# Exercises #1 – Signs and Assemblies

These were pulled from Exercises for $\S$1 of Chapter I of Bourbaki’s set theory book.

#### Exercise 1.

Let $\mathfrak{T}$ be a theory with no specific signs. Show that no assembly in $\mathfrak{T}$ is a relation and that the only assemblies in $\mathfrak{T}$ which are terms are assemblies consisting of single letters.

#### Solution.

Let $A$ be a relation in some formative construction in $\mathfrak{T}$. By definition, $A$ is neither a single letter nor an assembly beginning with $\tau$), but by the formative construction criteria, $A$ must be preceded by (1) a relation $B$ such that $A$ is $\neg B$, (2) relations $B$ and $C$ such that $A$ is $\vee BC$, or (3) terms $A_1,…,A_n$ such that $A$ is $sA_1A_2…A_n$ where $s$ is a relational sign. (3) is not possible since $\mathfrak{T}$ contains no specific signs. If it’s (1) (resp. (2)), then $B$ (resp. $B$ and $C$) has the property of either (1) or (2) since they are relations. We observe this argument can be continued on $B$ (resp. $B$ and $C$), hence inductively, until we end (by the finiteness of formative constructions) at some relation $A_0$, but $A_0$ cannot be a relation else we can continue the argument, so $A_0$ must be a term, a contradiction.

Now if $A$ is a term for some formative construction in $\mathfrak{T}$, $A$ is either a single letter or preceded by a relation $B$ such that $A$ is $\tau_x(B)$ for some letter $x$, but no relations exist in $\mathfrak{T}$, implying $A$ is a single letter.

$\blacksquare$

#### Exercise 2.

Let $A$ be a term or a relation in a theory $\mathfrak{T}$. Show that every sign $\square$ in $A$ (if there is any) is linked to a single sign $\tau$ situated to its left. Show that every sign $\tau$ in $A$ (if there is any) is either not linked at all or else is linked to certain signs $\square$ situated to its right.

#### Solution.

For $\square$, we argue step-by-step:

Step 1. If $A$ is a relation with $\square$ appearing, then using induction (similar to Exercise 1), for some assembly $B$ in the formative construction, there exists a relational sign $s$ of weight $n$ and $n$ terms $A_1,…,A_n$ such that $B$ is $sA_1A_2…A_n$. Each $\square$ is contained in some $A_i$, so on these $A_i$, proceed to Step 2.

Step 2. If $A$ is a term containing $\square$, then either $A$ begins with $\tau$ or with a substantific sign $s$ (of weight $n$). If its the former, then proceed to Step 3; else, there exists $n$ terms $A_1,…,A_n$ such that $sA_1A_2…A_n$ is in the formative construction where $\square$ is contained in some of the $A_i$. On each $A_i$, repeat Step 2.

Step 3. If $A$ is a term containing $\square$ beginning with $\tau$, for each $\square$ in $A$, $\square$ must link to the beginning $\tau$ or not. If not, then let $R$ be the relation such that $A=\tau_XR$. On the relation $R$, return to Step $1$.

This argument must end after a finite amount of steps since formative constructions are finite. We see every $\square$ of $A$ is linked to some $\tau$ at some point (to be precise, each time Step 3 occurs) of this inductive deconstruction of $A$.

We can argue similarly for $\tau$ (exchange $\square$ and $\tau$ in the previous argument).

$\blacksquare$

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