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Improved statement
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Noah Schweber
Noah Schweber
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Two comments on this:

First, it's not clear that one can formulate "there exists a transitive set $S\in V$ such that $S\prec V$" in first-order logic, so it's a bit tricky to phrase your question precisely.

More importantly, I claim that no theory $T$ can have the property you desire. Basically, suppose there were such a $T$. Then $T$ must prove "there exists a transitive model of $T$:" by elementarity, if $V\models T$ then $S\models T$. But then $T\models Con(T)$, which contradicts Goedel's theorem.

(One can also prove a weaker version of this without using Goedel: any theory $T$ containing Choice [EDIT: As Francois points out, Choice is unnecessary here] and Foundation cannot have the property you desire. Otherwise, $T$ would prove "there exists a transitive model of $T$." Now using Choice, we can build a sequence $S_1, S_2, . . . $ such that $S_i\models T$ and $S_{i+1}\in S_i$ for all $i$. But this contradicts Foundation.)

Two comments on this:

First, it's not clear that one can formulate "there exists a transitive set $S\in V$ such that $S\prec V$" in first-order logic, so it's a bit tricky to phrase your question precisely.

More importantly, I claim that no theory $T$ can have the property you desire. Basically, suppose there were such a $T$. Then $T$ must prove "there exists a transitive model of $T$:" by elementarity, if $V\models T$ then $S\models T$. But then $T\models Con(T)$, which contradicts Goedel's theorem.

(One can also prove a weaker version of this without using Goedel: any theory $T$ containing Choice and Foundation cannot have the property you desire. Otherwise, $T$ would prove "there exists a transitive model of $T$." Now using Choice, we can build a sequence $S_1, S_2, . . . $ such that $S_i\models T$ and $S_{i+1}\in S_i$ for all $i$. But this contradicts Foundation.)

Two comments on this:

First, it's not clear that one can formulate "there exists a transitive set $S\in V$ such that $S\prec V$" in first-order logic, so it's a bit tricky to phrase your question precisely.

More importantly, I claim that no theory $T$ can have the property you desire. Basically, suppose there were such a $T$. Then $T$ must prove "there exists a transitive model of $T$:" by elementarity, if $V\models T$ then $S\models T$. But then $T\models Con(T)$, which contradicts Goedel's theorem.

(One can also prove a weaker version of this without using Goedel: any theory $T$ containing Choice [EDIT: As Francois points out, Choice is unnecessary here] and Foundation cannot have the property you desire. Otherwise, $T$ would prove "there exists a transitive model of $T$." Now using Choice, we can build a sequence $S_1, S_2, . . . $ such that $S_i\models T$ and $S_{i+1}\in S_i$ for all $i$. But this contradicts Foundation.)

Source Link
Noah Schweber
Noah Schweber
  • 21.1k
  • 10
  • 110
  • 331

Two comments on this:

First, it's not clear that one can formulate "there exists a transitive set $S\in V$ such that $S\prec V$" in first-order logic, so it's a bit tricky to phrase your question precisely.

More importantly, I claim that no theory $T$ can have the property you desire. Basically, suppose there were such a $T$. Then $T$ must prove "there exists a transitive model of $T$:" by elementarity, if $V\models T$ then $S\models T$. But then $T\models Con(T)$, which contradicts Goedel's theorem.

(One can also prove a weaker version of this without using Goedel: any theory $T$ containing Choice and Foundation cannot have the property you desire. Otherwise, $T$ would prove "there exists a transitive model of $T$." Now using Choice, we can build a sequence $S_1, S_2, . . . $ such that $S_i\models T$ and $S_{i+1}\in S_i$ for all $i$. But this contradicts Foundation.)