We define the circle with the center point – $(a,b)$ and his radius – $r>0$ as:
$$B(a,b,r)=\{\langle x,y \rangle\ \in \mathbb{R}^2:(x-a)^2+(y-b)^2\le r^2\}$$
Let $A$ be the set of the plane's circles:
$$A=\{B(a,b,r):a,b\in \mathbb{R},r\in(0,\infty )\}$$
Let $\mathcal{L}$ be the vocabulary which contains a binary relation $S$, and in addition contains for each circle – $B(a,b,r)\in A$, a constant symbol $\bar c_{a,b,r}$. Let $M$ a model to interpret $\mathcal{L}$, defined by:$$M=\langle A,S\rangle$$ When $S$ defined to be the $\subseteq$ binary relation between the circles ($S(x,y)$ meaning that $x\subseteq y$). The model also interpret $\bar c_{a,b,r}$ as a circle $B(a,b,r)$. Let $T=Th(M)$ the theory of $M$. We define $T_1$, $T_2$ in the vocabulary $\mathcal{L}^+=\mathcal{L}\cup\{d\}$, where $d$ is a new constant symbol:
$$T_1=T\cup\{\forall x [S(x,d)]\}$$
$$T_2=T\cup\{S(d,c_{a,b,r}):a,b\in \mathbb{R},r\in(0,\infty )\}$$
Now, for $T_1, \ T_2$ I need to determine if:
- The theory is inconsistent
- The theory is consistent, and it has a model which is an expansion of $M$ to $\mathcal{L}^+$.
- The theory is consistent, but it does not have a model which is an expansion of $M$ to $\mathcal{L}^+$.
$\textbf{My attempt}$: I wanted to think about the meaning before diving into giving $d$ interpretation and so on.
Therefore, for $T_1$: the additional sentence to $T$ means that for every circle $x$ is within the biggest circle. Now in the world discussion of $B(a,b,r)$, there isn't the biggest one, so if I am appealing to the compatible theorem, I can choose a circle that will contain all finite groups of circles within him. Therefore $T_1$ is consistent but she doesn't have an expansion of $M$ to $\mathcal{L}^+$.
For $T_2$: the additional sentence to $T$ tell us that there is the smallest circle, which impossible, also if we applying the compatible theorem because finite groups of circles can scatter around the cartesian coordinate system, such that we won't be able to choose one that contained in all circles. Hence, we have that $T_2$ is inconsistent.
I am not sure if my way is correct. Thus, I will be glad to get some help from you. Thank you!
Best Answer
Both theories are inconsistent.
For $T_1$, here's a hint: If $N\models T_1$, then $N$ is a model of $T$ in which $d$ is an $S$-maximal element. But a model of $T$ cannot have an $S$-maximal element. Why not?
For $T_2$, your answer is correct, but I would expect a more rigorous explanation. You wrote:
What is the "compatible theorem"? Do you mean the compactness theorem? What does "finite groups of circles can scatter around the cartesian coordinate system" mean precisely?
One way to give a very clear explanation is to find finitely many sentences in $T$ and finitely many of the new axioms $S(d,c_{a,b,r})$ in $T_2$ which are inconsistent. (Because of the compactness theorem, we know that we can always explain inconsistency of $T_2$ by finitely many axioms.) In fact, in this case a single sentence of $T$ and $2$ of the new axioms in $T_2$ will be enough.
It's surprising to me that your question asks about $$T_2=T\cup\{S(d,c_{a,b,r}):a,b\in \mathbb{R},r\in(0,\infty )\}$$ instead of $$T_2'=T\cup\{S(c_{a,b,r},d):a,b\in \mathbb{R},r\in(0,\infty )\}.$$ In fact, $T_2'$ is consistent. Can you prove this?
Taken together, the fact that $T_1$ is inconsistent but $T_2'$ is consistent makes a nice point: The theory $T$ rules out having a circle which contains all other circles, but it's possible to find a model with a circle $d$ which contains all of the standard circles (the elements named by constants in the standard model $M$).