Empty theory is a sequence of isolated points 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!
 A: Remember that a theory in this context is just a set of sentences. The empty theory is $\{\}$, the theory containing no sentences at all. This is the weakest possible theory (so the answer to your question "wouldn't it imply everything?" is "no"), and the only things it proves are the logical tautologies, that is, the sentences true in every structure. Remember that $\mathcal{M}\models S$ iff every sentence in $S$ is true in $\mathcal{M}$; if $S=\{\}$, then $\mathcal{M}\models\mathcal{S}$ is vacuously true for every structure $\mathcal{M}$, so the consequences of $\mathcal{M}$ are merely the sentences true in all structures.
Meanwhile, the notion of "isolated point" is indeed a topological notion, but that doesn't mean it's irrelevant to logic; indeed, the whole point of considering the type spaces is that there is a connection between logical properties of theories and topological properties of the associated space.
You should break the exercise down into three parts:

*

*First, classify the models of $T$ up to isomorphism. (HINT: there's exactly one of each cardinality.)


*Next, determine the complete consistent theories extending $T$. Note that multiple non-isomorphic models may have the same theory. (Indeed, there will only be countably many such theories.)
Note that so far, topology hasn't entered the picture; only the final bulletpoint involves topology:

*

*Lastly, remember that points in $S_0(T)$ are exactly complete consistent extensions of $T$, so the previous bulletpoint tells you what they are; it only remains to determine the topology. Establishing isolatedness will be pretty easy (HINT: can you write down a sentence true in exactly those structures of cardinality $17$, for example?); understanding the non-isolated point is a bit more nuanced.

A: 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.
