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I'm working a bit with Heyting algebras (which are pseudocomplemented distributive lattives, right?) and I have a question about DeMorgan's laws. I know that, in general, it's not the case that $-(X \wedge Y) = -X \vee -Y$ in a Heyting algebra. However, it's also the case that, if $-X \vee --X = 1$ in such an algebra, then both DeMorgan's laws hold, right? Further, we also know that it's possible to interpret Heyting algebras as given topologies, since a topology will generally be a pseudocomplemented distributive lattice. Take then the usual topology (let's call it $\mathcal{T}$) on the real line. Define, for $O \in \mathcal{T}$, $-O = \mathrm{int}(\mathbb{R}\setminus O)$. Join is taken as union and meet as intersection. It seems then that DeMorgan's law outlined above doesn't hold: if one considers the intervals $(0, 1)$ and $(2, 3)$, we have

$-((0, 1) \wedge (2, 3)) = \mathrm{int}(\mathbb{R} \setminus ((0, 1) \cap (2, 3)))\\ = \mathrm{int}(\mathbb{R} \setminus \varnothing)\\ = \mathrm{int}(\mathbb{R}) = \mathbb{R}$.

On the other hand:

$-(0, 1) \vee -(2,3) = \mathrm{int}(\mathbb{R} \setminus (0, 1)) \cup \mathrm{int}(\mathbb{R} \setminus (2,3))\\ = \mathrm{int}((-\infty, 0] \cup [1, \infty)) \cup \mathrm{int}((-\infty, 2] \cup [3, \infty))\\ = (-\infty, 0) \cup (1, \infty)) \cup ((-\infty, 2) \cup (3, \infty))\\ = (-\infty, 0) \cup (1, \infty)$,

which is not equal to $\mathbb{R}$ (it's missing the interval $[0, 1]$).

But then, consider again a set such that $U \in \mathcal{T}$. By the above definitions, it follows that $-U \vee --U = \mathrm{int}(\mathbb{R} \setminus U) \cup \mathrm{int}\bar{U}$. We know that $U \subseteq \mathrm{int}\bar{U}$, so it follows that $\mathrm{int}(\mathbb{R} \setminus U) \cup U \subseteq \mathrm{int}(\mathbb{R} \setminus U) \cup \mathrm{int}\bar{U}$. However, unless I'm mistaken (and I may very well be, I haven't tried to prove this yet), $\mathrm{int}(\mathbb{R} \setminus U) \cup U = \mathbb{R}$. So $\mathbb{R} \subseteq \mathrm{int}(\mathbb{R} \setminus U) \cup \mathrm{int}\bar{U}$, whence $\mathbb{R} = \mathrm{int}(\mathbb{R} \setminus U) \cup \mathrm{int}\bar{U}$. So $-U \vee --U = 1$.

Hence, we have an example of a pseudocomplemented distributive lattice in which both DeMorgan's law doesn't hold and, for every $U \in \mathcal{T}$, $-U \vee --U = 1$. Obviously, I must have gone wrong somewhere, but I'm not exactly sure where. Can you guys help me locate the source of my mistake?

Nagase
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    "However, unless I'm mistaken (and I may very well be, I haven't tried to prove this yet), $\mathrm{int}(\mathbb{R}\setminus U)\cup U=\mathbb{R}$" – well, there's the catch: if $U$ is open, then neither $\mathrm{int}(\mathbb{R}\setminus U)$ nor $U$ contains the boundary of $U$, e.g. for $U = (0,1)$ you get $\mathbb{R} \setminus {0,1} \neq \mathbb{R}$. – dtldarek May 04 '14 at 08:22
  • Note that much simpler, i.e. finite examples can be given on failure of (one of) De Morgan's laws in a Heyting lattice; see https://math.stackexchange.com/questions/120187/do-de-morgans-laws-hold-in-propositional-intuitionistic-logic Also " However, it's also the case that, if −X∨−−X=1 in such an algebra, then both DeMorgan's laws hold, right?" True as you get a Stone algebra in either case; the proofs are slightly non-obvious though. See e.g. https://ncatlab.org/nlab/show/De+Morgan+Heyting+algebra but most books that a chapter/section of Stone lattices prove this too. – the gods from engineering Mar 20 '21 at 20:13

2 Answers2

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Take $U=(0,1)$. $\neg U=(-\infty, 0)\cup (1,\infty )$. $\neg \neg U=(0,1)$. Thus, $\neg U \cup \neg \neg U\ne 1$.

Ittay Weiss
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  • I'm not sure I understand. This is supposed to be a counter-example to which of my arguments? – Nagase May 04 '14 at 06:22
  • I edited my answer. Is it clearer now? – Ittay Weiss May 04 '14 at 06:24
  • Ahh, the edit made things clearer. As I suspected, the problem was in the second argument, not the first (the first still goes through, right?). – Nagase May 04 '14 at 06:28
  • yes, of course. De Morgen holds iff the algebra is complemented. – Ittay Weiss May 04 '14 at 06:29
  • Wait, a Heyting algebra, if I understood the idea correctly, need not be complemented for $\neg X \vee \neg\neg X = 1$. If the latter holds, every regular will be complemented, but if $X$ is not regular, it may not have a complement, right? – Nagase May 04 '14 at 06:57
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$(-\infty,2)$ contains $[0,1]$

Empy2
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  • Hmm. Got it. What if I change my example to the intervals (-1, 1) and (-1/2, 1/2)? Does the argument go through? – Nagase May 04 '14 at 06:35
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    Both of the answers are now $(-\infty,-1/2)\cup(1/2,\infty)$ – Empy2 May 04 '14 at 06:49
  • This was a little more difficult than what I had first thought. I think I got it, though. If I take the intervals (0, 1) and (-1/2, 1/2), then $\neg(0, 1) \cup \neg(-1/2,1/2) = ((-\infty, 0) \cup (1, \infty)) \cup ((-\infty, -1/2) \cup (1/2, \infty)) = (-\infty, 0) \cup (1/2, \infty)$. On the other hand, $\neg((0,1) \cap (-1/2, 1/2)) = \text{int}(\mathbb{R} \setminus (-1/2, 1)) = \text{int}((-\infty, -1/2] \cup [1, \infty)) = (-\infty, 1/2) \cup (1, \infty)$. Is that right? – Nagase May 04 '14 at 18:41