Countably Many Subdivisions of the Real Line into Open Intervals

[PingPong]

Homework Statement

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).

Homework Equations

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

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 :)

 

[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?

 

[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!

 

[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.

 

[mXCSNT]

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

 

[EnumaElish]

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!

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?

 

[mXCSNT]

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

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.

 

[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.