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Question: To show that 2 sets have the same cardinality, there needs to be atleast one bijective mapping between them. So given the below proof of a bijective mapping below, why can't we say that $\left | \mathbb{R} \right |=\left | \mathbb{N} \right |$? Where is the mistake in this proof?

Background: I am studying Real Analysis from the book "Real Analysis: A Long-Form Mathematics Textbook" by Jay Cummings. In Theorem 2.9 of $\left | \mathbb{R} \right |>\left | \mathbb{N} \right |$ through Cantor's Diagonaliation argument. Now, I get why the presented mapping isn't bijective but I don't why get there can't be any bijective mapping from $\left | \mathbb{R} \right | $ and $\left | \mathbb{N} \right |$? If there is proof of this please let me know. But, while thinking about this I made this proof which seems to tell that their cardinality is same, please do also let me know where the mistake is in the given proof.

Wrong Proof: The mapping here is from $\mathbb{N_{0}}$ to $\mathbb{(0,1)}$. Through something like sigmoid function, we can show that $|(0,1)| = |\mathbb{R}|$, and though a simple shift we can also show that $|\mathbb{N_{0}}| = |\mathbb{N}|$. And we also can prove that cardinality does show transitive property that is if $|A|=|B|$ and $|B|=|C|$ then $|A|=|C|$ (Composite of bijectives is also bijective function). This is the mapping: $$ ....a_{13}a_{12}a_{11} \rightarrow 0.b_{11}b_{12}b_{13}....\\ ....a_{23}a_{22}a_{21} \rightarrow 0.b_{21}b_{22}b_{23}....\\ ....a_{33}a_{32}a_{31} \rightarrow 0.b_{31}b_{32}b_{33}.... $$ where, $ a_{ij}, b_{ij} \in \{0,1,2,3,4,5,6,7,8,9\} $. To make the mapping we first have "$....a_{13}a_{12}a_{11} \rightarrow 0.b_{11}b_{12}b_{13}....$", through Cantor's diagonilization we can pop out 2 numbers from left and right side, both of which will then be mapped to each other and added to this relation. In this way this relation is being built. (better explanation in part of the proof where we prove this function is surjective)

To convince us that the left side is indeed $\mathbb{N_{0}}$. We can see that "$....a_{i3}a_{i2}a_{i1}$" this will always be postive (or zero) and will never have decimal points, and so it is indeed a representation of $\mathbb{N_{0}}$. (representation of $\mathbb{N_{0}}$ itself might be wrong)

To convince us that this relation is indeed a function:

  1. We can pick any element of $\mathbb{N_{0}}$ and there does exist some way in which we can represent it in the form "$....a_{i3}a_{i2}a_{i1}$" and so some mapping of every element of domain exists.(maybe I'm wrong here)
  2. The mapping of one $x$ in $\mathbb{N_{0}}$ will always land on the same $y$ in $\mathbb{(0,1)}$, due to how we have defined the mapping, no number on left side will ever repeat. This is because the relation is being built through Cantor's diagonalization process on both sides.

This relation is a function, now is it bijective?

  1. It is injective (one-to-one) because of its definition.
  2. Is it surjective (onto)? --> This is the main part of the proof. (most probably where I'm wrong)
    • Now, through Cantor's diagonalization (CD) on right side we can at any "step" pick out a number which isn't mapped to yet, but again through CD on the left side, we can pick out a natural number which is left to be mapped. Well we have a number on right side not mapped to yet and a number on left side which hasn't gotten its mapping yet, we simply map them to each other. In this way no matter at what "step" you look no element of $\mathbb{N_{0}}$ or $(0,1)$ aren't mapped. So, no element of $(0,1)$ is left unmapped to by $\mathbb{N_{0}}$.
    • So, no element of $(0,1)$ can be found which cannot be mapped to in this fashion. (raises some eye-brows)

Well this shows us that we have just constructed a bijective mapping between $\mathbb{N_{0}}$ and $(0,1)$, which implies that $|\mathbb{N_{0}}| = |(0,1)|$. And so we have: $$ |\mathbb{N_{0}}| = |(0,1)| \\ |(0,1)| = |\mathbb{R}| \\ |\mathbb{N}| = |\mathbb{N_{0}}| \\ $$ which imples => $$ |\mathbb{R}|= |\mathbb{N}| $$

Remarks: I think if someone can help me find one example in $(0,1)$ which isn't mapped to through this definition then we can confidently say that this proof is incorrect. Or if we can show that this relation cannot actually map all elements of $\mathbb{N_{0}}$ (which is it's domain) then also this proof is incorrect.

Edits:

  1. Realised that the representation itself might be wrong for this proof, which means that "$....a_{i3}a_{i2}a_{i1}$" could actually not be $\mathbb{N_{0}}$
Viraj Agarwal
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    There are a few issues. For instance, do you have a natural number with infinitely many nonzero digits? – Brian Moehring Nov 01 '23 at 07:04
  • This might be a reason but it's not a convincing reason to me, partly because I can't contemplate what infinity looks like and so I can only rely on my axioms to understand it, so dont the axioms say that this "natural number with infinitely many nonzero digits" is still a natural number and hence part of $\mathbb{N_{0}}$? Or am I wrong here? – Viraj Agarwal Nov 01 '23 at 07:12
  • But yeah the representation of the natural numbers itself might be the problem here in this proof. – Viraj Agarwal Nov 01 '23 at 07:12
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    You are wrong since every natural number has finite many digits although there are infinite many. – Peter Nov 01 '23 at 07:30
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    The usual axioms for the natural numbers don't say anything about their decimal representation. So if you are going to use notation like, say, $111111\dots$, then you better define what you mean before you use it. Otherwise, it's just a meaningless scribble on a page. – Joe Nov 01 '23 at 07:32
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    @Peter: It's a little imprecise to say that every natural number has finitely many digits. Rather, when we represent a natural numbers using decimal notation in the standard way, such a representation will only have finitely many digits. (Usually, I would feel that this is an overly pedantic comment to make, but it is very common for newcomers to a subject to conflate numbers with their representations, and so I feel that it helps to be precise in this context.) – Joe Nov 01 '23 at 07:36
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    You request a proof that no mapping between $\mathbb R$ and $\mathbb N$ can be bijective. But Cantor's diagonal argument is said proof. It can be applied to every such mapping, so the argument is essentially: Any mapping between $\mathbb R$ and $\mathbb N$ is not bijective. Which is equivalent to saying that no such mapping is bijective. – Vercassivelaunos Nov 01 '23 at 07:40
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    OP: you write "Now, I get why the presented mapping isn't bijective but I don't why get there can't be any bijective mapping". The standard proof begins with an arbitrary mapping and shows that it is not bijective. – Greg Martin Nov 01 '23 at 07:40
  • Thank you so much! I now understand why this proof is wrong. And also thank you for answering in a way which was accessible for a newbie like me, I'll add an answer to this question now. – Viraj Agarwal Nov 01 '23 at 07:42

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Ok, this proof is wrong because in the very start we cannot represent natural numbers by a infinite number of digits. They cant have infinte digits. Because as @Joe commented above "when we represent a natural number using decimal notation in the standard way, such a representation will only have finitely many digits." If you want to understand this better then here is a post which explains this very nicely. Essentially it was a mistake to assume the representation of the natural numbers to be their definition.

So, the representation "$....a_{i3}a_{i2}a_{i1}$" isn't valid and is actually bigger than $\mathbb{N_{0}}$.

Viraj Agarwal
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