How Does Subset Proof in Abstract Algebra Work?

[ktheo]

Homework Statement

Question 1. Let U be a universal set, A and B two subsets of U.
(1) Show that
B ⊆ A ∪ (B ∩ A^c).
(2) A = B if and only if there exists a subset X of U such that A ∪ X = B ∪ X and X\A^c = X\B^c.

The Attempt at a Solution

My attempt at a solution is as follows:

Part 1: Showing B ⊆ A ∪ (B ∩ A^c)

(A∪B)∩(A∪A^c)
(A∪B)∩(\bigcup)

Since A∪B are both in universe, it serves that B ⊆ A ∪ (B ∩ A^c).

Part 2
A = B if and only if there exists a subset X of U such that A ∪ X = B ∪ X and X\A^c = X\B^c.

So I claimed double inclusion proof here, letting X\inA

Case 1: X\inX

X\inA\cupX\RightarrowX\inB\cupX\RightarrowX\inB

Case 2: X\notinX\RightarrowX\inX\A^c\RightarrowX\inX\B^c\RightarrowX\inX or X\inB^c but X\notinX so X\inB

So I think the problem with this question is that I am not fully understanding the concept of the property known as the difference.

 

[jbunniii]

My attempt at a solution is as follows:

Part 1: Showing B ⊆ A ∪ (B ∩ A^c)

(A∪B)∩(A∪A^c)
(A∪B)∩(\bigcup)

Since A∪B are both in universe, it serves that B ⊆ A ∪ (B ∩ A^c).

I don't understand your proof. May I suggest something along these lines: if b \in B, and b \not\in A, then b \in B \cap A^c. Therefore...

So I think the problem with this question is that I am not fully understanding the concept of the property known as the difference.

You may find it helpful to use the following identity: A \setminus B = A \cap B^c for any sets A, B \subset U.

 

[ktheo]

I don't understand your proof. May I suggest something along these lines: if b \in B, and b \not\in A, then b \in B \cap A^c. Therefore...
Therefore B\inB and B\inA? Is that what you're implying? I'm confused where B\inB comes into play. I'm not sure I know how I'm supposed to approach this then... I thought I was supposed to manipulate the side B into the right side using the property for AUA^c equal to the universe. Or am I supposed to do some sort of proof using inclusion... I kind of use A=B as the signal that I am supposed to do that. I'm sorry bear with me I am quite new to all this...

You may find it helpful to use the following identity: A \setminus B = A \cap B^c for any sets A, B \subset U.

When A and B in this case are already compliments, that makes no difference? I can still switch using that identity?

 

[jbunniii]

When A and B in this case are already compliments, that makes no difference? I can still switch using that identity?

Yes, the identity is valid in all cases. If you have something like X \setminus A^c then that equals X \cap(A^c)^c = X \cap A.

 

[ktheo]

Hi jbunni, could you check out my second attempt at question part 1:

So B⊆A∪(B∩A^c)

To show this, we can say that when x∈B, implies there exists an x∈A∪(B∩A^c)

We will let x∈B,

Case 1: x∈ A

x∈ A--->x∈ A or x∈(B∩A^c). So clearly, x∈A.

Case 2: x∉A

x∈A or x∈(B∩A^c). We have declared x∉A, so x∈B and x∈A^c. Now we note that X∈A^c is = to X∉A.

Thus proving that B⊆A∪(B∩A^c)

 

[jbunniii]

Hi jbunni, could you check out my second attempt at question part 1:

So B⊆A∪(B∩A^c)

To show this, we can say that when x∈B, implies there exists an x∈A∪(B∩A^c)

We will let x∈B,

Case 1: x∈ A

x∈ A--->x∈ A or x∈(B∩A^c). So clearly, x∈A.

Case 2: x∉A

x∈A or x∈(B∩A^c). We have declared x∉A, so x∈B and x∈A^c. Now we note that X∈A^c is = to X∉A.

Thus proving that B⊆A∪(B∩A^c)

I think you have the right idea, but I would word it somewhat differently. See if this sounds cleaner to you:

Let x \in B. We consider two cases:

Case 1: x \in A. In this case x \in A \cup (B \cap A^c) because A \subset A \cup (B \cap A^c).

Case 2: x \not\in A. Then x \in (B \cap A^c). Therefore x \in A \cup (B \cap A^c), because (B \cap A^c) \subset A \cup (B \cap A^c)

In both cases, we have established that x \in B implies x \in A \cup (B \cap A^c). This shows that B \subset A \cup (B \cap A^c).