# Thread: countability of the reals

1. ## countability of the reals

We know that the real numbers $\bold{R}$ are uncountable. But what if we partitioned the real numbers into a infinite but countable number of closed intervals. So we would have for example: $[0,1] \cup [1,2] \cup [2,3], \dots$. Do the same thing for the negative reals. Then the number of intervals are countable right?

2. $[0,1]$ contains all the reals between $0$ and $1$ inclusive. Similarly, $[1,2]$ contains all the reals between $1$ and $2$ inclusive etc...

3. Originally Posted by particlejohn
Then the number of intervals are countable right?
Right. But each of these intervals is uncountable.

4. Hello,
Originally Posted by particlejohn
We know that the real numbers $\bold{R}$ are uncountable. But what if we partitioned the real numbers into a infinite but countable number of intervals. So we would have for example: $[0,1] \cup [1,2] \cup [2,3], \dots$. Do the same thing for the negative reals. Then the number of intervals are countable right?
Hmm I'd say yes, because there's an interval corresponding to an integer. And $\mathbb{Z}$ is equipotent to $\mathbb{N}$, infinite and countable set.

Let $S=\{\dots \cup [-2,-1) \cup [-1,0) \cup [0,1) \cup \dots \}$

$S=\{A_i ~|~ \forall x \in A_i, ~i \le x < i+1 \}$ (actually i is the least integer contained in $A_i$)

Every element of S can be associated to an integer of $\mathbb{Z}$. Thus S is countable.

5. Originally Posted by particlejohn
We know that the real numbers $\bold{R}$ are uncountable. But what if we partitioned the real numbers into a infinite but countable number of closed intervals. So we would have for example: $[0,1] \cup [1,2] \cup [2,3], \dots$. Do the same thing for the negative reals. Then the number of intervals are countable right?
You cannot find a countable partition of $\mathbb{R}$ by countable sets.
If that is what you are asking.