Given the above definition, I am having troubles with proving the following proposition:
My proof goes as follows:
Take an aritary subset of expressions from the union $\bigcup_{i\in \Bbb N}^{\infty} T_{i}$. Let M denote that set. Note that M must be finite by the definition of "deductive rule". For each expression in M, there exists some $T_{n}$ for some natural number $n$ which that expression belongs to. Therefore let us take the "largest" $i$ such that $T_{i}$ is a superset of M. Since $T_{i}$ is a theory and M is a subset of $T_{i}$, then every expression deduced from the expressions in M must be in the set $T_{i}$. Now since $T_{i} \subset \bigcup_{i\in \Bbb N}^{\infty} T_{i}$, then every expression deduced from the expressions in M must be in the union.
Is my above proof correct? Can someone have a look? If that is not correct, can I please ask for a proof for the proposition? Thanks so much.
This is indeed correct, the property of being in the deductive closure is proved by a finite set of expressions that it is a deduction of. So if we have a chain of theories, as such an increasing sequence is called, and we have a deduction from their union $\cup_i \mathcal{T}_i$, all statements "above the line" (the premises) come from some $\mathcal{T_i}$, and we can indeed take their maximal index and find a theory among them that contains all of the premises, and thus the conclusion is in this specific theory, and hence in the union.
This hinges essentially on the fact (that a "deductive rule" or "proof") only uses finitely many premises (no infinitary proofs).
This kind of fact (the union of a chain of "blahs" is a "blah") is often a prelude to an appliation of Zorn's lemma in the poset of "blahs" and inclusion. So in this case , Zorn's lemma would allow us to conclude that any theory is contained in a maximal theory, so that we cannot add a statement and still have a theory, so maximal in inclusion sense.