Here is my procedure: so we want to prove $\forall r\in \mathbb{R},$ there exists a sequence $q_n$ of rationals such that $\forall\epsilon\gt 0,$ there exists a $N$ such that $n\gt N\implies |q_n-r|\lt\epsilon$.

I believe we can apply Denseness of $\mathbb{Q}$, which states that if $a,b\in \mathbb{R}$ and $a\lt b$, then $\exists q\in\mathbb{Q}$ such that $a\lt q \lt b$.

By density theorem, we can claim: let $r\in\mathbb{R}$, since $r-\epsilon\lt r+\epsilon$ for any $\epsilon\gt 0$, $\exists q_n\in\mathbb{Q} $ (this is for any $n\in \mathbb{N}$) such that $r-\epsilon\lt q_n\lt r+\epsilon$. Then we can pick and ideal $N$ such that $|q_n-r|\lt \epsilon$.

First, I am not quite confident about the proof as I felt I skipped or missed lots of elements. Second, my process doesn't seem to follow the "let $r$ be..., let $q_n$ be..., let $\epsilon$ be..., ...". Could someone help me fix the proof?

  • 1
    $\begingroup$ The main idea you seem to have left out is the connection between the index $n$ that subscripts your sequence and the value $\epsilon$ for which you claim $r-\epsilon \lt q_n \lt r+\epsilon$. Your description of doing it once, relying on "Denseness of $\mathbb Q$", is sound, but you pass to having a sequence of such values $q_n$ with the slightly glib remark, "Then we can pick [an] ideal $N$ such that $|q_n - r| \lt \epsilon$." Actually we need to guarantee convergence of the sequence as a whole. $\endgroup$
    – hardmath
    May 7, 2017 at 15:56
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    $\begingroup$ Um, what's the definition of a real number? Whare ARE they? What axioms and definitions are you you assuming. There is utterly no way I can evaluate this proof if I don't know what context you are working in but a suspect this is hopelessly circular. The density theorem usually relies upon every real number being a limit of a sequence of rationals. Which is usually the very DEFINITION of the real numbers. $\endgroup$
    – fleablood
    May 7, 2017 at 16:00
  • $\begingroup$ How are you choosing epsilon or episilons. So I choose q1= 3 so that pi -.5 < 3 <pi + .5. I then choose q2= 3 so that pi -.25 <3 <pi+.25, and I choose q3=3 so that pi-.15 <3 <pi +.15 etc. Does that mean pi is the limit of 3. You may say I never chose the right epsilon but how was I supposed to know. And as for "picking an ideal N", how the heck do I know there is one much less what it is. $\endgroup$
    – fleablood
    May 7, 2017 at 16:13
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    $\begingroup$ It's as fleablood says: if you want to prove something "obvious", you have to be completely sure of the definitions you are starting with - the proof will vary wildly depending on them (this is in contrast to more high level theorems, where you would just use in the proof any "obvious" facts like equivalence of various definitions as a matter of fact). $\endgroup$
    – tomasz
    May 8, 2017 at 0:44

4 Answers 4


You have the right overall idea, but you haven't actually defined a sequence or proved it converges. I think you're getting these two steps confused: the procedure you use to prove the sequence converges is a useful guide to how you want to define the sequence, but defining the sequence and proving it converges are separate tasks.

To define a sequence $(q_n)$, you need to say how to find $q_n$ if I give you $n$. You've said to pick $q_n$ such that $r-\epsilon<q_n<r+\epsilon$, but what is $\epsilon$? You need to specify how $\epsilon$ is defined in terms of $n$. Keep in mind that we don't want to talk about an arbitrary $\epsilon$ at this point--that will only come up when we are proving that $(q_n)$ converges to $r$. In order to define $q_n$ in the first place, we need to be specific about exactly what properties we are defining it to have, and so have to specify all our numbers.

Intuitively, we want the $q_n$ to be getting closer and closer to $r$ as $n$ gets larger. This suggests we want to be making your "$\epsilon$" get smaller and smaller as $n$ gets larger. There are lots of ways to do this; a simple way is to take $\epsilon$ to be $1/n$. So to be precise, we are defining $q_n$ to be some rational number such that $$r-\frac{1}{n}<q_n<r+\frac{1}{n}$$ for each $n$. We know such a $q_n$ exists by the density of $\mathbb{Q}$ in $\mathbb{R}$.

Now we need to prove that the sequence $(q_n)$ really converges to $r$. Here's where the arbitrary $\epsilon$ comes in. Let $\epsilon>0$. We want to prove that there exists $N$ such that for all $n\geq N$, $r-\epsilon<q_n<r+\epsilon$.

So, we need to somehow choose $N$ in terms of $\epsilon$. What do we need to know about $n$ in order to say that $r-\epsilon<q_n<r+\epsilon$? Well, we know that $r-1/n<q_n<r+1/n$, so as long as $1/n\leq \epsilon$, we can conclude that $r-\epsilon<q_n<r+\epsilon$. So we just need to choose $N$ to be such that $1/n\leq\epsilon$ whenever $n\geq N$. For this, we can just choose $N$ to be any integer such that $N\geq 1/\epsilon$. And that completes the argument!

I encourage you to try writing up the argument above more formally as a proof. Hidden below is how I might write it:

Let $r$ be a real number. For each $n\in\mathbb{N}$, let $q_n$ be a rational number such that $$r-\frac{1}{n}<q_n<r+\frac{1}{n}.$$ Such a rational number exists because $\mathbb{Q}$ is dense in $\mathbb{R}$. These numbers form a sequence $(q_n)$ of rational numbers which I claim converges to $r$.

To prove that $(q_n)$ converges to $r$, let $\epsilon>0$. Let $N$ be any integer such that $N\geq1/\epsilon$. Then for any $n\geq N$, $1/n\leq1/N\leq\epsilon$. We thus have $$r-\epsilon\leq r-\frac{1}{n} < q_n < r+\frac{1}{n}\leq r+\epsilon.$$ That is, for any $n\geq N$, $$|q_n-r|<\epsilon.$$ Since $\epsilon>0$ was arbitrary, this proves that $(q_n)$ converges to $r$.

  • $\begingroup$ After the step r-1/n<$q_n$<r+1/n for all n...can I use squeeze theorem ? @ Eric wofsey sir $\endgroup$
    – Normal
    Aug 31, 2018 at 19:00
  • $\begingroup$ Excellent explanation, this is how any real analysis book should explain things $\endgroup$ May 8, 2023 at 6:03

You have not mentioned how the reals are defined. If they are defined as equivalence classes of Cauchy sequences of rationals, then the proof is almost immediate.


This works not just for the rationals but for any set of reals which is unbounded in both directions and has the property that there is a real number $0 < r \le 1/2$ such that between any two distinct elements of the set a and b there is a third element c such that $r \le |(c-a)/(b-a)| \le 1-r$,

The proof essentially involves repeated bisection, with the condition involving r ensuring that this converges.

I don't know if it enough that between any two elements there is a third element without the condition involving r.


If I remember it correctly, we usually don't prove that real numbers are limits of convergent sequences of rational numbers. We DEFINE real numbers that way.

However, I guess you started with a different definition of real numbers and now you are trying to prove that your definition implies the statement in question. But in that case we are missing the key part of your definition. What exactly are your real numbers? It has to be the set of rational numbers with some additional property, for example Least-upper-bound property.

Eric Wofsey already showed us how to formally deduce our statement from the density theorem. Now I would advise to take a step back and try to prove the density theorem again. Why is it true? You couldn't prove it without the unmentioned property of real numbers. This might help you to understand even better what real numbers are.


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