TITLE: Why would the category of sets be intuitionistic?
QUESTION [21 upvotes]: This question is probably really naive. And, I hope the title doesn't come off as too combative. I think that topoi of $\mathbf{Set}$-valued sheaves provide an excellent motivation for higher-order intuitionistic logic, because most such topoi aren't boolean. But, I have an extremely hard time accepting that $\mathbf{Set}$ is intuitionistic.
Let me elaborate on this a bit.
Suppose our intuition for the phrase "subset of $X$" comes from the idea of having an effective total function $X \rightarrow \{0,1\}$ that returns an answer in a finite amount of time. In this case, the subsets of $X$ ought to form a Boolean algebra.
Suppose on the other hand that our intuition for the phrase "subset of $X$" comes from the idea of having an effective partial function $X \rightarrow 1$, such that the program for $f(x)$ either halts in a finite amount of time, or else runs forever. Then, with a bit of thought, we see that subsets of $X$ should be closed under union and intersection. But also, there's really no way to take the complement of a subset in general. So, under this viewpoint, the subsets of $X$ form a bounded distributive lattice.
So, I'm willing to believe that $\mathbf{Set}$ is a Boolean topos (the classical viewpoint), and I'm also willing to believe that it's much, much less than that (the "algorithmic" viewpoint), because I think these are both well-motivated and reasonable positions.
The thing about trying to apply intuitionistic logic to $\mathbf{Set}$ is that it seems to straddle a very strange middle ground between these two extremes. It says: yes, you can take the complement of any subset. But no, it may not be the case that $A^c \cup A = X$ (where $A$ is a subset of $X$, and $A^c := X \setminus A.$). And, it may not be the case that $(A^c)^c = A.$ But nonetheless, you can be assured that $((A^c)^c)^c = A^c$. Huh? Look, I realize this makes perfect sense from a BHK perspective; for instance, the complement of the complement of $(0,1) \cup (1,2)$ is $(0,2)$. So, the double negation law fails. But, that's in a topological space! I mean, why would $\mathbf{Set}$ be subject to this kind of reasoning?

REPLY [5 votes]: There's a third intuition: our intuition for subsets comes from the notion of predicate — a subset $S \subseteq X$ corresponds to the condition on $x \in S$ (for $x$ of type $X$).
The idea of subsets as characteristic functions $X \to \{ 0, 1 \}$ comes from ideas about semantics:

Predicates on $X$ correspond to truth-valued functions on $X$
$\{0, 1\}$ is the set of truth values

Of course, this only holds water if you already believe that truth is two-valued, which you don't if you disbelieve the mathematical universe is classical.
The topological character of truth turns out to follow naturally from an idea about what implication means:

$P,Q \vdash R$ if and only if $P \vdash Q \to R$ 

Since we equate having two premises $P,Q$ with having a single premise $P \wedge Q$, it follows that the class of predicates on any object — and thus subsets of that object — is a Heyting algebra. 
Since it's the very nature of Set that's under discussion, the Heyting algebra is automatically a complete Heyting algebra, and thus a locale. 

All that said, my position is that foundations (and thus Set) are classical, and even if I were persuaded to someday go all-in with intuitionistic logic, it would be in the form of something like working exclusively in the internal language of a nonboolean topos.