# The Continuum Hypothesis Part 2 (The Constructible Universe)

Image Attribution:

Rimu Shuang. "Untitled Photo". Jan 1, 2014. Under a Creative Commons 3.0 Attribution License.

### January 2, 2014

### continuum_hypothesis, math

This post is going to be tackling the notion of the constructible (or definable) universe *L* and the axiom *V* = *L*, which intuitively states that all sets are “definable”, i.e. there is a first-order way of uniquely stipulating what each set is. The counter-axiom, *V* ⊋ *L*, states that some sets are not definable, that is there are some sets which may exist, but which we can never uniquely identify using first-order logic.

Thus, in a certain sense, the axiom *V* = *L* is a statement that all sets are “well-behaved” and can be known, rather than remain forever unknown. To begin, a subset of a model *M* is *definable in M* if there is a first-order statement

*ϕ*and a tuple of elements from the model $(a_1, a_2, \ldots, a_n) = \overline{a}$ such that $\phi(x, \overline{a})$ holds if and only if

*x*is in the subset. We say that an element

*x*of a model is definable if the subset {

*x*} is definable. Finally, given some set

*X*, the set of definable sets is defined (haha) as Def

^{X}= {

*Y*⊂

*X*∣

*Y*is definable in

*X*}

The constructible universe *L* is then constructed in stages *L*_{α} and is indexed by the ordinals where

*L*_{α + 1} = Def^{Lα}

and in the case of limit ordinals, we have

*L*_{λ} = ⋃_{α < λ}*L*_{α}

. Finally, *L*_{0} = ∅.

Then the axiom *V* = *L* states that ∀*x*∃*α*(*x* ∈ *L*_{α}).

It will turn out to be the case that *L*_{κ} for any limit ordinal *κ* models *V* = *L*. I will quickly skim over why such a *L*_{κ} satisfies ZF.

The fact that it satisfies the Emptyset axiom is immediate as is the fact that it satisfies Extensionality. It satisfies Pairing due to the fact that if *ψ* and *ϕ* each define separate sets in *x* ∈ *L*_{α} and *y* ∈ *L*_{β} where *α*, *β* < *κ*, the pair of those two sets can be defined via *φ* where

*φ*(*x*) := ∀*y* ∈ *x*(*ψ*(*y*) ∨ *ϕ*(*y*))

and hence {*x*,*y*} ∈ *L*_{max (α, β) + 1}.

To see that *L*_{κ} satisfies Union, take an arbitrary set *S*. Let *ϕ* define the set *S*. Then *ψ*(*x*) := *ϕ*(⋃*x*) defines ⋃*S* (note that this works because *a* = ⋃*x* is itself shorthand for ∀*z*(*z* ∈ *a* ↔︎ ∃*y*(*z* ∈ *y* ∈ *x*))).

Because *L*_{κ} is a subset of *V*, it must also obey Foundation. Because *ω* ∈ *L*_{κ}, it must obey Infinity.

Now Powerset and Replacement are slightly more involved to show and so I will save them for the next post (in addition to actually showing *V* = *L*)

For those people who might wonder why I’ve used ZF instead of ZFC, it turns out that we get Choice if we assume that *V* = *L*, which I will explain in a later post as well, so I will not go into showing Choice here.

Now assuming we’ve shown that *L*_{κ} ⊨ *Z**F*, if we could show that *L*_{κ} ⊨ *V* = *L*, then we would have shown that *V* = *L* is consistent with *Z**F*. This (*L*_{κ} ⊨ *Z**F*) may seem immediately apparent, but there is a little bit of subtlety here. In particular, if it were the case that *L*_{α}^{Lκ} = *L*_{α} we would be done immediately. However, this is not necessarily the case; what happens if *L*_{κ} leaves out some of the elements that would otherwise be in *L*_{α}? I will save this final step for showing that *V* = *L* is consistent with ZF in the next post as well.