Infinitely countable subset of infinite set

[Anonymous No.9867549]

Let [math]B[/math] be infinitely uncountable.
Then there's [math]A\subset B[/math] infinitely countable.

Let [math]x \in B[/math], [math]f(1)=x[/math] which we call [math]x_{1}[/math].
Let [math]y \not \in \{x_{1},...,x_{n}\}[/math], then [math]f(n+1)=y[/math] *.
This defines [math]f:\mathbb{N}\rightarrow \{x_{1},....\}[/math] by recursion.
[math]B=\{x_{1},....\}[/math].

(*) if there's not such [math]y[/math] then [math]\{x_{1},...,x_{n}\}=A[/math].

[Anonymous No.9867551]

I switched [math]A[/math] and [math]B[/math] in the last two lines.

[Anonymous No.9867556]

>>9867549
What's the point of this thread?

[Anonymous No.9867558]

>>9867556
Something I had to solve.
I was waiting for someone to point out a mistake or to show his proof.
Can you point out where the axiom of choice has been used?

[Anonymous No.9867559]

>>9867549
Your proof is invalid as it assumes ACC

[Anonymous No.9867561]

>>9867556
Looks like he's trying to disprove uncountability but all he's shown is that there exists an injective function into B but it's not surjective. But the way he chooses f(n+1) really bugs me, just picking a random member in the complement.

[Anonymous No.9867562]

>>9867558
>Let [math]y \not \in \{x_{1},...,x_{n}\}[/math], then [math]f(n+1)=y[/math] *
here

[Anonymous No.9867564]

>>9867562
>This defines [math]f:\mathbb{N}\rightarrow \{x_{1},....\}[/math] by recursion.
Or here, to be more precise

[Anonymous No.9867566]

>>9867561
I have to prove this.
>But the way he chooses f(n+1) really bugs me, just picking a random member in the complement.
How would you do?

[Anonymous No.9867574]

>>9867566
>Let [math]y \not \in \{x_{1},...,x_{n}\}[/math], then [math]f(n+1)=y[/math]
This implication is not true. You want to pick y such that...

[Anonymous No.9867578]

>>9867566
Dunno, maybe take the countable union over n of all finite subsets of B with size n? The procedure seems less ambiguous than your weird function but maybe it's just as messy.

[Anonymous No.9867579]

>>9867564
>Or here, to be more precise
Sure about that?
Axiom of choice here seems like picking [math]y[/math] from [math]B\backslash \{x_{1},...,x_{n}\}[/math].
i.e. there's a function [math]g:\{B\backslash \{x_{1},...,x_{n}\}\}\rightarrow B\backslash \{x_{1},...,x_{n}\}[/math]
such that [math]g(B\backslash \{x_{1},...,x_{n}\})\in B\backslash \{x_{1},...,x_{n}\}[/math].
Then, [math]f(n+1)=g(B\backslash \{x_{1},...,x_{n}\})[/math].

The proof of recursion theorem, which says that [math]f[/math] is a function, uses induction.

[Anonymous No.9867581]

>>9867578
>The procedure seems less ambiguous than your weird function but maybe it's just as messy.
Check this for more detail:
>>9867579

[Anonymous No.9867586]

>>9867579
>Axiom of choice here seems like picking [math]y[/math] from [math]B\backslash \{x_{1},...,x_{n}\}[/math].
If you already have the set [math]{x_{1},...,x_{n}\}[/math], picking a y isn't a problem as proven in the last line of the OP. The problem is in making an infinite number of such choices so that [math]\{x_{1},....\}[/math] is defined.

[Anonymous No.9867591]

>>9867586
But the recursion theorem guarantees the existence of such function: https://en.wikipedia.org/wiki/Recursion#The_recursion_theorem..
What is there problematic about it?

[Anonymous No.9867597]

>>9867591
>https://en.wikipedia.org/wiki/Recursion#The_recursion_theorem
How would you define the f for this? It cannot be any arbitrary function. It has to be so that F is injective. Which means that the EACH [math]{x_{1},...,x_{n}\}[/math] should correspond to a unique y.

[Anonymous No.9867600]

I am trying to prove the line "choose a sequence [...]".
The book says that the axiom of choice is required for such proof.

[Anonymous No.9867604]

>>9867600
>The book says that the axiom of choice is required for such proof.

[Anonymous No.9867605]

>>9867597
You didn't understand the proof.
[math]f[/math] is clearly bijective.

[Anonymous No.9867609]

>>9867605
The f in the OP is bijective but for the F (in your link) to be injective, the f (in your link) should satisfy stricter conditions.

[Anonymous No.9867653]

>>9867609
>The f in the OP is bijective
That's what matters.
[math]F[/math] in the proof in the URL is to show that what is defined in OP is indeed a function.

[Anonymous No.9867662]

literally none of the tex has shown up in this thread for me

[Anonymous No.9867672]

>>9867662
Disable uBlock Origin/AdBlock.

[Anonymous No.9867776]

>>9867561
How come randomly picking the first point from the set doesn't bug you?
Fucking summer casuals.