Consider the universe U={x|x∈Z,1≤x≤10}U=\{x | x \in \mathbb{Z}, 1 \le x \le 10\} and the following subsets of UU

A={2,5,9} B={1,4,7,8,10} C={1,2} In the following questions we denote A′ the complement of A. i.e. A′=U−A A′ A∪B∪C B′−A A△B I can’t access my notes and finding this hard to work out. Thanks. Answer A′=U−A. A∪B∪C={x|x∈A∨x∈B∨x∈C}. B′−A=(U−B)−A. AΔB={x|(x∈A∨x∈B)∧x∉(A∩B)}={x|x∈(A∪B)∧x∉(A∩B)}. I will do 1, you can probably do the rest from here. U={1,2,3,4,5,6,7,8,9,10}. So A′=U−A=U−{2,5,9}={1,3,4,6,7,8,10}. Here are … Read more

Jaccard dissimilarity and the triangle inequality

Suppose that δ(A,B)=|AΔB||A∪B|, where Δ represents symmetric difference. Then how does one prove the triangle inequality, viz that δ(A,B)+δ(B,C)≥δ(A,C)? Answer If I understand you correctly, the objects you are considering are finite sets and your dissimilarity function is defined as d(A,B)=|A⊖B||A∪B|, where ⊖ denotes the symmetric difference. Consider the Venn diagram for three sets A, … Read more

Is the Cartesian product of two uncountable sets uncountable? [duplicate]

This question already has answers here: Is the set of all pairs of real numbers uncountable? (2 answers) Closed 6 years ago. Is Cartesian product of two uncountable sets uncountable? Suppose we have a set of real numbers R, Can’t it be shown that R is uncountable by Cantor’s diagonalization method, so it follows that … Read more

A∖(B∩C)=(A∖B)∪(A∖C)A\backslash (B\cap C) = (A\backslash B)\cup (A\backslash C); only one inclusion seems to work

I encountered the following problem: A∖(B∩C)=(A∖B)∪(A∖C). So I need to prove two things: A∖(B∩C)⊆(A∖B)∪(A∖C) (A∖B)∪(A∖C)⊆A∖(B∩C) The first one is fairly easy to me; what I don’t understand is 2. My approach was to suppose x∈A∧¬(x∈B) or x∈A∧¬(x∈C) and try to derive x∈A∧¬(x∈B)∧¬(x∈C), i.e. ∀x ((x∈A∧¬(x∈B))∨(x∈A∧¬(x∈C))→(x∈A∧¬(x∈B)∧¬(x∈C))) But this doesn’t seem possible. For all I know there are … Read more