# On an intuitive level, density/countability of Q

1. Jun 4, 2013

### johnqwertyful

Something I've thought about for a bit on an intuitive level is about the countability/density of Q. It seems almost contradictory. I know what countable means, I know what dense means. I know how to prove Q is dense. I know how to prove Q is countable. But on an intuitive level, it's hard to wrap my mind around it.

Intuitively, if something is countable, given any element in the set and enough time, you can count until you reach it. Like for N, given ANY natural number, no matter how big, and infinite time, and I could count until I reach it.

But with Q, lets say I wanted to count to 1. There's an infinite number of elements between 0 and 1, so I would be there forever, never being able to reach 1? What about 1/2? I shouldn't be able to reach there.

The way to reconcile it in my head is that the bijection with N isn't order preserving, right? In the way you count Q on [0,1], you would superficially go through it, then again, and again and again. It's still a little unclear though. Anyone able to reconcile them intuitively?

2. Jun 4, 2013

### johnqwertyful

After three quarters of undergrad analysis, going on to grad analysis next quarter, I feel like I should know this lol.

3. Jun 5, 2013

### deluks917

You are correct. There is no order preserving bijection from N to Q. Why is the density of Q combined with countability confusing you? You did not explain this.

One issue many students I know learnign analysis have said confuses them is the following:

Consider the Ball of size [itex]\frac{1}{2^n}[\itex]. The union of all such balls have length at most 1 (this is made rigourous with measure theory but it works as you would guess). However since all any real is arbitarily close to a rational. So doesn't this say the length of the real line is less than 1?

Do you see what is wrogn with their understanding of density.

4. Jun 5, 2013

### lurflurf

You are pretty close. We start counting the rationals (pick your favorite bijection). We reach any particular rational in a finite number of steps. We do not (obviously) count any infinite set (such as the rationals in (0,1)) in a finite number of steps. This is not unlike the integers. We reach any integer n a finite number of steps, but we cannot reach all even numbers in a finite number of steps. This seems to be more about countability than density, but think of several examples, not just Q. I like m/2^n and x^(m/n) among others.