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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.

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  • $\begingroup$ Your reasoning seems sound to me. Essentially $A \Delta B$ consists of all elements that are in exactly one of $A$ and $B$ (like XOR). So it contains $271$ if and only if either $A$ or $B$ contains it, but not both. Write your proof down in terms of $A \cup B$ and $A \cap B$: When do those sets contain $271$, and when do they not? $\endgroup$ – Brian Tung Jul 30 '15 at 1:14
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    $\begingroup$ This question is very well-asked: You've shown us a lot of your thought processes, which helps people give you appropriate answers. Kudos! $\endgroup$ – pjs36 Jul 30 '15 at 1:18
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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.

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This is partly a question of quantifiers: You are trying to prove "For all A, there exists a B such that ...".

From the definition of $A \Delta B$, you should be able to prove that

$B = (A \Delta B) \Delta A$

So for any $A$, you are free to choose an $A \Delta B$ that contains $271$ (or any element, or elements, you choose), and the appropriate $B$ is guaranteed to exist.

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If 271 belongs to A ,let B be the empty set.If 271 does not belong to A, let B=X.

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