Exercises #4 – Equalitarian Theories

These were pulled from Exercises for [latex]\S5[/latex] of Chapter I in Bourbaki’s set theory book.
(Assume [latex]\mathfrak{T}[/latex] has no constants, i.e. no letters in explicit axioms; thus we can abuse I.2.3.C3 without repercussions)

Show the relation [latex]x=y[/latex] is functional in [latex]x[/latex] in [latex]\mathfrak{T}[/latex].

Denote the relation [latex]x=y[/latex] by [latex]R[/latex]. Let [latex]a[/latex] and [latex]b[/latex] be distinct letters from [latex]x[/latex] and [latex]y[/latex] and do not appear in any explicit axioms of [latex]\mathfrak{T}[/latex]. If [latex](a|x)R[/latex] and [latex](b|x)R[/latex], then [latex]a=b[/latex] by Theorem I.5.2.2, so [latex]x=y[/latex] is single-valued. By Theorem I.5.2.1, [latex]y=y[/latex] is a theorem in [latex]\mathfrak{T}[/latex], that is, [latex](y|x)R[/latex] is a theorem in [latex]\mathfrak{T}[/latex], hence [latex](\exists x)R[/latex] is a theorem in [latex]\mathfrak{T}[/latex]. It follows by I.3.4.C20, [latex]R[/latex] is functional in [latex]x[/latex] in [latex]\mathfrak{T}[/latex].

[latex]\blacksquare[/latex]

Show that if the scheme [latex](\exists x)R\implies R[/latex] provides implicit axioms in [latex]\mathfrak{T}[/latex], then [latex]x=y[/latex] is a theorem in [latex]\mathfrak{T}[/latex]. (See previous exercise)

Let [latex]R[/latex] be the relation [latex]x=y[/latex]. Applying the scheme to [latex]R[/latex] and the letter [latex]x[/latex], [latex](\exists x)R\implies R[/latex] is an implicit axiom in [latex]\mathfrak{T}[/latex] (which we shall denote by [latex]A[/latex]). Since [latex](\exists x)R[/latex] is a theorem in [latex]\mathfrak{T}[/latex] using Theorem I.5.2.1 and I.4.2.S5, [latex]R[/latex] must be a theorem in [latex]\mathfrak{T}[/latex] by the implicit axiom [latex]A[/latex].

[latex]\blacksquare[/latex]

Show that if the scheme [latex](R\iff S)\implies (\tau_X(R)=\tau_X(S))[/latex] provides implicit axioms in [latex]\mathfrak{T}[/latex], then [latex]x=y[/latex] is a theorem in [latex]\mathfrak{T}[/latex].

Let [latex]x[/latex] and [latex]y[/latex] be distinct letters, and let [latex]R[/latex] be the relation [latex]x=x[/latex] and [latex]S[/latex] be the relation [latex]x=y[/latex]. Then applying the given scheme to [latex]R[/latex] and [latex]S[/latex] and using I.2.3.C3 and I.1.2.CS5,

[latex]
((y=y)\iff (y=y))\implies ((y|x)\tau_XR=(y|x)\tau_XS)
[/latex]

is a theorem in [latex]\mathfrak{T}[/latex]. [latex]y[/latex] does not appear in neither [latex]\tau_XR[/latex] nor [latex]\tau_XS[/latex], so [latex] ((y|x)\tau_XR=(y|x)\tau_XS)[/latex] is equivalent to [latex]\tau_XR=\tau_XS[/latex]. Clearly [latex](y=y)\iff (y=y)[/latex] is a theorem in [latex]\mathfrak{T}[/latex], so [latex]\tau_XR=\tau_XS[/latex] is a theorem in [latex]\mathfrak{T}[/latex]. But [latex]x[/latex] satisfies [latex]R[/latex] and [latex]y[/latex] satisfies [latex]S[/latex] by Theorem I.5.2.1, so [latex]x=y[/latex] is a theorem in [latex]\mathfrak{T}[/latex].

[latex]\blacksquare[/latex]

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