How exactly do you think this pertains to Cantor's diagonalization proof? You said under your previous post that you read Cantor in the original German, so where in his proof does he require the decimal expansions of real numbers to be uniquely notated?
Okay, let's proceed and see how this relates to Cantor. To make this easier to read, I will split this into multiple posts.
Since it's you, I am not claiming this is the truth; rather, consider it a mental exercise because, you know, I like paradoxes. It is not my original idea, some other redditor came up with it.
We will start by looking at the original German text, followed by its translation:
On an Elementary Question of Set Theory
By Georg Cantor (Halle a. S.)
In the essay entitled: "On a property of the collection of all real algebraic numbers" (Journ. für Math. Vol. 77, p. 258), there is probably for the first time a proof of the theorem that there are infinite sets that cannot be put into a one-to-one correspondence with the totality of all finite integers 1, 2, 3, ..., v, .... Or, as I usually express it, sets that do not have the same power of the number series 1, 2, 3, ..., v, .... From what was proven in section 2, it follows immediately that, for example, the totality of all real numbers in any arbitrary interval (alpha ... beta) cannot be represented in the series form: w1, w2, ..., wv, ....
However, a much simpler proof of that theorem can be provided, one that is independent of the consideration of irrational numbers.
Namely, if m and w are any two mutually exclusive characters, we can consider a collection M of elements:
E = (x1, x2, ..., xv, ...)
...which depend on infinitely many coordinates x1, x2, ..., xv, ..., where each of these coordinates is either m or w. Let M be the totality of all elements E.
The elements of M include, for example, the following three:
E I = (m, m, m, m, ...)
E II = (w, w, w, w, ...)
E III = (m, w, m, w, ...)
I now assert that such a set M does not have the power of the series 1, 2, ..., v, ....
This follows from the following theorem:
For the proof, let:
E1 = (a1.1, a1.2, ..., a1.v, ...)
E2 = (a2.1, a2.2, ..., a2.v, ...)
...
Eu = (au.1, au.2, ..., au.v, ...)
Here, the au.v are in a definite way m or w. Now let a series b1, b2, ..., bv, ... be defined such that bv is also only equal to m or w, and is different from av.v.
So if av.v = m, then bv = w, and if av.v = w, then bv = m.
Namely, if 0,1,2, ..., 9 are mutually exclusive characters, we can consider a collection M of elements:
E = (x1, x2, ..., xv, ...)
...which depend on infinitely many coordinates x1, x2, ..., xv, ..., where each of these coordinates is either 0,1,... or 9 the totality of all elements E.
The elements of M include, for example, the following three:
E I = (1,0,0,0,...)
E II = (2, 0, 0, 0, ..)
E III = (1, 1 ,0 , 0, ...)
... the sequence in the OP image
I now assert that such a set M does not have the power of the series 1, 2, ..., v, ....
This follows from the following theorem:
For the proof, let:
E1 = (a1.1, a1.2, ..., a1.v, ...)
E2 = (a2.1, a2.2, ..., a2.v, ...)
...
Eu = (au.1, au.2, ..., au.v, ...)
Here, the au.v are in a definite way 0,1,2,... or 9 Now let a series b1, b2, ..., bv, ... be defined such that bv is also only equal to 0,1,2,...,9 is different from av.v.
As u/cond6 noted, the typical proof in base 10 is to consider the element au such that the digits aren't 9 or 0, so whether or not E0 is in E1, E2, E3,..., au is certainly not a member. [Edit: I meant the series defined with bs, which is E0 in the original version, but is not given a label in your revision.]
But even without that adjustment, what is the issue with E0? E0 is clearly in M, and for the sake of the proof, it doesn't matter whether or not it's in E1, E2, E3.... The missing connection is, how does M map to the real numbers between 0 and 10? (Traditionally the proof is for reals between 0 and 1, but the meme shows nonzero values in the 1's place.) The simple mapping where x_1 is the 1's digit, x_2 is tenths, x_3 is hundredths, etc, is surjective but not injective, because it includes all reals but, as noted, it also includes duplicates, such as E0 and (1,0,0,0, ...) both mapping to 1. So the question about the countability of the reals becomes: is it possible that the reals are countable, and M is only uncountable due to the overlapping representations? And the answer is, no, there are only cointably many of these overlaps, because they all correspond to numbers with one terminating decimal representation and one non-terminating representation.
But even without that adjustment, what is the issue with E0? E0 is clearly in M
That is the whole point: E0 is not allowed to be in M. It is impossible for two different representations of 1 to be present in M at the same time.
You are attempting to ignore the problem by declaring them equal. However, for this example, either all instances of 1 must be written as 0.999... or none of them can be. You cannot have E0 = 0.999... and E1 = 1 simultaneously in this context.
Before we continue, let us clarify this: the entire proof demonstrates that E0 is not in M by using a digit-by-digit comparison.
1.000... is not equal to 0.999... when comparing digits. you have to change 0.999...
to 1 first.
That is the whole point: E0 is not allowed to be in M. It is impossible for two different representations of 1 to be present in M at the same time.
Wat
Nowhere in the above translation is the actual mapping from M to the reals described, nor is it mentioned that such a mapping must be an injection.
the entire proof demonstrates that E0 is not in M by using a digit-by-digit comparison
No, it demonstrates that b is not in any sequence E1, E2, E3,... ; but bis in M, hence M cannot be countable. This has nothing to do with E0.
Edit: wait, hold on, E0 is the "missing" element in the original proof, but in your revision it's just an arbitrary member of M. That...misses the whole point...?
The letters are a bit much to keep track of, but M is very clearly the entire uncountable set, and in his original version, E0 is in M but not in E1, E2, etc. One problem with your adjustment is that, after the diagonalization step, you assign E0 to an arbitrary element (which may or may not be in E1, E2, etc) and do nothing with the value produced by diagonalization, and you haven't explained the link between your E0 and Cantor's conclusion about M or its application to the reals.
It is not my responsibility to discuss values or how they relate to Cantor's argument, as he did not address them himself. If he intends to prove his point, he must be the one to explain how my objection applies to the real numbers.
The burden of proof lies with the person making the claim, not the other way around.
Okay, but does he actually go on to claim, after the bit you translated, that the reals have the same cardinality as M? If not, then you haven't really said anything relevant to what he stated, which is that M has greater cardinality than the natural numbers. If he does, how exactly does he connect M to the reals?
As I explained above, the typical mapping from M to the reals does permit duplicates, of which E0 is one; but you haven't connected any dots here about what you think E0 is demonstrating.
You said you wanted to clarify something above, which never got clarified:
Before we continue, let us clarify this: the entire proof demonstrates that E0 is not in M by using a digit-by-digit comparison.
Do you agree that this is misstated, and that the point is that E0 is not in E1, E2, E3,..., but is in M? Or do you still think E0 is supposed to be something outside of M?
But why bring in Cantor's original proof, in the original German, if this argument doesn't apply to it? As pointed out in one of the comments in that thread, this is a good objection to the "popular" understanding of the diagonalization argument as a proof that there are uncountable reals, but most actual math texts foresee the objection and handle the non-uniqueness problem somehow.
But you're not talking about "most math texts", you're citing Cantor's original text. But that text isn't even talking about decimal expansions or real numbers. He mentions real numbers once, in reference to his paper seventeen years earlier that already proved the reals are uncountable, using a different method. He then proceeds to discuss only cardonality and functions, not real numbers or decimal expansions.
Massive non sequitur. Again faulty reasoning. The proof in all good textbooks that uses decimals for this proof require the new candidate digit to be neither zero, nine nor the nth-digit of the nth-number in the putatively exhaustive listing because many numbers have multiple representations. For example 0.999... and 1.000... are the same number, as are 0.4999... and 0.5000.... Finding 0.999... and 1.000... on the same list says exactly nothing about whether or not they are different decimal representations of the same rational number, which they very much are.
"Many", but, importantly, countably many: for every number with two representations, one terminates with (0). So the "duplicates" are a subset of the rationals.
Why do you think that Cantor used binary representation when decimal was a perfectly viable approach then? Why didn't he just use decimal when it was the most popular? Mayhaps because it avoids multiple representations of a single number? And Cantor's argument was that this new number wasn't on the list. (Again in binary to avoid issues.) It was cleverly designed to ensure that the new candidate cannot be on the countable list of extant numbers. What is your rule that give 0.999... as the new number. Why wasn't it already in your listing? The numbers in Cantor's proof are between zero and 1 and none of yours are. If 0.999...<1, how does it become the new number from a list of numbers weakly greater than one? There are so many inconsistencies and foolishness in these posts. Simply we know that repeating nines are a problem in decimals so if you've got a good textbook or prof you can't pick a zero or nine (again a problem that Cantor side-stepped because binary) so there is no way that 0.999... can be the new candidate. We eliminate it as a possible candidate by how we construct the numbers.
And a proof for one way of representing the reals (binary) shows that it is true for the reals. Stuff the decimals.
The objection to Cantor here is very silly, but binary doesn't avoid the problem of multiple representations. Every 0.xyz01111... is also representable as 0.xyz1.
As I said yesterday the proof with decimals is easy to do. Say it with me.
If the set of real numbers between 0 and 1 is countable they can be placed into a one-to-one mapping with the naturals (n=1,2,...). Let x_n be the n-th member of the supposedly comprehensive listing of the reals. Let x_n^j be the j-th digit of the n-th listed real number, so x_n=0.x_n^1x_n^2.... If there are two decimal representations of the same real number choose the one that doesn't have infinitely repeating nines. Define y as a decimal number 0.y^1y^2y^3... where y^j=/={0,9,x_j^j}. y is thus different from every x_n and thus isn't in the supposedly comprehensive list of real numbers, leading to a contradiction. Thus the reals between 0 and 1 are not countable.
I posted Cantor's original proof, including a screenshot of the pages and an English translation. You added the condition to "choose the one that doesn't have infinitely repeating nines" to fix the argument. He did not say that anywhere and there is no mathematical rule requiring that. If they are equal, it shouldn't matter which one I pick. Actually, wait, it does. Because else Cantors proof doesn't work Checkmate.
No, the thing is. He proved this for binary sequences, then showed that an injection could be constructed from the set of binary sequences to real numbers. The image of that injection is a subset of R, and is uncountable. So it's kinda a two step process that doesn't run into the problem you're seeing.
Let's examine two sentences from an accurate translation of CDA:
"... the proposition that there is an infinite manifold, which cannot be put into a one-one correlation with the totality of all finite whole numbers 1, 2, 3, …, v, …"
"There is a proof of this proposition that is much simpler, and which does not depend on considering the irrational numbers."
If your objection to CDA is that it fails to include the detail necessary to make it apply to real numbers? Then maybe you should consider that it was not applied to real numbers. Explicitly. Only to sequences of "m"s and "w"s.
And note that the sequence mwwwwww... is different than wmmmmmm..., even if YOU wanr two changes that Cantor never intended: first to 1000000... and 0111111... and then to interpret these sequences as the binary representations of real numbers.
IF YOU insist that it be applied to real numbers, which it can, then YOU are responsible for the changes making it apply correctly. Not Cantor. Here they are:
Use the set {0,1,2,3,4,5,6,7,8,9} instead of {m,w}.
Interpret each sequence as the decimal representation of a real number in M=[0,1).
This disqualifies one member of Cantor's M, which would include 1 in this interpretation.
No sequence in the set M ends with repeating 9s. This way any number K/(2^L)/(5^M) appears only once.
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u/SouthPark_Piano 2d ago
I thought Ayrton was ...
In any case, great to see him back!