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The notation $A\sqcup B$ (and phrase "disjoint union") has (at least) two different meanings. The first is the meaning you suggest: a union that happens to be disjoint.
$\begingroup$ @MarcvanLeeuwen From the second sentence in the Wikipedia article: "disjoint sets are sets whose intersection is the empty set." If this definition was solely meant for pairs of sets, I would expect it to say "two sets are disjoint if..." rather than "disjoint sets are...". $\endgroup$ –
Typically, pairwise disjoint means that no two sets in the collection have a member in common. The answer to your first question is no, in fact each pair of distinct subsets has a common element. For your second question, you may as well ask if {b} {b} and {g} {g} are disjoint, which they are (unless, of course, b = g b = g). – Chris Leary.
0. The term disjoint refers to a collection of subsets, it means that its subsets are disjoint. The term pairwise disjoint refers to a familly of collections of subsets. It not only means that any two collections of that family are disjoint, i.e. they share no common element (that is, a common subset), but that in addition, if you take one set ...
With that definition, disjoint sets are necessarily mutually exclusive, but mutually exclusive events aren't necessarily disjoint. Consider points in the square with each coordinate uniformly distributed from $0$ to $1$ .
So in particular, the empty set is a subset of every set, and also that the empty set is disjoint with every set. Almost every time you prove something regarding the empty set, the best route to go is usually to use a proof by contradiction, since working with the empty set is almost always counter intuitive.
That means A can be contained in B and A and B can be distinct at same time! Two sets are equal if and only if they both have EXACTLY the same elements. Two sets are disjoint if they do NOT have commons elements. So no, to be DIFFERENT/DISTINCT is not the same that being DIsJOiNT. A = {0, 1} and B = {0} are DIFFERENTS (DISTINCT) (1 ∈ A but 1 ...
$\mathcal{M}$ contains an infinite sequence of disjoint sets. b. $\text{card}(\mathcal{M}) \ge \mathfrak{c}$. This is the problem I'm totally stuck at. First, I think there is a missing condition in (a). For (a) to be meaningful, (a) should be corrected : "M contains an infinite collection of disjoint and nonempty sets."
Sorted by: 8. Let A A be a family of sets. We say that the sets in A A are mutually disjoint if no two of them have any elements in common. In other words, if A, B ∈A A, B ∈ A, and A ≠ B A ≠ B, then A ∩ B = ∅ A ∩ B = ∅. Another term that means the same thing is pairwise disjoint. In pictorial terms, if you make a Venn diagram of ...
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