I can't find anything on this topic elsewhere. I'd like to know what keywords/sites I should be using to find what I'm looking for if this is to elementry of a question. (been using discrete math, set theory, proofs, symmetric difference, for a given number.)
Given sets A and B,
the symmetric difference A Δ B is defined by A Δ B = ( A – B ) ∪ (
B – A ) = ( A ∪ B ) – ( A ∩ B ).let X = {1,2,3, …271}
a) ∀A∈P(X), ∃B∈P(X), 271 ∈ AΔB. Prove it true or false.
So this is my understanding is as follows with what I think my mistake might be in brackets.
"∀A∈P(X)" For any possible A set, it will be a member of the power set of X.
( So I view A as all possible P(X), otherwise it would be written "A∈P(X)" right ? )
"∃B∈P(X)" There is a B set, that is a member of the power set of X.
( I view B as a single possible set from P(X). )
"271 ∈ A Δ B" 271 is a member of the symmetric difference between A and B.
( So, "when all possible sets of P(x) are compared to a specific set of P(X), the result would be "271 ∈ AΔB" every time." is the statement? )
I think it's false.
Because regardless of whether the B set does or doesn't have 271, there is going to be a possible A set that also does or doesn't have 271 as well.
Unless you separate A into subsets of "has 271" and "doesn't have 271", then it would be true for half, and false for the other half.
But I don't understand how to put my answer (if it's even right) into written form.
I'm doing several questions but I think if I can figure out how this is done, I can do the rest.
Best Answer
The order of the quantifiers is important. The statement as you have interpreted it would be written
$$(\exists B\in P(X))(\forall A\in P(X))(271\in A\Delta B).$$
That is, there exists a fixed set $B$ such that for every set $A$, we have $271\in A\Delta B$. In other words, first you fix $B$, then you have to check it against all $A$. But the statement as written is
$$(\forall A\in P(X))(\exists B\in P(X))(271\in A\Delta B).$$
That is, for every set $A$, there exists a set $B$ (possibly depending on $A$) such that $271\in A\Delta B$. In other words, $A$ can be chosen arbitrarily, but you don't have to pick a fixed $B$ - you just need to come up with a rule that, given an $A$, selects an appropriate $B$.
If that distinction makes sense, then hopefully you can see why the statement as written is true.