# Countably Many Subdivisions of the Real Line into Open Intervals

1. Feb 10, 2008

### PingPong

1. The problem statement, all variables and given/known data
Let $$A=\{(a_\alpha,b_\alpha),\alpha\in I\}$$ be a family of mutually non-intersecting intervals on the real line. Here $$I$$ is an arbitrary set of indexes. Prove that this family contains at most countably many elements (intervals).

2. Relevant equations

None, other than all of the stuff on countability of a set.

3. The attempt at a solution

I've tried to find an enumeration but I haven't gotten anywhere. I'm guessing that this is a property of the real line - it can only be split up into a countable number of open intervals, but I can't see how to prove it. A hint to start may be all I need.

At least I've been able to get the other five problems on this homework :)

2. Feb 10, 2008

### quasar987

To show a set is countable, it is sufficient to show there is an injection from the set to some countable set... such as, oh I don't know... the rationals?

3. Feb 10, 2008

### PingPong

Okay, (correct me if I'm wrong then), I can define an injection $$f:A\rightarrow Q$$ as $$f((a_\alpha,b_\alpha))=\frac{a_\alpha}{b_\alpha}$$ (in my class we've used bijections, but I can see that an injection will also work - surjectivity is not necessary for a set that may not be infinite). My only qualm about that is, what if $$b_\alpha$$ is zero for some alpha? Also, I'm assuming that we can ignore the uniqueness of a rational number (that is, ignore the fact that 1/3 and 9/27 are in the same class). Afterall, we showed in my class that $$Q^+ \sim J\times J$$, which also ignores differentiation of elements within classes.

Thanks for your help!

4. Feb 10, 2008

### Dick

a0/b0 isn't necessarily rational! And you can't ignore your uniqueness problem. quasar987 is likely suggesting you just pick a rational in each interval and assign it to f((a0,b0)). So your intervals can be placed in 1-1 correspondence with a subset of Q.

5. Feb 10, 2008

### mXCSNT

You could start by counting all the intervals of length at least 1/2.

6. Feb 10, 2008

### EnumaElish

What if you used $f_\alpha = f((a_\alpha,b_\alpha), k_\alpha)=k_\alpha a_\alpha + (1-k_\alpha)b_\alpha$ for some $0 < k_\alpha < 1$ (e.g. "some middle point" in each interval)?

A remaining question is, what if either of a or b is irrational for some alpha, so $f_\alpha$ is irrational? What property of the Rationals can you use to find a rational number "near" $f_\alpha$ and within $(a_\alpha,b_\alpha)$ for each $\alpha$?

Last edited: Feb 10, 2008
7. Feb 10, 2008

### mXCSNT

My mistake, I misread it... first map the real line to the interval (0, 1) then count all the intervals whose image under that map have length at least 1/2, of which there are at most 1.

8. Feb 10, 2008

### quasar987

The argument is very simple. Every interval of R contains rationals. Create a map that assigns to every alpha in I a rational contained in $$(a_\alpha,b_\alpha)$$. Since your intervals are disjoint by hypotheses, this map in injective. Done. You've created an injection from I to Q.

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