Countable/Uncountable collections

Question

I'm asked to produce an example of a countable collection of disjoint open intervals.

At first I had trouble seeing how this is possible since open intervals are not countable. My idea is to have my collection be: $\bigcup^{\infty}_{n=0}$$A_n$, where $A_n$ is the open interval $(n, n+1)$. Then, considering the function $f:\bigcup^{\infty}_{n=0}$ $A_n$ $\rightarrow$ $\mathbb{N}$ where $f(A_n)=n$, is this a bijection? Basically I'm asking if I'm missing anything in my example or if my example is correct.

I am then asked to give an example of an uncountable collection of disjoint open intervals, or to argue that no such collection exists.

My instinct when thinking about it wants to say that yes, this is possible... yet now I'm at a loss for where to start thinking of an example of such a collection.

Answer 1

You can't write $\bigcup \limits_{n = 0}^\infty A_n$, as this is the union of all intervals. You actually want the set that consists of all these intervals, i.e. $\{A_n \;|\; n \in \mathbb{N}\}$. Then your example works.

Hint for the second question: Observe that there are always at most countably many disjoint open intervals with length > 1. There are also at most countably many disjoint intervals of length > 1/2, countably many disjoint intervals of length > 1/3 etc..

Answer 2

There is no uncountable collection of disjoint open intervals.

Suppose there is and call it $\{I_{\alpha} \ \mid \ \alpha \in J\}$. An easy argument is to see that each of these open intervals contains a rational number $r_\alpha$ which would disburse an uncountable amount of rational numbers. This is a contradiction since $\mathbb Q$ is countable.

Answer 3

Every collection of disjoint open intervals in $R$ is countable because you can choose a rational number (by density theorem) in each of them and rationals are countable.

For first your $\bigcup^{\infty}_{n=0}A_n$ is a countable collection. It is a union of $\aleph_0$ open disjoint intervals.

Answer 4

There is no uncountable collection of open sets in $\Bbb R$. Let $\{A\}$ be a collection of disjoint open sets. For each $A$, let $x_A\in A$. Because each $A$ is open, there is an $\epsilon_A>0$ such that $(x_A-\epsilon_A,x_A+\epsilon_A)\subseteq A$. Because there is always a rational between any two real numbers, there is a rational $r_A\in(x_A-\epsilon_A,x_A+\epsilon_A)$. Since all $A$ are disjoint, each $r_A$ is distinct. Hence, there are countably many $A$.

Answer 5

The question is about constructing countable union of disjoint sets so dont think about the uncountable interval because any interval is uncountable obviously.Think interval as a unit and then take union of finite units (intervals). UAi where i is from 1 to n and n belongs to set of natural numbers.

Answer 6

You are actually running into a somewhat subtle issue here having to do with the difference between the cardinality of a set and the cardinality of its elements.

You are being asked to collect together a countable number of intervals, so the collection you make will be countable despite the fact that each of the intervals in your collection will itself be uncountable, since it is open and nonempty in $\mathbb{R}$.

This is related to the dichotomy between the cardinality of a given set and the 'hereditary cardinality' of that same set. The cardinality of a set is the standard measure of how many elements are in the set -- the hereditary cardinality of a set is (intuitively) the largest cardinality present when selecting from the set itself and all its elements, and all their elements, etc. (this is also called the 'transitive closure' of the set, so the heredetary cardinality of a set is the cardinality of its transitive closure)

For a simple example, the singleton $\{\mathbb{R}\}$ has cardinality $1$ but hereditary cardinality $\mathfrak{c}$. In your example, you will construct a collection $\{(a_n,b_n)\}_{n<\omega}$ with cardinality $\omega$ but heredetary cardinality $\mathfrak{c}$.