# Proving Sets Intersections/Unions

1. Jul 23, 2009

### roam

1. The problem statement, all variables and given/known data

Let A,B,C be sets.

(a) Show that:

$$A \cup B = (A\B)\cup(B\A)\cup(A \cap B)$$

(b) Show that:

$$A \times (B\C) = (A \times B) \ (A \times C)$$

2. Relevant equations

3. The attempt at a solution

For part (a) I need to prove the definition of a union. I think in order to prove that both sides of $$A \cup B = (A\B)\cup(B\A)\cup(A \cap B)$$ are equal each other I must show that:

$$A \cup B \subseteq (A\B)\cup(B\A)\cup(A \cap B)$$ and

$$A \cup B \supseteq (A\B)\cup(B\A)\cup(A \cap B)$$

I'm stuck here and don't know how to prove that the two are subsets of one another. Can anyone help me please?

2. Jul 23, 2009

### Office_Shredder

Staff Emeritus
I checked your latex and it looks like some of it didn't come out (you had some A\B's that only came out as A's that probably were intended to be set subtraction).

The basic idea is to show that if x is an element of one side, x is an element of the other side. For example, if the question is show $A = (A-B)\cup B$ if B is a subset of A, then I would do:

Suppose x is contained in A. Then either x is an element of B, or x is not. If x is an element of B, x is contained in $(A-B)\cup B$ by definition of union. If x is not an element of B, then x is in A-B and hence in $(A-B)\cup B$. In either case, x is an element of $(A-B)\cup B$

Now, suppose x is in $(A-B)\cup B$. Then either x is in A-B or x is in B. If x is in A-B, then x is in A necessarily. If x is contained in B, then as B is a subset of A, x is contained in A. In either case, x is an element of A

Hence x is contained in A if and only if x is contained in $(A-B)\cup B$ assuming B is a subset of A

Try using logic like that for your problem

3. Jul 23, 2009

### roam

I'm sorry, I don't know why the latex code didn't work, I will use the "-" sign instead of the "\" in my posts. Anyway here's the question:

http://img198.imageshack.us/img198/6299/74721788.gif [Broken]

So for the first one:

Suppose $$x \in A \cup B$$, then $$x \in A$$ or $$x \in B$$ or $$x \in A$$ and $$x \in B$$, thus $$x \in (A-B) \cup (B-A) \cup (A \cap B)$$ and hence:

$$A \cup B \subseteq (A-B) \cup (B-A) \cup (A \cap B)$$

Conversely, suppose $$x \in (A-B) \cup (B-A) \cup (A \cap B)$$, then $$x \in A$$ or $$x \in B$$ or $$x \in A$$ and $$x \in B$$, so by the definition of union $$x \in A\cup B$$.

$$(A-B) \cup (B-A) \cup (A \cap B) \subseteq A \cup B$$

Is this proof correct?

Last edited by a moderator: May 4, 2017
4. Jul 23, 2009

### HallsofIvy

Staff Emeritus
You shouldn't just jump form "if $x\in A\cup B$" to "hence". Show more detail.
If x is in $A\cup B$, the x is in A or in B (or both). If x is in A and NOT B then x is in $A- B$. If x is B but NOT A, then $B- A$. If x is in both A and B, then x is in $A\cap B$. In any case x is in $(A-B)\cup(B-A)\cup(A\cap B)$.

5. Jul 26, 2009

### roam

OK here's what I've done for the second one:

$$A \times (B-C) = (A \times B) - (A \times C)$$

Suppose $$x \in A \times (B-C)$$, it means that x belongs to $$A \times B$$ but not $$A \times C$$ (is this the explanation that is required?)

So $$x \in (A \times B) - (A \times C)$$ $$\Rightarrow A \times (B-C) \subseteq (A \times B) - (A \times C)$$

Now suppose that x is in $$(A \times B) - (A \times C)$$, that means $$x \in (A \times B)$$, $$x \notin (A \times C)$$ (is this sufficient?) therefore:

$$x \in A \times (B-C)$$

$$(A \times B) - (A \times C) \subseteq A \times (B-C)$$

So is it correct?

6. Jul 26, 2009

### Office_Shredder

Staff Emeritus
I would demonstrate that if x is in Ax(B-C) then x is in AxB but not in AxC, since that basically is what the question is asking

7. Jul 26, 2009

### roam

And if $$x \in (A \times B) - (A \times C)$$ then $$x \in (A \times B)$$ but $$x \notin A \times C$$ which means $$x \in A \times (B-C)$$.

8. Jul 26, 2009

### n!kofeyn

Instead of letting the point x be in a cross product of two sets, I might suggest the following.
Let $(x,y)\in A\times (B-C)$
$x\in A$, and $y\in B$ and $y\not\in C$
$x\in A$ and $y\in B$, and $x\in A$ and $y\not\in C$
$(x,y)\in A\times B$ and $(x,y)\not\in A\times C$
$(x,y)\in (A\times B) - (A\times C)$
Therefore, $A\times(B-C)\subseteq (A\times B) - (A\times C)$

Here I'm letting (x,y) be in the cross product, as a point in a cross product of two sets is an ordered pair of two points. For me, that clears things up a bit, as you are directly using the definition of each set function: the cross product and set minus. Now can you do the same for the other direction? For these set theory proofs, I think it is good to overkill them by including every single step so as to not make any leaps, which can often lead to mistakes. Although, it looks like you are doing okay, just include more steps.

9. Jul 26, 2009

### roam

So here's the converse (please correct me if I'm wrong):

Suppose $$(x,y) \in (A \times B) - (A \times C)$$

$$x \in A$$, and $$y \in B$$, $$x \notin A$$ $$y \notin C$$

So, $$x \in A$$, and $$y \in B$$, & $$y \notin C$$

$$(x,y) \in A \times (B-C)$$

Hence: $$(A\times B) - (A\times C) \subseteq A\times(B-C)$$

Does this show enough detail/steps?

10. Jul 26, 2009

### n!kofeyn

I would be careful right here. Recall that the definition of a Cartesian product is $A\times C = \{(a,c) \mid a\in A \text{ and } c\in C\}.$
For a point $(a,c)\not\in A\times C$, then $a\not\in A$ or $c\not\in C$.

So if $(x,y)\in(A\times B)-(A\times C)$,
then $(x,y)\in A\times B$ and $(x,y)\not\in A\times C$.
Then $x\in A$ and $y\in B$, and $x\not\in A$ or $y\not\in C$.
But we know that x is in A from the first part, so y must not be in C.
So then $x\in A$ and $y\in B$, and $x\in A$ and $y\not\in C$.
Then $x\in A$, and $y\in B$ and $y\not\in C$.
Then $x\in A$, and $y\in B-C$.
Thus $(x,y)\in A \times (B-C)$.
Therefore, $(A\times B)-(A\times C)\subseteq A\times (B-C)$.

By the way, I said cross product in my previous post, but I meant Cartesian product. Make sure you go through all my steps in both posts so that you understand. For these proofs, you have to very explicit with your "ands" and "ors", as well as with where your points are at. Hope this helps.

11. Jul 27, 2009