Oh no. Take Set×Set - this boolean topos has 4 points already. Regarding conjunction, it classifies (true, true) inclusion to Ω. In Grothendieck topologies, classifying maps are built by Yoneda lemma.
Ok, say, we have a four-point Ω (and ignore stuff that are not points).
We can have a variety of logics.
First, 4-valued booleans. a = (0,1), b=(1,0), T = (1,1), ⊥=(0,0), and operations are defined bitwise.
Second, numbers ⊥=0, a=1/3, b=2/3, T = 1. This linear order is a bounded distributive lattice, so everything is well-defined, conjunction is just minimum. Trying to find another four-point logic...
Well, because the definition comes from the classifying arrow, and this classifying arrow depends on how the topos works. If we have Set^2, that's one thing; if we have Set^[0,1,2,3], it's the other thing. In both cases Ω has 4 points.
Maybe I don't get something basic. What it looks like to me, since the square has arrows (true,true) and true, it defines only the behaviour of the classifying arrow for one pair, not all possible pairs.
Right. It classifies just one point, (true, true). But it's defined on the whole ΩxΩ. How is it defined? It maps every (generalized) arrow T → ΩxΩ to a value in Ω, and it evaluates how close is this arrow to (true, true).
no subject
Date: 2020-06-30 05:40 pm (UTC)Regarding conjunction, it classifies (true, true) inclusion to Ω. In Grothendieck topologies, classifying maps are built by Yoneda lemma.
no subject
Date: 2020-06-30 05:49 pm (UTC)no subject
Date: 2020-06-30 05:59 pm (UTC)We can have a variety of logics.
First, 4-valued booleans. a = (0,1), b=(1,0), T = (1,1), ⊥=(0,0), and operations are defined bitwise.
Second, numbers ⊥=0, a=1/3, b=2/3, T = 1. This linear order is a bounded distributive lattice, so everything is well-defined, conjunction is just minimum. Trying to find another four-point logic...
no subject
Date: 2020-06-30 06:04 pm (UTC)But my question is how does the definition reflect this fact? It seems to only say that T ∧ T = T - nothing about a ∧ b = ⊥, and not, say, a ∧ b = a.
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Date: 2020-06-30 08:01 pm (UTC)no subject
Date: 2020-06-30 08:17 pm (UTC)Maybe I don't get something basic. What it looks like to me, since the square has arrows (true,true) and true, it defines only the behaviour of the classifying arrow for one pair, not all possible pairs.
no subject
Date: 2020-07-01 12:08 am (UTC)