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...


*

*$1\to 0.1$

*$2\to 0.2$

*$3\to 0.3$
...

*$10\to 0.10$

*$11\to 0.11$


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


*

*0.12345...

*0.32145...

*0.87654...

*0.28374...

*0.67676...

*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. 
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A: How come there's supposedly at least one more real number than you can map to a member of $\mathbb N$?
Well, suppose there isn't - that Cantor's conclusion, his theorem, is wrong, because our enumeration covers all real numbers.  Wonderful, but let us see what happens when we take our enumeration and apply Cantor's diagonal technique to obtain a real number that can't be in this sequence.  But that contradicts our supposition!  Hence our supposition - that we can have an enumeration of all reals - is false.  That the argument can be applied to any enumeration is what it takes for Cantor's theorem to be true.
Wilfred Hodges wrote an excellent survey of wrong refutations of Catonr' argument, his An Editor Recalls Some Hopeless Papers (Postscript).  Section 7, dealing with how the counterfactual assumption confuses, might be of particular interest.
A: Cantor's diagonal is a trick to show that given any list of reals, a real can be found that is not in the list. 
First a few properties:


*

*You know that two numbers differ if just one digit differs. 

*If a number shares the previous property with every number in a set, it is not part of the set.


Cantor's diagonal is a clever solution to finding a number which satisfies these properties. The number which is the diagonal is transformed s.t. it doesn't share the first digit of the first number nor the second digit with the second and so on. Thus the number is unique to the list. 
This is why Cantor's diagonal as a method proves the result that the reals are uncountable.
