- #1
pandaBee
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Homework Statement
Suppose A is a set, and for every family of sets F, if ∪F = A then
A ∈ F.
Prove that A has exactly one element. (Hint: For both the existence
and uniqueness parts of the proof, try proof by contradiction.)
Homework Equations
The Attempt at a Solution
Let A be an arbitrary set
Suppose ∀F(∪F=A⇒A∈F)
Want to prove:
∃!x(x∈A) (A has a single, unique element.)
Existence:
Want to prove:
∃x(x∈A)
I use the Hint and use proof by contradiction, we add the assumption that
∀x(x∉A) into our list of givens.
But this means that A = ∅; then plugging this into our other given:
∀F(∪F=∅⇒∅∈F)
Let F' be the family set F' = ∅
then ∪F'=∅ and thus ∅∈F'
but this is a contradiction, since ∅∈F'=∅ is False.
Is this a legal move - setting F to the empty set?
If I don't set F to the empty set, then ∀F(∪F=∅⇒∅∈F) is true as far as I can tell, and there is no contradiction in assuming the negated goal.
is F is a family of empty sets, then since the empty sets contain no elements the union of the family set of empty sets is just the empty set.
However, is F IS an empty set then ∅∈F'=∅ and we've found our contradiction.
Any insight on this?