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Let $\mathcal{I}$ be the family of open subsets of $\mathbb{R}$ comprising of $\mathbb{R}$ and all open intervals with rational endpoints.

Prove that each open interval in $\mathbb{R}$ is a union of countable many of its members.

I am a bit confused by this, it seems to me that each open interval could be a union of three of its members as follow:

Take the interval $(a,b)\in\mathbb{R}$ where $a,b\in\mathbb{Q}$.

Then from the deinsity of the rationals we have that $c\in(a,b)$ where $c\in\mathbb{Q}$

We then have the following two intervals in $\mathcal{I}$ $(a,c)$ and $(c,b)$

Then again from the density of the rationals we have $c_1\in (a,c)$ and $c_2\in(c,b)$ with $c_1,c_2\in\mathbb{Q}$

We then have that $(a,b)=(a,c)\cup(c_1,c_2)\cup(c,b)$

However I'm sure this cant be right, could someone tell me whats wrong with this argument?

Thanks for any help.

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I think you've misunderstood the question. Presumably, you are asked to show that every open interval in $\mathbb{R}$, whether or not it has rational endpoints, is a union of countably many members of $\mathcal{I}$. Of course, your argument will not work for intervals whose endpoints are not rational, so you have to do something else. – Nate Eldredge Nov 9 '11 at 20:21
You are correct, I have misunderstood the question, thanks very much (sorry about that) – hmmmm Nov 9 '11 at 20:34
up vote 2 down vote accepted

Your method is not true if $a$ or $b$ (or both) are irrationals. However it is the prelude to the correct answer.

Consider the interval $(a,b)$. Take two monotonous rational sequences $a_n\to a, b_n\to b$ such that $a<a_n<b_n<b$ for all $n$.

Now we can prove that $\bigcup (a_n,b_n)=(a,b)$:

If $x\in\bigcup (a_n,b_n)$ then for some $n\in\mathbb N$ we have $a<a_n<x<b_n<b$, in particular $x\in (a,b)$. So this is the simple part.

On the other hand, if $x\in(a,b)$ we have that $x-a>\varepsilon$ for some $\varepsilon>0$, and by the fact that $a_n\to a$ we have that for some $k\in\mathbb N$, if $n>k$ then $a_n-a<x-a$, and in particular $x\in(a_n,b_n)$.

A similar consideration shows that $x<b_n$ for some $b_n$. Since the intervals are increasing in inclusion we have that for some number $k\in\mathbb N$ it holds that $x\in(a_k,b_k)$, as wanted.

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