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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!

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1 Answer 1

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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:

"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."

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 $M$).

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  • $\begingroup$ First of all, thank you! Second, I meant to use the compactness theorem, I'm sorry. Now your $T'$ of yours looks the same as $T_1$. I meant that if I use the compactness theorem, I can limit the number of circles, and then I can pick a circle by giving an interpretation of $d$ such that will contain any circle of the finite group of circles. Why for $T'$ is good but for $T_1$ it doesn't? $\endgroup$
    – Chopin
    Feb 22, 2021 at 17:34
  • $\begingroup$ There is a crucial difference between $T_1$ and $T_2'$. $T_1$ adds a single sentence to $T$ ("$d$ contains every circle"), while $T_2'$ adds infinitely many sentences to $T$ ("$d$ contains the circle $c_{a,b,r}$" for each standard circle $c_{a,b,r}$). In a model $N\models T_1$, $d$ must contain every circle in $N$. In a model $N'\models T_2'$, $d$ must contain all of the circles $c_{a,b,r}$, but $N'$ may have other circles which are not named by constants, and which are not forced to be contained in $d$. $\endgroup$ Feb 22, 2021 at 17:37
  • $\begingroup$ Ohhhhhhh. I haven't seen that it is a single sentence! So now we cannot use the compactness theorem on a single sentence, so it is not consistent, while $T'$ of yours contains plenty of sentences, so we can limit the amount of the sentences, and pick the biggest circle that will contain any circle. Right? $\endgroup$
    – Chopin
    Feb 22, 2021 at 17:51
  • $\begingroup$ @aasc232 that's correct! $\endgroup$ Feb 22, 2021 at 17:52
  • $\begingroup$ Wow! What a great question! isn't it? thank you so much you have helped me beyond all my expectations. $\endgroup$
    – Chopin
    Feb 22, 2021 at 17:54

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