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1) A and B are sets of natural numbers. They are the same, so they have the same cardinality.

2) Set B is transformed into set of rationals. This can be done only by adding new numbers e.g. 1/2, 1/4, 5/6, ... to this set.

3) Set B now contains the same amount of numbers as set A, with extra numbers from step 2. Therefore, set of rationals has greater cardinality than set of naturals.

user10342566
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  • Well, it's far too vague to be a proof. How exactly is $B$ transformed into the set of rationals? If it's just adding elements one-by-one, then the cardinality never changes. It's also not necessarily true that the cardinality of $C \cup D$ is greater than the cardinality of $C$ when $C$ is infinite. –  Sep 18 '18 at 20:14
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    This line:

    "Set B now contains the same amount of numbers as set A, with extra numbers"

    That is not a valid or relevant observation for cardinalities of infinity. Adding more/different numbers to an infinite set doesn't mean the cardinality is any "bigger".

     – fleablood Sep 18 '18 at 20:45
  • Dang I had just written a bloody brilliant answer and was putting a final touch on it ... when the question was closed. – fleablood Sep 18 '18 at 20:47
  • @fleablood I am curious why exactly are cardinalities of infinity useful? I also have doubts about this hotel with infinite rooms... Such a thing couldn't exist in reality, and there is really no proof how those rooms and guests would act when they limit to infinity. – user10342566 Sep 19 '18 at 09:18
  • I'm not interested in your doubts and I couldn't care less about whether math is useful or not and I have no desire to discuss it with you. – fleablood Sep 19 '18 at 15:10

3 Answers3

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What's wrong with your proof is that you are using vague, undefined terms like 'transform' and 'with extra numbers' to reason about cardinality. It seems that you are saying that if $X \subsetneqq Y$ then $|X| < |Y|$, which is true when $X$ and $Y$ are finite, but not if they are infinite.

A more general point: with proof, it should never be a question of 'what's wrong with it'. The burden is on the proof-writer to convince the proof-readers that it is a valid proof. In fact, there are lots of examples of well-reputed mathematicians who claim to have proofs of open problems, but the problems are considered by the mathematical community still to be 'open' because the proofs are not widely accepted.

Clive Newstead
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  • I'd argue that the terms "with extra numbers" isn't vague and that it is no more or less than $X \subset Y$. And we could say $|X| \subset |Y|$ implies $Y$ is "bigger" than $X$. It's just that that is NOT what mathematicians actually say. Because it isn't useful or even interesting to say that. – fleablood Sep 18 '18 at 20:52
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First of all, this is not a complete proof, because you have not explained step 2 properly. But I think I can answer your question, and the real problem is in step 3. You did not prove that the cardinality of the rationals is greater than that of the natural numbers. You merely showed that you can embed the naturals into the rationals so that there are missing elements (which is not very surprising: the set of naturals is a proper subset of the rationals). But this has nothing to do with the cardinality being smaller: check the definition of the ordering of cardinalities again, and you will see that it is not directly related to what you observed (correctly).

A. Pongrácz
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For infinite sets you may add infinite new elements without changing cardinality.

For example the set of odd integers has the same cardinality as the set of all integers.

It is amazing, but true that for example the set of $\{1000,2000,3000,....\}$ has the same cardinality as $\{1,2,3,....\}$

Mohammad Riazi-Kermani
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