}'\}$. Then you have a good definition for $x+y=[\{x_{n}+y_{n}\}]$ that doesn't depend on what members $\{ x_{n}\}\in x$ and $\{ y_{n}\}\in y$ of the equivalence classes $x$ and $y$ that you use.

A couple of things you'll need to be able to show: The sum of two Cauchy sequences is also a Cauchy sequence. And, a Cauchy sequence is bounded. Most everything else is just slogging through the details.

For $(i)$ if $C(\mathbb{Q})$ is the set of all cauchy sequences, we can see that the $0$ element would be the sequence $(0,0,0,...)$, and if we take two cauchy sequences $f_n,g_n\in C(\mathbb{Q})$, then $f_n+g_n=(f_1+g_1,f_2+g_2,...)$, and we can see the addition is still cauchy since: $|(f_n+g_n)-(f_m+g_m)|\le|f_n-f_m|+|g_n-g_m|\lt 2\epsilon$.

And $|\alpha f_n -\alpha f_m|=|\alpha||f_n-f_m|\lt |\alpha|\epsilon$, thus scalar multiples are cauchy, thus $C(\mathbb{Q})$ is a vector space.

• You should put the parts you have done/attempted in your question, and you will probably get more help. Is there any bit you're particularly stuck on? Apr 25, 2014 at 7:13

For (vi): Injectivity: Call this inclusion map $i$. Suppose $i(q)=i(p)$ for some $p,q\in \mathbb{Q}$. This means $[(p,p,\dots)]=[(q,q,\dots)]$ and therefore $(p,p,\dots)$ and $(q,q,\dots)$ are represantatives of the same equivalence class, which means by definition of the equivalence relation...

Field structure: You have shown in (iii) that the equivalence classes form a field, so check that $i(\mathbb{Q})$ has also the same field structure inherited from $\mathbb{R}$ (as defined above) and can hence be considered as copy of $\mathbb{Q}$ in $\mathbb{R}$ (as defined above).

• Thanks! I was a little confused on how to check $i(\mathbb{Q})$ has the same structure as $\mathbb{R}$? I don't really understand what to apply to $\mathbb{Q}$ and then check? Apr 27, 2014 at 21:05