Exercises #1 – Signs and Assemblies

These were pulled from Exercises for [latex]\S[/latex]1 of Chapter I of Bourbaki’s set theory book.

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

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

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

[latex]\blacksquare[/latex]

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

For [latex]\square[/latex], we argue step-by-step:

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

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

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

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

We can argue similarly for [latex]\tau[/latex] (exchange [latex]\square[/latex] and [latex]\tau[/latex] in the previous argument).

[latex]\blacksquare[/latex]

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