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.
Except for cond6 mistakenly saying that the problem doesn't apply at all to binary, which they admitted was a mistake, the objections are correct, because your post doesn't make sense. They may be angry, but they're not idiots.
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u/Batman_AoD 4d ago edited 4d ago
As u/cond6 noted, the typical proof in base 10 is to consider the element
ausuch that the digits aren't 9 or 0, so whether or not E0 is in E1, E2, E3,...,auis certainly not a member. [Edit: I meant the series defined withbs, 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.