1

By Zermelo, we know that $\mathbb{R}$ is well ordered. Let's make the following process: Take $x_0 \in \mathbb{R}$. Consider $\mathbb{R}\setminus \{x_0\}$. Take the minimum $x_1$ of $\mathbb{R}\setminus \{x_0\}$. To construct $x_m$ take it as the minimum of $\mathbb{R} \setminus\{x_0, \ldots, x_{m-1}\}$. Then we can get all the real numbers via this process for if $A$ is the set of ${x_n} 's$ and $\mathbb{R}\setminus A \neq \varnothing$ then we can take the minimum of $\mathbb{R}\setminus A$ and this minimum belongs to $A$ because that's the way $A$ was constructed. So $A = \mathbb{R}$.

What is the mistake?

HeMan
  • 3,257
  • 21
  • 48

1 Answers1

3

Then we can get all the real numbers via this process...

Okay, why is that?

...for if $A$ is the set of $x_{n}$'s and $\mathbb{R}\setminus A\ne \varnothing$ then we can take the minimum of $\mathbb{R}\setminus A$...

That's true.

...and this minimum belongs to $A$...

No.

...because that's the way $A$ was constructed.

Just because you keep choosing the smallest elements and you do this forever, doesn't mean you're going to reach all the elements above.

I think a good counterexample would be the set $\mathbb{N}\cup\{\omega\} = \{1,2,3,\ldots,\omega\}$ where we define $\omega$ to be greater than all the integers. Here, $x_{1} = 1$, and $x_{2}=2$, and $x_{3} = 3$, and etc. This gives $A = \{1,2,3,\ldots\}$, but alas $\omega\not\in A$.

MaximusIdeal
  • 2,949