Before starting reading this note please go through my note on uncountability of \(R\). There I have shown that \(R\), the set of real numbers, is uncountable. Now in this note, I will show a that there exists a bijection between \(R\) and \(N\). Well, obviously such bijection doesn't exists and my method has a fallacy, but I want you people to find out the fallacy yourself.

I will construct a function \(f:N\rightarrow R\) as follows:

From \(N\) choose an arbitrary element \({ n }_{ 1 }\) and map it to an arbitrary element \({ x }_{ 1 }\) from \(R\).

Next choose an arbitrary element \({ n }_{ 2 }\) from \(N-\{ { n }_{ 1 }\} \) and map it to arbitrary element \({ x }_{ 2 } \) of \(R-\{ { x }_{ 1 }\} \) and continue the process.

Thus in the i-th step map the arbitrary element \({ n }_{ i }\in N-\{ { n }_{ 1 },{ n }_{ 2 },...,{ n }_{ i-1 }\} \) to an arbitrary element \({ x }_{ i }\in R-\{ { x }_{ 1 },{ x }_{ 2 },...,{ x }_{ i-1 }\} \).

Then, from the construction of \(f\), \(f\) is injective. Choose any two different elements from \(N\), and observe that they are mapped to two elements of \(R\) under \(f\), which must be different (different because whenever in a step an element of \(N\) is mapped to that of \(R\), the latter is deleted from \(R\) before proceeding to the next step). Again \(f\) is surjective because choose any arbitrary element from \(R\), see that it is the image of an element of \(N\) (Since our process of construction of \(f\) is infinite, this means at some step of our process we must have mapped an element of \(N\) to that element of \(R\)).

So \(f\) turns out to be bijective. Clearly, there is a strong fallacy in my considerations. Can anybody point it out?

## Comments

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TopNewestThis doesn't follow. Just consider the simple function \(f : \mathbb{N} \to \mathbb{N}\) where \(f(n) = n+1\). The process is infinite, yet 1 (or 0, if your natural numbers start with 0) has no pre-image. – Ivan Koswara · 11 months ago

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– Kuldeep Guha Mazumder · 11 months ago

Yes that's exactly the fallacy. The fallacy lies in the fact that the method by which the above function is being constructed doesn't guarantee surjectivity. Assume it to be surjective, then consider a closed subset of \(R\) and apply the logic that I used while proving the uncountability of \(R\).Log in to reply

Well, you need to ensure that\(f\) is indeed a function. Where have you done that? – User 170039 · 10 months, 3 weeks ago

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– Ivan Koswara · 10 months, 3 weeks ago

Every natural number is mapped to exactly one real number. That's enough to prove that it is a function.Log in to reply

– User 170039 · 10 months, 3 weeks ago

Consider the same argument with the position of \(\mathbb{N}\) and \(\mathbb{R}\) reversed. Will you say that in this case also \(f\) (if we retain the same notation) is a function?Log in to reply

EDIT: Actually, \(f\) won't be a function, exactly because \(|\mathbb{R}| > |\mathbb{N}|\). First, because \(\mathbb{R}\) is uncountable, so the process won't finish even with infinite time. Second, since \(|\mathbb{R}| > |\mathbb{N}|\), at some point there's no more element of \(\mathbb{N}\) that hasn't been paired with.; the process of making \(f\) is halted in this way. If we allow to pick any element of \(\mathbb{N}\) and we can handle the "and continue arbitrarily" portion, then yes, \(f\) becomes a function. – Ivan Koswara · 10 months, 3 weeks ago

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– User 170039 · 10 months, 3 weeks ago

Yes, precisely that is the point.Log in to reply