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Feb 23, 2020 at 20:10 history edited Valery Isaev CC BY-SA 4.0
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Feb 23, 2020 at 17:58 history became hot network question
Feb 23, 2020 at 16:27 comment added Valery Isaev @TimCampion I'm asking about tight apartness relations because every apartness relation on $X$ determine a tight one on a quotient of $X$. For example, the relation "$(x - y)$ is a unit" on a local ring is an apartness relation. The corresponding ideal is (the unique) maximal ideal consisting of non-units, the quotient is a Heyting field, and the tight apartness relation on this field is the same as before. Thus, I believe that, to understand apartness relations, it is enough to understand tight ones.
Feb 23, 2020 at 16:11 comment added Valery Isaev @TimCampion Yes, they both carry a tight apartness relation. Also, my second comment was completely wrong. First, the condition "$\neg \neg x = y \implies x = y$" is equivalent to the diagonal being $\neg \neg$-closed and not $\neg \neg$-dense. Second, as Ingo pointed out, it is actually equivalent to the object being $\neg \neg$-separated.
Feb 23, 2020 at 16:04 comment added Valery Isaev @IngoBlechschmidt Yes, you're right. I just was confused. I deleted my comment.
Feb 23, 2020 at 16:01 comment added Tim Campion @ValeryIsaev Surely you're right. Just to clarify (since it's the "tight"-ness part that has me confused more so than the "apartness" part), you're saying that in a topos, $\mathbb N^{\mathbb N}$ and $\mathbb R$ have tight apartness relations, right? BTW is there a name for objects $X$ with $\neg\neg$-dense diagonal? I didn't realize there was a condition like this on the equality relation intermediate between decidability and non-decidability...
Feb 23, 2020 at 15:48 comment added Ingo Blechschmidt @ValeryIsaev: Can you clarify? The condition "$\forall x,y : M. \neg\neg(x = y) \Rightarrow x = y$", interpreted in the internal language, is exactly the condition for the object $M$ to be separated with respect to the $\neg\neg$-topology.
Feb 23, 2020 at 15:41 answer added Ingo Blechschmidt timeline score: 12
Feb 23, 2020 at 13:46 comment added Valery Isaev @TimCampion It does not imply decidability of equality, it only implies that it is $\neg \neg$-separated. There are plenty of examples of objects with an apartness relation and without decidable equality. E.g., functions $\mathbb{N} \to \mathbb{N}$ and Dedekind reals.
Feb 23, 2020 at 11:48 comment added Bas Spitters The relation with the Hausdorff topology may be helpful.
Feb 23, 2020 at 6:48 comment added Tim Campion I'm a little out of my depth here, but I think the first bullet point tells you that in a topos, any object admitting an apartness relation must have decidable equality. I think the first two toposes I tend to think of besides $Set$ are $sSet$ and sheaves on a topological space $X$. In $sSet$, I think maybe only discrete objects have decidable equality? And in $Sh(X)$, maybe it's just coproducts of representables on clopen sets?
Feb 23, 2020 at 3:12 history asked Valery Isaev CC BY-SA 4.0