I am working on the proof of the Lindenbaum's lemma and there are some passages which are not very clear for me. Here is the statement:
Let $\mathbb{L}$ a countable signature, $T$ a consistent set of $\mathbb{L}$ sentences and $\sigma_0$ a sentence which is not provable by $T$. Then there exists a maximal consistent set $T^*$ of $\mathbb{L}$-sentences, which contains all $\mathbb{L}$-sentences from $T$ and $\neg\sigma_0$.
To prove this, the first step is to consider a universal list of $\mathbb{L}$-sentences $U=[\sigma_1,...]$ and to extand it to $U_0=[\neg\sigma_0,\sigma_1,...]$. This should not be something special but if $U$ is the universal list of all $\mathbb{L}$-sentences how can it be extended again to another universal list?
In the proof it will be constructed a sequence of sets with "limit" $T^*$ as follows.
$T_0=[\ ], \ T_{n+1}=\{T_n+[\sigma_n] \text{for} \ Con(T+T_n+\sigma_n)\} \text{and} \{T_n \ \text{otherwise}\}$
then it will be shown that $T^*$ contain$\neg\sigma$, Con($T^*$), $T^*$ contains all sentences of T an it is maximal.
Moreover it will be shown that $T^*+T$ is consistent and I don't get why we do need it.
Any suggestion would be appreciate thanks
The sentence $\neg\sigma_0$ is already in the list $U$, so it appears twice in $U_0$; we’ll see that this is not a problem. The point of this step is to make sure that $\neg\sigma_0$ appears first in the list. Since $T\not\vdash\sigma_0$, we know that $T+\neg\sigma_0$ is consistent and hence that $T_1=\{\neg\sigma_0\}$.
The sentences $\sigma_0$ and $\neg\sigma_0$ were in the universal list $U$, so there are $k,\ell\in\Bbb Z^+$ such that $\sigma_k=\sigma_0$ and $\sigma_\ell=\neg\sigma_0$. When we construct $T_{k+1}$, we’ll find that $T+T_k+\sigma_k$ is inconsistent, since $\neg\sigma_0\in T_k$, so we’ll have $T_{k+1}=T_k$. When we construct $T_{\ell+1}$, on the other hand, we’ll find that $T+T_\ell+\sigma_\ell$ is consistent, because $\sigma_\ell=\neg\sigma_0$ is already in $T_\ell$, so we’ll ‘add’ $\sigma_\ell$ — except that since it’s already there, we’ll actually have $T_{\ell+1}=T_\ell$.
This isn’t quite the usual statement of Lindenbaum’s lemma; this one emphasizes the fact that we can extend any theory $T$ to a maximal consistent theory that includes any given sentence consistent with $T$.