I'm working my way through Henson and van den Dries Model Theory lecture notes doing odd-numbered exercises. Yet I am stuck on the exercise 4.17:
Show that if $T$ is the empty theory in the language $L$ of equality, then the space $S_0(T)$ consists of a sequence of points $(T_n | n \geq 1)$ that are isolated, together with a point $T_\infty$ to which this sequence converges.
Nowhere in the notes do they define what an "empty theory" is (I am only aware of an empty language). How would such a thing even look like? What are the consequences of it? In particular, wouldn't it imply everything?
Then the "isolated points" are also not defined anywhere. How could I possibly start proving the statement? It seems more of a topology question to me than model theory...
Any help or hints would be appreciated!
Let me fill in a few details of Noah's excellent answer. The empty theory $T$ in the language $\mathcal{L}=\{=\}$ is just that – the theory that contains no $\mathcal{L}$-sentences. Any set is a model of $T$. You ask if $T$ implies everything; this is not the case. Indeed, the sentences derivable from $T$ are precisely the $\mathcal{L}$-sentences that hold in every set. These are also called the tautologies of the predicate calculus. For instance, the following is a theorem of $T$: $$\forall v\forall w(v=w\vee v\neq w),$$ since it is true in every set (ie, true in every model of $T$). On the other hand, consider the $\mathcal{L}$-sentence $\phi:\equiv\exists v\exists w(v\neq w)$. $\phi$ is not a theorem of $T$, since there are models of $T$ in which it does not hold. For example, consider a model $\{\star\}$ consisting of a single element. Note that $\neg\phi$ is not a theorem of $T$ either, since $\phi$ holds in the model $\{\star,\bullet\}$ consisting of $2$ elements. So, in other words, both $T\cup\{\phi\}$ and $T\cup\{\neg\phi\}$ are consistent. Hopefully this clarifies the question a bit.
Now, recall that $S_0(T)$ is the set of all complete (and consistent) $\mathcal{L}$-theories that contain $T$. Since every $\mathcal{L}$-theory contains $T$ (because the empty set is a subset of any set), this means $S_0(T)$ is just the set of all complete $\mathcal{L}$-theories. Topologically, a basis of clopen sets for $S_0(T)$ is parametrized by the set of all $\mathcal{L}$-sentences: for any $\mathcal{L}$-sentence $\phi$, we define $[\phi]\subseteq S_0(T)$ to be the set of all elements of $S_0(T)$ that contain $\phi$. So, if $\phi\equiv\forall v\forall w(v=w\wedge v\neq w)$, then $[\phi]=S_0(T)$, since $\phi$ holds in any model of $T$. On the other hand, $[\neg\phi]=\emptyset$. (Why?)
For a less vacuous example, consider the sentence $\phi\equiv\exists v\exists w(v\neq w)$. Then both $[\phi]$ and $[\neg\phi]$ are non-empty. Indeed, if $M$ is a model of $\phi$ (say $M=\{\star,\bullet\}$), then the set of all $\mathcal{L}$-sentences that hold in $M$, denoted $\operatorname{Th}_\mathcal{L} M$, is a complete and consistent $\mathcal{L}$-theory containing $\phi$. So $\operatorname{Th}_\mathcal{L}M\in [\phi]$. Likewise, if $N$ is a model of $\neg\phi$ (say $N=\{\star\}$), then $\operatorname{Th}_\mathcal{L} N\in[\neg\phi]$.
So, this is the topology we're working with. Recall that a point $x$ in an arbitrary topological space $X$ is "isolated" if $\{x\}$ is an open subset of $X$. In particular, a complete $\mathcal{L}$-theory $S$ is isolated in $S_0(T)$ if $\{S\}$ is open. As an exercise, show that this holds if and only if there exists an $\mathcal{L}$-sentence $\phi$ such that $[\phi]=\{S\}$. Furthermore, can you show that, for such $\phi$, we have $T\models\phi\to\psi$ for every $\psi\in S$? We then say that $S$ is isolated by $\phi$. As a final hint, to complement Noah's answer, consider the sentence $$\phi_n\equiv\exists v_1\dots\exists v_n\bigwedge_{i\neq j}v_i\neq v_j\wedge\forall w\bigvee_{i=1}^n w=v_i.$$ Up to isomorphism, what does a model of $\phi_n$ look like? (Can there be two non-isomorphic models of $\phi_n$?) In particular, what can you conclude about the open set $[\phi_n]\subseteq S_0(T)$? (Remember that two isomorphic $\mathcal{L}$-structures $M\cong N$ are elementarily equivalent: $\operatorname{Th}_\mathcal{L}M=\operatorname{Th}_\mathcal{L}N$.) I'll leave my comments at this; hopefully you can now work out the details yourself.
Edit: Okay, here's a complete solution to the question. At each stage, try to read a small bit at a time and work out the rest for yourself! For each $n\in\mathbb{N}$, let $\phi_n$ be the sentence given above, which says that there exist precisely (and no more than) $n$ elements. Now, because $\mathcal{L}$ consists of only equality, any function is an $\mathcal{L}$-embedding, and a function between $\mathcal{L}$-structures is an isomorphism if and only if it is a bijection. In particular, suppose we have models $M\models\phi_n$ and $N\models\phi_n$ for some $n\in\mathbb{N}$. Then $|M|=|N|=n$ (why?) and so we can find a bijection – ie an $\mathcal{L}$-isomorphism – $M\to N$. In particular, since isomorphic structures are elementarily equivalent, this means $\operatorname{Th}_\mathcal{L}M=\operatorname{Th}_\mathcal{L}N$, and so (since $M$ and $N$ were arbitrary) this shows that any two models of the sentence $\phi_n$ have the same $\mathcal{L}$-theory. Denoting this theory by $T_n$, we then have $[\phi_n]=\{T_n\}$, and so in particular $T_n$ is isolated.
Now, there are $\mathcal{L}$-structures in which $\phi_n$ does not hold for any $n\in\mathbb{N}$ – just consider any infinite set. To deal with these models, first fix any infinite cardinal $\kappa\geqslant\aleph_0$, and suppose that $M$ and $N$ are two models of $T$ of size $\kappa$. (Ie, two sets of size $\kappa$.) Then there exists a bijection $M\to N$, which is an $\mathcal{L}$-isomorphism by the comments above, and so $\operatorname{Th}_\mathcal{L}M=\operatorname{Th}_\mathcal{L}N$. Thus any two models of cardinality $\kappa$ have the same complete $\mathcal{L}$-theory, for any $\kappa\geqslant\aleph_0$. In fact, this means that any two infinite models have the same complete $\mathcal{L}$-theory! Indeed, suppose $|M|=\kappa$ and $|N|=\lambda$ for $\kappa,\lambda\geqslant\aleph_0$. Without loss of generality assume $\kappa\geqslant\lambda$. Since $\mathcal{L}$ is finite, the Löwenheim-Skolem theorem tells us that there exists an elementary substructure $M'\preccurlyeq M$ with $|M'|=\lambda$. Then $\operatorname{Th}_\mathcal{L}M=\operatorname{Th}_\mathcal{L}M'$, and, by the argument above, since $|M'|=\lambda=|N|$, we also have $\operatorname{Th}_\mathcal{L}M'=\operatorname{Th}_\mathcal{L}N$, so $\operatorname{Th}_\mathcal{L}M=\operatorname{Th}_\mathcal{L}N$ as desired. Thus any two infinite models of $T$ have the same complete $\mathcal{L}$-theory; call this theory $T_\infty$.
Now, we first claim that $S_0(T)=\{T_n\}_{n\in\mathbb{N}}\cup\{T_\infty\}$. Indeed, let $M$ be any model of $T$ (ie any set). If $M$ is finite, then $\operatorname{Th}_\mathcal{L}M=T_{|M|}$, and if $M$ is infinite, then $\operatorname{Th}_\mathcal{L}M=T_\infty$, by the arguments above. In particular, every complete $\mathcal{L}$-theory appears in $\{T_n\}_{n\in\mathbb{N}}\cup\{T_\infty\}$, as desired. By the argument above we've shown that each $T_n$ is isolated, by the formula $\phi_n$, so it remains only to show that the sequence $(T_n)_{n\in\mathbb{N}}$ converges to $T_\infty$. This amounts to the following statement: for any $\mathcal{L}$-sentence $\phi$ with $T_\infty\in[\phi]$, there exists some $k\in\mathbb{N}$ such that $T_n\in[\phi]$ for any $n\geqslant k$.
To see this, suppose $\phi$ is an $\mathcal{L}$-sentence such that $T_n\notin[\phi]$ for arbitrarily large values of $n$. This means that, for every $k\in\mathbb{N}$, we can find $n\geqslant k$ such that $\phi$ does not hold in any $\mathcal{L}$-structure of size $k$. By the compactness theorem, this means that the following family of sentences is consistent: $$\Sigma=\{\exists v_1\dots\exists v_k\bigwedge_{i\neq j}v_i\neq v_j\}_{k\in\mathbb{N}}\cup\{\neg\phi\}.$$ Indeed, for any finite subset $\Delta\subset\Sigma$, there will be a largest value of $k$ such that the sentence $\exists v_1\dots\exists v_k\bigwedge_{i\neq j}v_i\neq v_j$ appears in $\Delta$. We will then have $M\models\Delta$ for any set $M$ such that $k\leqslant|M|\in\mathbb{N}$ and $T_{|M|}\notin[\phi]$, meaning that $\Delta$ is consistent. So, by compactness, $\Sigma$ is consistent, and hence we can find a model $M\models\Sigma$. Then $M$ is infinite (why?), so $\operatorname{Th}_\mathcal{L}M=T_\infty$ (why?), but $M\models\neg\phi$, whence $T_\infty\notin[\phi]$. This proves the desired result, and so we are done.