# Homework Help: Proving Two Sets Are Equivalent

1. Aug 23, 2014

### Bashyboy

To prove that two sets are in fact the same, do I actually have to prove that the two are subsets of each other; or could I prove that they are equivalent by some other means, such as invoking the definitions of the sets?

For instance, the I am trying to show that the binary set operator $+$ is commutative; that is,

$A + B = B + A$,

where

$A + B = (A-B) \cup (B-A)$.

By using the commutative property of the $\cup$ operator, I was able to prove the equality:

$A + B = (A-B) \cup (B-A) = (B-A) \cup (A-B) = B+A$

Is this a valid proof?

2. Aug 23, 2014

### micromass

Yes, as long as you've already proven that union is commutative.

3. Aug 23, 2014

### Bashyboy

°Yes, I did prove that the union is commutative.

I have another question; it comes from the same problem. I am asked to determine whether $A + \emptyset = A$ is a true statement. Here is my proof:

$A + \emptyset = (A - \emptyset) \cup (\emptyset - A)$

Before we proceed, let us determine the nature of $(A - \emptyset)$ and $(\emptyset - A)$.

$(A - \emptyset) = \{ x~| x \in A \wedge x \notin \emptyset \}$. The statement $x \notin x \emptyset$ is always true by definition. Therefore,

$(A- \emptyset) = \{x~| x \in A \wedge T \} = \{x~| x \in A\} = A$.

Here is the portion of my proof that I am unsure of:

$\emptyset - A = \{x~| \underbrace{x \in \emptyset} \wedge x \notin A \}$. The underlined portion is always a false statement, as the empty set never houses any elements. As such,

$\emptyset - A = \{x~| F \wedge \underbrace{x \notin A}\}$. The truth value of the underlined portion is irrelevant to the truth value of the entire statement. Thus,

$\emptyset - A = \{x~| F \}$...

I would interpret this as being the empty set, but I do not have any basis for such an inference. How might I justly proceed from this last step in my proof?

EDIT:

What if I wrote $\emptyset - A = \{x~|\forall xp(x) = F\}$, where $p(x)$ is the condition that must be satisfied in order for the element $x$ to be a member. What $p(x)=F$ states is, that every $x$ makes the statement false, meaning that it can contain no elements.

Would this be sufficient reasoning?

Last edited: Aug 23, 2014
4. Aug 24, 2014

### HallsofIvy

B- A is, by definition, "All members of B that are not in A". If B is empty, B- A is the empty set no matter what A is.

5. Aug 24, 2014

### vela

Staff Emeritus
I think you're fine simply saying that $\emptyset-A = \emptyset$, but if you're still worried, you can show that $\emptyset \subset \emptyset-A$ and $\emptyset-A \subset \emptyset$.