# Countably Many Subdivisions of the Real Line into Open Intervals

• PingPong
In summary, the conversation discusses proving that a family of mutually non-intersecting intervals on the real line contains at most countably many elements. The method proposed is to show that there exists an injection from the set of intervals to the set of rational numbers, which can be done by assigning a rational number to each interval. The conversation also considers potential issues with this method, such as the possibility of irrational numbers being assigned, but a solution is proposed using the property that every interval of the real line contains rationals.
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 :)

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?

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.

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.

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

PingPong said:
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.

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:
mXCSNT said:
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.

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.

## What does it mean for the real line to be divided into open intervals?

Dividing the real line into open intervals means that the entire real numbers are partitioned into smaller sections, with each section being an open interval. This allows for a more organized and systematic approach to studying the properties of the real numbers.

## Why is it important to have countably many subdivisions of the real line into open intervals?

Countably many subdivisions of the real line into open intervals allow for a more precise and detailed analysis of the real numbers. It also helps in understanding the structure and behavior of the real numbers, as well as in solving mathematical problems that involve the real numbers.

## What is the significance of open intervals in this context?

Open intervals are important in this context because they allow for the inclusion of all real numbers between two distinct points. This means that no real number is left out and the entire real line is covered without any overlap or repetition.

## How are countably many subdivisions of the real line into open intervals related to the concept of infinity?

Countably many subdivisions of the real line into open intervals are closely related to the concept of infinity because the real numbers are infinite and the process of dividing them into smaller intervals is also infinite. This concept becomes even more intriguing when considering that there are different sizes of infinity, and the real numbers are an example of an uncountable infinity.

## Can the real line be divided into uncountably many open intervals?

No, the real line cannot be divided into uncountably many open intervals. This is because the real numbers themselves are uncountable, and if we were to divide them into uncountably many intervals, there would be an infinite number of intervals with no real numbers in them. This goes against the definition of open intervals, which must contain at least one real number.

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