A "bijection" from $\mathbb{N}$ to $\mathbb{R}$?

Question

We know that there doesn't exist a bijection from $\mathbb{N}$ to $\mathbb{R}$. But the following 'argument' proposes to find such an "bijection".

First Fallacious Argument

Let $a_1\in \mathbb{N}$ and $b_1\in\mathbb{R}$. Define $f_{a_1}:\mathbb{N}\to\mathbb{R}$ such that $f_{a_1}({a_1})=b_1$. Now consider $\mathbb{N}\setminus\{a_1\}$ and $\mathbb{R}\setminus \{b_1\}$ and choose $a_2\in\mathbb{N}\setminus\{a_1\}$ and $b_2\in\mathbb{R}\setminus \{b_1\}$. Now define a mapping $f_{a_2}:\mathbb{N}\to\mathbb{R}$ such that $f_{a_2}({a_2})=b_2$ and so on.

Finally define, $$f:=\bigcup_{a\in \mathbb{N}}f_a$$which is the required "bijection".

Second Fallacious Argument

Consider the reverse mapping, i.e., as before let $a_1\in \mathbb{N}$ and $b_1\in\mathbb{R}$. Define $f_{b_1}:\mathbb{R}\to \mathbb{N}$ such that $f_{b_1}({b_1})=a_1$. Now consider $\mathbb{N}\setminus\{a_1\}$ and $\mathbb{R}\setminus \{b_1\}$ and choose $a_2\in\mathbb{N}\setminus\{a_1\}$ and $b_2\in\mathbb{R}\setminus \{b_1\}$. Now define a mapping $f_{b_2}:\mathbb{N}\to\mathbb{R}$ such that $f_{b_2}({b_2})=a_2$ and so on.

Finally define, $$f:=\bigcup_{b\in \mathbb{R}}f_b$$which is the required "bijection".

So what is(are) the flaw(s) in the above argument(s)?

Answer 1

First argument: As Michael says, you need to prove that $f$ is surjective in order to say that it is a bijection. As it happens, this is impossible to do since $\mathbb R$ is uncountable, but this is beside the point - you have provided no proof that $f$ is surjective, so it is invalid to claim that $f$ is a bijection.

Second argument: Here the flaw lies in what you mean by 'and so on'. You pick distinct elements $b_1,b_2,\dots$ and define an appropriate function $f_{b_i}$ for each one. But then you assume that for every real number $b$ there is an associated function $f_b$, without any proof.

Indeed, since $\mathbb R$ is uncountable, this will not be the case, since the sequence $b_1,b_2,b_3,\dots$ contains only countably many real numbers, and therefore cannot run over the whole of $\mathbb R$.

Answer 2

You've found an injection from $\mathbb N$ to $\mathbb R$, but $f$ is not a surjection. (To construct $f$ you pick countably many elements in $\mathbb R$.)

Edited to address the second argument:

It's not clear what you mean by $$ \bigcup_{b\in \mathbb R}f_a $$ since $(b_n)$ is a sequence in $\mathbb R$ in the way you've defined it. You've constructed countably many functions $(f_{b_n})$, so if your assumption is that every element of $\mathbb R$ is equal to some $b_n$, then you are using circular reasoning and proving the countability of $\mathbb R$ using the countability of $\mathbb R$.