how do you construct the natural numbers?

[Anonymous No.10390958]

brainlet here, i dont get how the ZFC axioms can generate the set off all natural numbers.

[Anonymous No.10390992]

>>10390958
Let 0 = {} and n+1 = n U {n}.
https://en.wikipedia.org/wiki/Set-theoretic_definition_of_natural_numbers

[Anonymous No.10391008]

>>10390992
i dont understand how intuiting the number 0 and the n+1 recursive function is any more rigorous then just intuiting the set of all naturals.... can you elaborate?

[Anonymous No.10391059]

>>10391008
>intuiting the set of all naturals
Just claiming that all the naturals exist seems less rigorous than defining zero as a set and then defining a function that generates the other numbers.

[Anonymous No.10391070]

https://www.youtube.com/watch?v=S4zfmcTC5bM

[Anonymous No.10391074]

https://www.youtube.com/channel/UCs4aHmggTfFrpkPcWSaBN9g/videos

[Anonymous No.10391075]

>>10390958
false premise. you can't construct an infinite set

[Anonymous No.10391145]

You're wasting your time here honestly.

[Anonymous No.10392537]

>>10390992
>n+1
Hmm. What is this symbol "1"?

[Anonymous No.10392824]

>>10391059
to define such a function you wold have to know that the naturals form a set (the function would be a subset of N x N)
You cannot construct a infinite set from the other axioms of ZF, so you have to axiomatically declare one infinite set/sucessor set (which would be omega) and then use this set as N.

[Anonymous No.10393084]

>>10390958
If you have K things, then you really have K+1 things, because the set containing those original K things, is itself another thing.

[Anonymous No.10393092]

nigga look at your hand an count your fingers lmao natural numbers right in front of you simple

[Anonymous No.10393139]

>>10391075
Fucking based

[Anonymous No.10393203]

>>10392824
To elaborate, this is because the set of hereditarily finite sets (finite sets of finite sets of..., i.e. sets you can write out explicitly as a finite string of brackets and commas) is a model of the rest of the ZF axioms, with no infinite set. So we need to take "there is an infinite set" as a base axiom.

[Anonymous No.10394161]

1st method
Say [math]X[/math] is equipotent to [math]Y[/math] iff there exists a bijection between both. Show this relation is an equivalence relation and define cardinals as the equivalence classes. Then, define "[math]X[/math] is finite" as "[math]X[/math] isn't equipotent to [math]X \,\cup\, \left\{X\right\}[/math]." Define natural numbers as cardinals of finite sets.

2nd method
Define [math]0\;=\;\varnothing[/math] and [math]s: X \;\longmapsto\; X\,\cup\,\left\{X\right\}[/math]. Then define [math]\mathbf N \;=\; \bigcap \mathrm{Inf}[/math] where [math]\mathrm{Inf}[/math] is the set of all sets that are postulated by the axiom of infinity. Then, prove the Peano axioms.

[Anonymous No.10394170]

>>10393139
GO AWAY REALS

[Anonymous No.10394177]

>>10392537
It's just convenient notation. Really it should be S(n) since we haven't defined 1 yet.

[Anonymous No.10394180]

>>10392537
>>10394177
(continued) And of course it happens that S(n) = n+1. Proof:
n + 1
= n + S(0) [definition of 1]
= S(n + 0) [definition of +, case a + S(b) = S(a + b)]
= S(n) [definition of +, case a + 0 = a].