# Why is the set of all real numbers uncountable?

I understand Cantor's diagonal argument, but it just doesn't seem to jive for some reason.

Lets say I assign the following numbers ad infinitum...

1. $1\to 0.1$
2. $2\to 0.2$
3. $3\to 0.3$ ...
4. $10\to 0.10$
5. $11\to 0.11$

and so on... How come there's supposedly at least one more real number than you can map to a member of $\mathbb{N}$?

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Because... Cantor explicitly tells us what that "one" real number is... ??? –  The Chaz 2.0 Nov 14 '12 at 16:39
@agent154: By the way, where is $\frac13$ in your list? –  Dejan Govc Nov 14 '12 at 16:40
@agent154: Not exactly; $\frac13=0.333333333\ldots$ has an infinite number of 3's in its decimal expansion. –  Dejan Govc Nov 14 '12 at 16:44
...are you sure you understand Cantor's diagonal argument? –  camccann Nov 14 '12 at 16:45
@MakotoKato That's a bit of a useless argument. Presumably, if he knew what the Lebesgue measure was, he would already be comfortable with Cantor's argument. –  Thomas Andrews Nov 14 '12 at 16:45

If we apply the Cantor diagonal argument to the list you have there, and assign the $n$th digit in our constructed number to be 0 if the $n$th digit of the $n$th number in the list is 1, and 1 if it isn't, we construct the real number $0.0111111\ldots$ (which happens to be equal to $\frac1{90}$) and in fact is not anywhere in your list, since there is no integer that looks like $0111111\ldots$. So the construction worked fine: you presented what you claim is a list of all real numbers, and then the construction produced a real number that you omitted. Where's the problem?

(If you're not happy with that, perhaps it's enough to point out that your list not only omits all the repeating decimals such as $\frac 13, \frac23, \frac 16, \frac 17, \frac 27\ldots$, but also all the numbers less than 1/10. So it's not even a list of all the rational numbers.)

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the fact that you can inject $\mathbb R$ into $\mathbb N$ doesn't prove that there are more real numbers then natural ones. It just shows that there are no less real numbers then natural ones. –  Golob Nov 14 '12 at 21:26
@mjd That actually helps make a bit more sense. At least my assignment doesn't work since it doesn't account for numbers less than 1/10 as you state. But I guess my issue is with the fact that I don't understand this well enough yet. I only just learned about countability, and even then very poorly, since my prof is not a good teacher. –  agent154 Nov 15 '12 at 19:41

Let's look at the interval $0 < x < 1.$ Let's assume all of its numbers are countable, i.e. they are listable. Let's say the list goes like:

1. 0.12345...
2. 0.32145...
3. 0.87654...
4. 0.28374...
5. 0.67676...
6. etc...

From the list, I make a new number. Its first decimal place is the same as the first number's, its second decimal place is the same as the second number's, its third decimal place is the same as the third number's, etc. The new number I get is:

$$0.12676...$$

Using this number, I create a new number. If a digit is a $1$ then I change it to a $2$. If a digit is anything else then I change it to a $1$. Thus:

$$0.12676... \mapsto 0.21111...$$

This new number, $0.21111...$ is not on my list. It's not the first number because it's different in the first place, it's not the second number because it's different in the second place, it's not the third number because it's different in the third place, it's not the fourth number because it's different in the fourth place, etc. It follows that my new number is not in my list, and so I could not have listed all of the numbers $0 < x < 1$. Thus, by contradiction, the interval is not countable.

How come there's supposedly at least one more real number than you can map to a member of $\mathbb N$?