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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...

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    $\begingroup$ It's simply not true that natural numbers are allowed to have infinite decimal expansions to the left. This isn't a matter you get to disagree with; it's just what a natural number is. You can consider "numbers" which have infinite decimal expansion to the left but then these are not what everyone else calls "natural numbers". $\endgroup$ – Eric Wofsey Oct 28 '18 at 20:49
  • $\begingroup$ The representation being used in the comment you linked to only looks like decimal expansion. What the bits they were using represented were positive powers of $2$. So $1000...=2^0$, $01000...=2^1$, $001000...=2^2$, and so on. You can represent every natural in this form, however, your representation must "terminate", else it would be infinite in size, not just in terms of bit length. Also, consider that $10000....$, $0100...$, $001000....$, $0001000...$,... are all "finite" in bit length, but the sequence is infinite. As we can always add another $0$ to the left of our $1$. $\endgroup$ – Melody Oct 28 '18 at 20:54
  • $\begingroup$ @EricWofsey if they cannot have infinite digits to the left, how can there be infinite many of them? Also, wouldn't that end with a point at which you cannot multiply it by 10? (for base 10) Is there anything out there that proves that a natural number will be something different if you increase it infinite many times? (again: for dummies.... i doubt that) $\endgroup$ – Dirk Reichel Oct 28 '18 at 21:10
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    $\begingroup$ So you're saying that if all natural numbers are finite, then there are only finitely many of them? $\endgroup$ – Asaf Karagila Oct 28 '18 at 21:35
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    $\begingroup$ @AsafKaragila "infinity" fooled me :( Having no "limit" isn't the same as "infinite". There is no "limit" to the size of a natural number, which makes them infinite in count, but each number itself needs to be finite (else it would be infinity).... the longer I think about it, the more correct it looks to me $\endgroup$ – Dirk Reichel Oct 28 '18 at 22:01
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This won't answer all of your questions, but here is a quick proof that a set of elements, each of which has finite length, can have infinite cardinality. Let $\newcommand\N{\mathbb{N}}\N$ be the set of natural numbers where each natural numbers are defined as having a finite decimal expansion. Suppose that the cardinality of $\N$ is finite. Then it must have some cardinality $k$ where $k$ is some finite number. By definition, that implies that there is a bijection $f$ from $S=\{1,2,\dots,k\}$ to $\N$. Now, let $a_1=f(1)$, $a_2=f(2)$, and so on until we have $a_k=f(k)$. In addition, let $a_{max}$ be the maximum of all of $a_1$ through $a_k$. Now, let $b=a_{max}+1$. It should be clear that $b$ isn't equal to any of $a_1$ through $a_k$ as it is larger than each of them. As such, there is no $x\in S$ where $f(x)=b$ as then $b=a_x$. On the other hand, $b$ is some number with a finite decimal expansion ($a_{max}$) plus $1$. Its decimal expansion can be at most $1$ longer than that of $a_{max}$, and a finite number plus $1$ is still finite. As such, $b\in\N$. We, therefore, have a contradiction as $f$ isn't subjective and so no such $k$ can exist. That means that $\N$ can't have a finite size.

To put this argument in less formal language, $\N$ is infinite not because any of it's elements is infinite, but because it has elements that get arbitrarily large. Suppose that we did have some finite size $k$. Then what would we say about $k+1$? It's still a natural number. The problem with putting a size on $\N$ isn't the size of the elements, but that they are unbounded.

This, in turn, with your proof is that the position of $\pi$ would not be that of a natural number as natural numbers have finite length. Your proof is actually correct that the cardinality of reals is equal to the cardinality of the set of all sequences with infinite digits.

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  • $\begingroup$ Looks like that isn't solvable "for dummies". As there is no point at which you cannot multiply a natural number by 10 it can be done infinitely... still the notation cannot be infinitely long. Will probably take some time to accept this. Simply put: the answer for the third question is: a real number does have a size and having infinite digits to the left makes it infinite and infinity is not a number thus this approach isn't possible :( While writing, it starts to make sense... $\endgroup$ – Dirk Reichel Oct 28 '18 at 21:52
  • $\begingroup$ The second part is also the answer to the first. If you multiply a number by 10 an infinite number of times, it will be infinite and so will no longer be a number. You can only multiply a finite number of times. $\endgroup$ – memerson Oct 28 '18 at 21:57
  • $\begingroup$ Unfortunately I assumed that having "no limit" allows to do it "infinite" many times. There is no limit in multiplying by 10 and still getting another natural number, but doing it "infinitely" is something different... now I will check "Cantor's diagonal argument" another 100 times as the solution to my approach of mapping being wrong looks so simple to me as it is probably also not the big deal... thx again for everything $\endgroup$ – Dirk Reichel Oct 28 '18 at 22:10
  • $\begingroup$ Also, look up $p$-adic numbers: these give the extension of natural numbers allowing infinitely many digits to the left, written in number system of base $p$. Note that Cantor's argument does apply to them, and their cardinality indeed equals to that of reals (continuum). $\endgroup$ – Berci Oct 29 '18 at 9:46
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"If natural numbers can't be infinite in length, then there wouldn't be infinite in numbers." You are extrapolating properties of the natural numbers to what is called "potential infinity." But that is not what Cantor means by Aleph0.$\newcommand\N{\mathbb{N}}$

"Potential Infinity" is a hypothetical number that represents the hypothetical end to the sequence 1,2,3,... . Nobody treats it as a real object, but only in hypothetical evaluations of expressions where its impact on the result diminishes to nothing. For example, f(x)=3/(2+x) approaches 3/2 "as x goes to potential infinity." The point is that it is treated like a number with the same properties as "regular" numbers, but you can never reach it. "Actual infinity" is a well-defined mathematical concept, but it is nothing like a "regular number."

The "cardinality" of the set {1} is the cardinal number 1. That is, the set has one member. The cardinality of the set {1,2} is the cardinal number 2. In general, the cardinality of the finite set {1,2,3,...,n} is the largest number in it, n.

The Axiom of Infinity says that there is a set $\N$ with a similar form, but that has no largest member. Its cardinality is Aleph0, which can be called "Actual Infinity." It is not a member of the set it describes. It is not a "regular number" in any way. Even though we can extend the definition of numbers to include it as an "irregular number," it does not have the all the same properties. One that is different, is that you can't represent it with a string of digits. Not even an infinite string whose value you might think you could claim is "Potential Infinity."

Diagonalization does not work on natural numbers because it requires a digit for every member of $\N$, and that does not represent a natural number.

About your second problem: you can't consider an $\N$ by $\N$ array of characters to be square. "Square" means that you can only pair rows with columns one a 1:1 basis. While you can make such a pairing with an $\N$ by $\N$ array, you can also pair them on a 1:2 basis. Or 3:5. Or 14:9. The point of using Aleph0 for a cardinality, is that it represents any countably infinite set the same way.

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  • $\begingroup$ Thanks for the extended answer. I got multiple things wrong here to start with. By better understanding (at least a part) of the whole "infinity"-problematic a lot makes sense now. $\endgroup$ – Dirk Reichel Nov 19 '18 at 9:35

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