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There is a Wikipedia article about interior algebras. An interior algebra is a Boolean algebra with an additional unary operator, the interior operator, satisfying certain additional axioms. The axioms are dual to the Kuratowski closure axioms.

A Boolean algebra has a unary operator called complementation. Any nonzero element x of a (non-degenerate) Boolean algebra is distinct from (not equal to) its complement. (Proof: the meet of x with itself is x, while the meet of x with its complement is 0.)

One might think that in an interior algebra, we would have a similar proof that the interior of x is distinct from x, if only for elements satisfying certain conditions. But there appears to be no such proof.

Is there a proof I've overlooked? Is there a proof there's no proof?

And in general this seems odd; if anyone can explain why it's not that would be helpful.

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    @Qi$a$ochuYua$n$ let us [co$n$ti$n$ue this discussion in chat](http://chat.stackexchange.com/rooms/911/discussion-between-michael-carroll-and-qiaochu-yuan)2011-07-24

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Look at interior algebras derived from topological spaces, the $\boldsymbol{A(X)}$ of the Wikipedia article: the elements that are their own interiors are precisely the open sets in the space $X$, which can range from just $\{\varnothing, X\}$ to $\mathscr{P}(X)$. The predicate ‘$x$ is not its own interior’ can be true of no element, of almost all elements, or of an enormous range of intermediate possibilities, depending on the algebra; it’s simply the negation of ‘$x$ is open’, and there’s no reason to expect it to hold for any naturally defined set of elements other than the obvious one, namely, the set of elements that are not open. There’s also no reason to expect an analogy with complementation, since the two operations are very different: $(x^C)^C = x$, while $(x^I)^I = x^I$ (where the superscript $I$ is the interior operator).