# countability of the reals

• Sep 6th 2008, 12:53 PM
particlejohn
countability of the reals
We know that the real numbers $\displaystyle \bold{R}$ 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: $\displaystyle [0,1] \cup [1,2] \cup [2,3], \dots$. Do the same thing for the negative reals. Then the number of intervals are countable right?
• Sep 6th 2008, 12:58 PM
particlejohn
$\displaystyle [0,1]$ contains all the reals between $\displaystyle 0$ and $\displaystyle 1$ inclusive. Similarly, $\displaystyle [1,2]$ contains all the reals between $\displaystyle 1$ and $\displaystyle 2$ inclusive etc...
• Sep 6th 2008, 01:02 PM
Laurent
Quote:

Originally Posted by particlejohn
Then the number of intervals are countable right?

Right. But each of these intervals is uncountable.
• Sep 6th 2008, 01:04 PM
Moo
Hello,
Quote:

Originally Posted by particlejohn
We know that the real numbers $\displaystyle \bold{R}$ are uncountable. But what if we partitioned the real numbers into a infinite but countable number of intervals. So we would have for example: $\displaystyle [0,1] \cup [1,2] \cup [2,3], \dots$. 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 $\displaystyle \mathbb{Z}$ is equipotent to $\displaystyle \mathbb{N}$, infinite and countable set.

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

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

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

(Speechless)
• Sep 6th 2008, 05:04 PM
ThePerfectHacker
Quote:

Originally Posted by particlejohn
We know that the real numbers $\displaystyle \bold{R}$ 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: $\displaystyle [0,1] \cup [1,2] \cup [2,3], \dots$. Do the same thing for the negative reals. Then the number of intervals are countable right?

You cannot find a countable partition of $\displaystyle \mathbb{R}$ by countable sets.
If that is what you are asking.