What are $L_1$ and $L_2$ in the Gödel Constructible Hierarchy

Question

I'll start by defining the constructible hierarchy for the sake of proper prefacing to the question. So we begin by defining:

$$ \operatorname{Def}(X) := \big\{ \{ y\mid y \in X \text{ and } (X,\in) \vDash \Phi(y, z_1, \ldots , z_n) \; \} \;\big|\; \Phi \text { is a first order formula and } z_1, \ldots, z_n \in X \big\}. $$

The Gödel Constructible Hierarchy then follows with transfinite recursion as:

$$ L_0 := \emptyset \\ L_{\alpha + 1} := \operatorname{Def}(L_\alpha). $$

Or if, $\lambda$ is a limit ordinal, then we can define:

$$ L_\lambda := \bigcup_{\alpha < \lambda} L_\alpha. $$

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Note. This is not for school, I'm just reading Paul Cohen's book on the Continuum Hypothesis for funzies.

My question is: what is $L_1$ and $L_2$ in the hierarchy? Based on my working things out, it seems to me that:

$$ L_1 := \{\emptyset\}, \text{ and } \\ L_2 := \big\{\{\emptyset\}\big\}. $$

But, I feel like that may be incorrect. And, given that the last time I was looking over the material was 10 years ago, I figured that I'd ask the community.

Answer 1

Here is a slightly more general way to approach this issue.

If $X$ is a finite set such that every element of $X$ is definable, then $\operatorname{Def}(X)=\mathcal P(X)$.

This is not very hard, but now we combine this with the following observation:

If $X$ is a finite transitive set, then every member of $X$ is definable.

To see why, we can prove this by recursion on the rank of the members of $X$: if $x$ is such that all the members of $x$ are definable, then $x$ is the unique set such that $y\in\leftrightarrow\bigvee_{u\in x}\varphi_u(y)$, where $\varphi_u$ is a formula defining $u$.

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Now we get the full power of the construction, up to $\omega$:

1. $L_0=\varnothing$, certainly a finite transitive set;
2. $L_{n+1}=\operatorname{Def}(L_n)$, but since $L_n$ is finite and transitive, this is just $\mathcal P(L_n)$, and in particular a transitive finite set.

So indeed, $L_1=\{\varnothing\}$ and $L_2=\{\varnothing,\{\varnothing\}\}$.

Answer 2

As Wojowu said above, for any arbitrar $n\in\mathbb{N}$, $$ \text{Def}(L_n) = \mathcal{P}(L_n), $$ where $\mathcal{P}(X)$ represents the power set of $X$. So,

$$ L_1 := \{\emptyset\} \\ L_2 := \left\{ \emptyset, \{\emptyset\} \right\} $$