Abstract Algebra, Sets Proof

[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$\in$A

Case 1: X$\in$X

X$\in$A$\cup$X$\Rightarrow$X$\in$B$\cup$X$\Rightarrow$X$\in$B

Case 2: X$\notin$X$\Rightarrow$X$\in$X\A^c$\Rightarrow$X$\in$X\B^c$\Rightarrow$X$\in$X or X$\in$B^c but X$\notin$X so X$\in$B

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$\in$B and B$\in$A? Is that what you're implying? I'm confused where B$\in$B 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 $B \subset A \cup (B \cap A^c)$.