Given a set $S$, a tight apartness relation on $S$ is a relation $\#$ which is tight, irreflexive, symmetric, and a comparison, or more specifically, a relation $\#$ such that


* for all elements $a \in S$ and $b \in S$, $a = b$ if and only if $\neg a \# b$
* for all elements $a \in S$, $\neg a \# a$
* for all elements $a \in S$ and $b \in S$, $a \# b$ implies that $b \# a$
* for all elements $a \in S$, $b \in S$, and $c \in S$, $a \# c$ implies that $a \# b$ or $b \# c$.


Assume that the category of sets is an elementary topos. This means that it contains a subobject classifier $\Omega$.


If the category of sets is Boolean, the subobject classifier $\Omega$ automatically has a tight apartness relation given by negation of equality, because in such a case, $\Omega$ is just the set with two elements and thus has decidable equality.


Is the converse of the above statement true? Namely, if $\Omega$ has a tight apartness relation $\#$, then is the category of sets Boolean?



Does a tight apartness relation on a subobject classifier imply the elementary topos is Boolean?
Yes. The tightness axiom, $(a=b)\iff \neg (a\#b)$, implies
$$\neg\neg(a=b) \iff \neg\neg\neg(a\#b) \iff \neg(a\#b) \iff (a=b).$$
Taking $b=\top$, we find that $\neg\neg a \iff a$ for all $a: \Omega$. This is the law of double negation, which is equivalent to excluded middle.


