4
$\begingroup$

In the following let $\mathbf{V} = \bigcup_{\alpha \in \mathbf{ON}} V_\alpha$ denote the cumulative hierarchy. Let $\{\varphi_0, \dots, \varphi_n, \dots \}$ denote a list of all $ZF$ axioms. I am reading the following sentence:

"Given $n \in \omega$, the symbols "$\mathbf{V} \models \{\varphi_0, \dots, \varphi_n\}$" and "$\varphi_0 \land \dots \land \varphi_n$" stand for exactly the same thing." (Just/Weese, p 192)

I don't understand how this is true. The first expression says that $\varphi_i$ are all true in $\mathbf{V}$ given any valuation. The second expression seems to be a formula that may or may not be true but there is nothing saying that it is true in $\mathbf{V}$. Thank you for shedding light into my confusion.

1 Answers 1

1

In the presence of the axiom of Foundation/Regularity, $\mathbf V$ is exactly the class of all sets. Thus being "true in $\mathbf V$" is just the same as being true, period.

Technically, relativizing to $\mathbf V$ (which is what I think "$\mathbf V \vDash$" must mean here) is a no-op.


.. or put differently: The symbols "$\mathbf V\vDash\{\varphi_0,\ldots,\varphi_n\}$" is an instruction at the metalevel for you do do various manipulation that results in a wff in the language of set theory. The resulting formula contains no $\mathbf V$, no $\vDash$ and so forth, because these symbols are not in the language of set theory.

Similarly the symbols "$\phi_0\land\cdots\land\phi_n$" are an instruction for you to construct a certain wff in the language of set theory. The manipulations you do here are simpler, but at the least you need to do something about the "$\cdots$" which is not in the language, and unfold the "$\phi$"s to the formulas they denote.

The point is that these two sets of instructions result in wffs that are logically equivalent if only the formula (that results from unfolding) $\forall x.x\in\mathbf V$ is assumed to be true.

  • 0
    Or put differently: If $\psi = \varphi_0 \land \dots \land \varphi_n$ and $\psi' = \mathbf{V} \models \{\varphi_0, \dots ,\varphi_n \}$ then $ZF \vdash \psi \leftrightarrow \psi'$.2012-12-18