Cantor's diagonal argument:
As a starter I got 2 problems with it (which hopefully can be solved "for dummies")
First: I don't get this: Why doesn't Cantor's diagonal argument also apply to natural numbers? If natural numbers cant be infinite in length, then there wouldn't be infinite in numbers. By using a randomly ordered list, you wouldn't end with an endless sequence of 0's you have to change. Also it initially goes for "set of numbers". It is applied to the "right" side (fractional part) to prove "uncountability" but can't be used for the "left" side (integer part) because of "reasons" (I simply do not get it).
Second: the way is is used so many times, would just work in the case that the length and width of the list equals. Just listing all natural numbers where $0<=x<100$ will have a width of 3 but a length of 100. At base x each increase of digits will increase the length by x times. At base 10, all 4 digit numbers will create a list with 1000 entries. The length increases exponentially while the width does linearly. This is wrong in so many ways but: Doing this infinitely makes it a square?!? (Natural numbers are a "part" of integers but as you can map both with each other they are considered being the same size aka 2 different countable infinities always have the same "size" while one can be just a part of the other)
As the last part: lets assume we divide all real numbers in 2 parts. The integer part which defines the "set" we use. (there will be "countable" infinite of them)
Now, all we need to do is mapping the fractional part. Just use the list of natural numbers and flip it over for their position (numeration). Ex 0.629445 will be at position 544926. You could argue that this isn't possible for numbers like $\sqrt{2}$.
Lets pick $\pi$:
3.1415926535897932384626433832795… will be
3rd set at position: …5972383346264832397985356295141
There is no reason you cannot pick the next digit and put it in front for the position. There is no limit in length for natural numbers -> you can write a natural number which is the index for just that fractional part. Simply put: you can map EVERY number with $0<=x<100$ with a natural number. And a countable infinite amount of sets containing countable infinite entries still is countable.
So there are 3 Questions (I probably need to split this question):
- Why doesn't Cantor's diagonal argument also apply to natural numbers? (for dummies: why you can't simply use it to the left)
- If the count of digits equals the the length of the list, doesn't it just proves that this construction cannot contain all possibilities?
- Wouldn't the construction of a set like in "the last part" be a prove that all real numbers are countable infinite? (Which part can't be done / is invalid?)
The most important part would be the third question. (If it only qualifies for one answer) Thanks in advance.
An additional big "thanks" in advance for correcting all the spelling, orthography and typos...