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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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What do you mean by "certain conditions"? –  Qiaochu Yuan Jul 24 '11 at 18:03
    
@Qiaochu Yuan: Well, I would not expect a proof that every element was distinct from its interior, if only because the interior of 0 must be 0. I would expect a proof along the lines of: the interior of x is distinct from x provided that x is ..., where "certain conditions" fills in the "...". For example we might say that x is open if it's equal to its interior. Then a proof that open elements are distinct from their interiors would be a proof that elements satisfying certain conditions are distinct from their interiors. (But the axioms do not require this. That example doesn't fly.) –  MikeC Jul 24 '11 at 19:11
    
That example is a bit silly but I think it answers the question. –  MikeC Jul 24 '11 at 19:14
    
It seems to me that "x is equal to its interior" is already a fundamental notion here, and that one should describe other things in terms of this condition, rather than the other way around. So I still don't really understand the question. –  Qiaochu Yuan Jul 24 '11 at 19:17
    
@Qiaochu Yuan: The complement of x is always different from x, while its interior need never be. That seems peculiar. But perhaps there's no good question here. Thanks for commenting. –  MikeC Jul 24 '11 at 19:39

1 Answer 1

up vote 3 down vote accepted

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).

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