# Exercise 7.72#

Let $$π$$ be the sheaf of people as in Section 7.4.3, and let $$π: \Omega β \Omega$$ be βassuming Bob is in San Diego β¦β

1. Name any predicate $$π: π β \Omega$$, such as βlikes the weather.β

2. Choose a time interval $$π$$. For an arbitrary person $$π β π(π)$$, what sort of thing is $$π(π )$$, and what does it mean?

3. What sort of thing is $$π(π(π ))$$ and what does it mean?

4. Is it true that $$π(π ) β€ π(π(π ))$$? Explain briefly.

5. Is it true that $$π(π(π(π ))) = π(π(π ))$$? Explain briefly.

6. Choose another predicate $$π: π β \Omega$$. Is it true that $$π(π β§ π) = π(π) β§ π(π)$$? Explain briefly.

1. Letβs use βlikes the weatherβ as suggested.

2. Weβll let $$s$$ equal Bob to make this all simpler, although we could write it all for an arbitrary person. So $$p(s)$$ is a particular subset of times Bob likes the weather.

3. We translate this predicate to βBob likes the weather or he is not in San Diegoβ using the βor notβ translation discussed elsewhere. We expect this sheaf to return a subset of times, and this predicate should be able to return a subset of times.

4. Or more precisely $$π(π ) β€_S π(π(π ))$$. While $$p(s)$$ is a particular set of times not qualified by anything, $$j(p(s))$$ is qualified by the requirement that Bob not be in San Diego. Most precisely, the subset of times that βBob likes the weatherβ given by $$p(s)$$ should be a subset of the times that itβs true that βBob likes the weather or he is not in San Diegoβ.

5. Yes; adding an additional qualifier of βBob is in San Diegoβ should make no difference because weβve already qualified the statement on that proposition. If Bob was not in San Diego, heβs still not in San Diego.

6. Letβs assign $$q$$ to ββ¦ is wearing redβ. Then βBob both likes the weather and is wearing red, and is not in San Diegoβ should be equivalent to βBob likes the weather and is not in San Diego, and Bob is wearing red and is not in San Diegoβ.

## Authorβs solution#

1. Take π(π ) to be βπ  likes the weather.β

2. Let π be the interval 2019/01/01 β 2019/02/01. For an arbitrary person $$π β π(π)$$, $$π(π )$$ is a subset of $$π$$, and it means the subset of $$π$$ throughout which $$π$$ likes the weather.

3. Similarly $$π(π(π ))$$ is a subset of $$π$$, and it means the subset of $$π$$ throughout which, assuming Bob is in San Diego, $$π$$ liked the weather. In other words, $$π(π(π ))$$ is true whenever Bob is not in San Diego, and it is true whenever $$π$$ likes the weather.

4. It is true that $$π(π ) β€ π(π(π ))$$, by the βin other wordsβ above.

5. It is true that $$π(π(π(π )) = π(π(π )$$, because suppose given a time during which βif Bob is in San Diego then if Bob is in San Diego then π  likes the weather.β Then if Bob is in San Diego during this time then π  likes the weather. But that is exactly what π(π(π )) means.

6. Take π(π ) to be βπ  is happy.β Suppose βif Bob is in San Diego then both π  likes the weather and π  is happy.β Then both βif Bob is in San Diego then π  likes the weatherβ and βif Bob is in San Diego then π  is happyβ are true too. The converse is equally clear.