# Exercises #4 – Equalitarian Theories

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

#### Exercise 1.

Show the relation $x=y$ is functional in $x$ in $\mathfrak{T}$.

#### Solution.

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

$\blacksquare$

#### Exercise 2.

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

#### Solution.

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

$\blacksquare$

#### Exercise 3.

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

#### Solution.

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

$((y=y)\iff (y=y))\implies ((y|x)\tau_XR=(y|x)\tau_XS)$

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

$\blacksquare$

This site uses Akismet to reduce spam. Learn how your comment data is processed.