Proving Sets Intersections/Unions

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Also, for your first proof, you might want to make sure you are using the right symbols. In the first part, it looks like you might be using the correct symbols, but in the second part (the converse), it looks like you are using the symbols for the converse of a statement. Just make sure you are using the correct symbols. I would also suggest using words like "Suppose" and "then" to make it easier to read.
  • #1
roam
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Homework Statement



Let A,B,C be sets.

(a) Show that:

[tex]A \cup B = (A\B)\cup(B\A)\cup(A \cap B)[/tex]

(b) Show that:

[tex]A \times (B\C) = (A \times B) \ (A \times C)[/tex]

Homework Equations





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 [tex]A \cup B = (A\B)\cup(B\A)\cup(A \cap B)[/tex] are equal each other I must show that:

[tex]A \cup B \subseteq (A\B)\cup(B\A)\cup(A \cap B)[/tex] and

[tex]A \cup B \supseteq (A\B)\cup(B\A)\cup(A \cap B)[/tex]

I'm stuck here and don't know how to prove that the two are subsets of one another. Can anyone help me please?
 
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  • #2
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 [itex] A = (A-B)\cup B[/itex] 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 [itex](A-B)\cup B[/itex] by definition of union. If x is not an element of B, then x is in A-B and hence in [itex](A-B)\cup B[/itex]. In either case, x is an element of [itex](A-B)\cup B[/itex]

Now, suppose x is in [itex](A-B)\cup B[/itex]. 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 [itex](A-B)\cup B[/itex] assuming B is a subset of A

Try using logic like that for your problem
 
  • #3
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 [tex]x \in A \cup B[/tex], then [tex]x \in A[/tex] or [tex]x \in B[/tex] or [tex]x \in A[/tex] and [tex]x \in B[/tex], thus [tex]x \in (A-B) \cup (B-A) \cup (A \cap B)[/tex] and hence:

[tex]A \cup B \subseteq (A-B) \cup (B-A) \cup (A \cap B)[/tex]

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

[tex](A-B) \cup (B-A) \cup (A \cap B) \subseteq A \cup B[/tex]

Is this proof correct?
 
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  • #4
You shouldn't just jump form "if [itex]x\in A\cup B[/itex]" to "hence". Show more detail.
If x is in [itex]A\cup B[/itex], the x is in A or in B (or both). If x is in A and NOT B then x is in [itex]A- B[/itex]. If x is B but NOT A, then [itex]B- A[/itex]. If x is in both A and B, then x is in [itex]A\cap B[/itex]. In any case x is in [itex](A-B)\cup(B-A)\cup(A\cap B)[/itex].
 
  • #5
OK here's what I've done for the second one:

[tex]A \times (B-C) = (A \times B) - (A \times C)[/tex]

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

So [tex]x \in (A \times B) - (A \times C)[/tex] [tex]\Rightarrow A \times (B-C) \subseteq (A \times B) - (A \times C)[/tex]

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

[tex]x \in A \times (B-C)[/tex]

[tex](A \times B) - (A \times C) \subseteq A \times (B-C) [/tex]

So is it correct?
 
  • #6
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
And if [tex]x \in (A \times B) - (A \times C)[/tex] then [tex]x \in (A \times B)[/tex] but [tex]x \notin A \times C[/tex] which means [tex]x \in A \times (B-C)[/tex].
 
  • #8
roam said:
OK here's what I've done for the second one:

[tex]A \times (B-C) = (A \times B) - (A \times C)[/tex]

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

So [tex]x \in (A \times B) - (A \times C)[/tex] [tex]\Rightarrow A \times (B-C) \subseteq (A \times B) - (A \times C)[/tex]

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

[tex]x \in A \times (B-C)[/tex]

[tex](A \times B) - (A \times C) \subseteq A \times (B-C) [/tex]

So is it correct?

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

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
n!kofeyn said:
Instead of letting the point x be in a cross product of two sets, I might suggest the following.
Let [itex](x,y)\in A\times (B-C)[/itex]
[itex]x\in A[/itex], and [itex]y\in B[/itex] and [itex]y\not\in C[/itex]
[itex]x\in A[/itex] and [itex]y\in B[/itex], and [itex]x\in A[/itex] and [itex]y\not\in C[/itex]
[itex](x,y)\in A\times B[/itex] and [itex](x,y)\not\in A\times C[/itex]
[itex](x,y)\in (A\times B) - (A\times C)[/itex]
Therefore, [itex] A\times(B-C)\subseteq (A\times B) - (A\times C)[/itex]

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.

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

Suppose [tex](x,y) \in (A \times B) - (A \times C)[/tex]

[tex]x \in A[/tex], and [tex]y \in B[/tex], [tex]x \notin A[/tex] [tex]y \notin C[/tex]

So, [tex]x \in A[/tex], and [tex]y \in B[/tex], & [tex]y \notin C[/tex]

[tex](x,y) \in A \times (B-C)[/tex]

Hence: [tex](A\times B) - (A\times C) \subseteq A\times(B-C)[/tex]

Does this show enough detail/steps?
 
  • #10
roam said:
So here's the converse (please correct me if I'm wrong):

Suppose [tex](x,y) \in (A \times B) - (A \times C)[/tex]

[tex]x \in A[/tex], and [tex]y \in B[/tex], [tex]x \notin A[/tex] [tex]y \notin C[/tex]

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

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

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
Thanks very much your post was very helpful.
 

What is the definition of sets intersection and union?

The intersection of two sets A and B is the set of all elements that are common to both A and B. The union of two sets A and B is the set of all elements in either A or B (or both).

How can sets intersection and union be represented?

Sets intersection can be represented using the symbol ∩ while sets union can be represented using the symbol ∪.

What is the relationship between sets intersection and union?

Sets intersection and union are complementary operations. This means that the intersection of two sets is equal to the complement of the union of the complements of the two sets.

How can sets intersection and union be proven?

In order to prove sets intersection, you need to show that the elements in the intersection set belong to both original sets. To prove sets union, you need to show that all elements in the original sets belong to the union set.

What are some practical applications of sets intersection and union?

Sets intersection and union are commonly used in mathematics and statistics, but they also have practical applications in computer science, data analysis, and decision making. For example, sets intersection can be used to find commonalities between two data sets, while sets union can be used to combine data from multiple sources.

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