272 reputation
19
bio website
location Los Angeles, CA
age 33
visits member for 3 years, 10 months
seen Jul 21 at 0:08

Jul
2
awarded  Curious
Sep
1
awarded  Nice Question
Aug
30
revised What explains the asymmetry here?
added 96 characters in body
Aug
29
comment What explains the asymmetry here?
It seems that well-definedness has no effect on the distributivity of the image operation over unions or intersections. However it does disrupt the distributivity of the inverse image operation over intersections, i.e. for a "multivalued" function $f: X \rightarrow \mathcal P \left(Y\right) \setminus \{\emptyset\}$, define the inverse image of $A$ under $f$, $f^{-1}(A)$ as $\{ x \mid f(x) \cap A \neq \emptyset \}$. Then $f^{-1}(A \cup B) = f^{-1}(A) \cup f^{-1}(B)$ but $f^{-1}(A \cap B) \subseteq f^{-1}(A) \cap f^{-1}(B)$, similarly to the asymmetry of the image operation.
Aug
29
awarded  Critic
Aug
29
comment What explains the asymmetry here?
@Clive, can you comment? How does well-definedness correspond to distributivity over unions?
Aug
29
revised What explains the asymmetry here?
edited tags
Aug
29
awarded  Nice Question
Aug
29
revised What explains the asymmetry here?
edited tags
Aug
29
comment What explains the asymmetry here?
Intuitively I would expect that if somewhere there appears the qualification that $f$ must be injective to make an equality, then elsewhere $f$ must be surjective.
Aug
29
comment What explains the asymmetry here?
Another interesting thing is that with inverse image sets there is no asymmetry; $f^{-1}(A \cup B) = f^{-1}(A) \cup f^{-1}(B)$ and $f^{-1}(A \cap B) = f^{-1}(A) \cap f^{-1}(B)$, no ifs ands or buts.
Aug
29
comment What explains the asymmetry here?
Thanks Brian, as I mentioned in my comment on Peter's answer, I understand the proof, I'm just wondering why there is the apparent asymmetry that the image operation distributes over unions but not intersections (in general).
Aug
29
comment What explains the asymmetry here?
It's really an aesthetic quality I'm noticing. $\cup$ and $\cap$ behave in most ways dually, so that you can get from one theorem about $\cup$ to another about $\cap$ by just "reversing everything". This seems like an exception to that rule. I'm thinking maybe it's due to the inherent asymmetry of the notion of function. Maybe this isn't a question that has a rigorous answer :)
Aug
29
comment What explains the asymmetry here?
I understand that $f$ being injective makes it an equivalence, but my question is why the apparent asymmetry of the situation? There are no qualifications on the first equation.
Aug
29
asked What explains the asymmetry here?
Jul
31
awarded  Yearling
Sep
15
comment How can I determine the cardinality of a set of polymorphic functions?
You're right, System F is the right framework in which to answer this sort of question, and specifically using the idea of Parametricity. Thanks!
Sep
15
accepted How can I determine the cardinality of a set of polymorphic functions?
Aug
18
comment Linear functions with rounding
@Olivier - Hm, I don't see how? For what initial value(s) would that be a problem?
Aug
18
revised Linear functions with rounding
added 180 characters in body