Interpreting V=L

Question

In Set Theory there could be two ways to interpret the axiom of constructibility $V=L$.

1st - The statement that every set is constructible. In other words these are the only type of set that exist.

2nd - As an inference rule, if $\varphi$ is true of constructible sets then it is true of all sets.

Which of these is meant by $V=L$. I think the 1st is restrictive but the 2nd has the benefit of being unrestrictive while still settling many mathematical questions.

Smullyan for instance says many mathematicians reject $V=L$ because they cannot intuitively accept the restrictions that puts on the membership of $\wp x$.

More formally the two options can be put this way. Allowing classes for simplicity and knowing $L\subseteq V$.

1st - $V\subseteq L$.

2nd - For all first order set statements $\varphi$ if $L\vDash\varphi$ then $V\vDash\varphi$.

In the second we allow for $L\subset V$ so it does not restrict the universe but it carries all the "nice" properties of $L$ over to $V$.

Why is this important? If you are a realist who feels ZFC is obviously under-specified, who doesn't want the restriction of constructibility, who thinks most large cardinals are propably fictions, and who thinks GCH is obvious the 2nd option is a good one.

Godel and Cohen showed GCH is independent of ZF, as are many other statements. Rather than peicemeal add these statements or their negations to ZF, just adopt the 2nd inference principle which seems to give all the right answers.

So my question asked which is the standard interpretation. The answer seems to be the 1st. This leads to the question, has the 2nd interpretation been investigated and what are the conclusions?

Answer 1

Stefan Mesken has already given a great answer to this question (+1).

This answer is merely a supplementary answer to address your edit. I'm not a set theorist, so this is primarily based on what I understand of the answer and comments.

You've presented two possible interpretations for $V=L$,

1. Every set is constructible. I.e., the statement $\forall x \exists \alpha : \alpha\in\mathrm{Ord}\wedge x\in L_\alpha$ is true for on the class of all sets.
2. If a first order statement $\phi$ is true of $L$, then it is true of $V$.

Your edit then claims that the consensus seems to be that the first is what is meant.

This seems to be incorrect. The consensus seems to be that, while the first is the usual phrasing of the statement, the two statements are equivalent.

I'm not sure what people have said that you're not understanding, but since the comments appear clear to me, I'll try to expand on the comments of Noah Schweber and Andres E Caicedo above.

First, note that statement (1) is first order. Thus if (2) is true, the fact that (1) trivially holds for all constructible sets implies that (1) holds for all sets.

Conversely if (1) is true of the class of all sets, then any first order statement true of all constructible sets is necessarily true of all sets. Thus (2) holds.

If you still think that the two statements are distinct, you should probably revise the question again and address the comments this time.

Lastly, the point of the comments is that by accepting (2), you cannot have $L\subsetneq V$. In this light, your edit doesn't make much sense, since accepting (2) forces you to accept $V=L$.

Answer 2

Like Andrés has already pointed out in a comment $$ V = L $$ is the statement "every set is constructible" in the sense that $$ \forall x \exists \alpha \in \mathrm{Ord} \colon x \in L_{\alpha}. $$ This is a $\Pi_2$-statement but that's not immediately apparent in this form (because of the "$x \in L_\alpha$" part).

What you need to prove that $V =L$ is expressible as a $\Pi_2$-statement is the fact that $$ \beta \mapsto (L_\alpha \mid \alpha < \beta) $$ is $\Sigma_1$ definable, namely by the formula $$ \begin{align*} \phi(\beta, f) &\equiv \beta \in \mathrm{Ord} \wedge f \colon \beta \to V \\ & \wedge \forall \alpha < \beta ((\alpha = 0 \implies f(\alpha) = \emptyset) \\ & \wedge (\alpha + 1 = \beta \vee f(\alpha+1) = \mathrm{Def}_{\{\in\}}(f(\alpha)) \\ & \wedge (\alpha \text{ limit } \implies f(\alpha) = \bigcup_{\gamma < \alpha} f(\gamma)). \end{align*} $$

(It's still not entirely obvious that this is $\Sigma_1$ because of the $f(\alpha+1) = \mathrm{Def}_{\{\in\}}(f(\alpha))$-part. But that can be expressed in a $\Delta_1$-fashion in the parameter $f(\alpha)$.