= Solution
If some $T_i$ were inconsistent, it would prove every formula by the classical principle of explosion. Because $T_i$ is a <deductively closed set of formulae>, it would then contain every formula and could have no proper extension. This contradicts $T_i\subsetneq T_{i+1}$. Thus every $T_i$ is consistent.
Let
$$
T=\bigcup_{i=0}^{\infty}T_i.
$$
If $T$ were inconsistent, the <finite character of formal proofs> would give a finite inconsistent subset $R\subseteq T$. The chain is increasing, so all finitely many members of $R$ lie in some common $T_N$. Then $T_N$ would be inconsistent, a contradiction. Hence $T$ is consistent.
If $T\vdash\varphi$, a proof again uses only finitely many assumptions from $T$, all lying in some $T_N$. Thus
$$
T_N\vdash\varphi.
$$
Since $T_N$ is deductively closed, $\varphi\in T_N\subseteq T$. Therefore $T$ is deductively closed.
Finally, suppose that $T$ were finitary. Part (b) would provide a finite $R\subseteq T$ such that $R\vdash T$. Choose $N$ with $R\subseteq T_N$. Then $T_N\vdash T$; since $T_N$ is deductively closed, this implies $T\subseteq T_N$. But
$$
T_N\subsetneq T_{N+1}\subseteq T,
$$
a contradiction. Consequently $T$ is not finitary, proving the <increasing union of deductively closed sets>.
Back to article page
