Sets

• Nov 2nd 2008, 03:48 AM
Showcase_22
Sets
Quote:

Let A,B,C be sets.

(i)Which of the following are always true?

1). A \ ( B \ C) = ( A \ B) U C
2). A \ (B U C) = (A \ B) \ C
3). A \ (B <intersection> C)= (A \ B) U (A \ C)
I have the first one as not always true and the second and third as always true. I just wanted someone to check this because I think i'm doing them wrong.

Is there a difference between A \ B \ C and (A \ B) \ C?

Sorry I don't know how to get unions and intersections to work so I used a U for the union and wrote <intersection> for intersections.
• Nov 2nd 2008, 04:02 AM
Moo
Hello,
Quote:

Originally Posted by Showcase_22
I have the first one as not always true and the second and third as always true. I just wanted someone to check this because I think i'm doing them wrong.

Is there a difference between A \ B \ C and (A \ B) \ C?

Sorry I don't know how to get unions and intersections to work so I used a U for the union and wrote <intersection> for intersections.

You can rewrite :
$\displaystyle A \backslash B=A \cap B^c$

It's \cup for union and \cap for intersection

Quote:

A \ B \ C
I don't think this is a correct writing, unless you can prove (A\B)\C=A\(B\C) :p
• Nov 2nd 2008, 04:24 AM
Plato
$\displaystyle A\backslash \left( {B\backslash C} \right) = A \cap \left( {B \cap C^c } \right)^c = A \cap \left( {B^c \cup C} \right)$
• Nov 2nd 2008, 04:28 AM
Showcase_22
Quote:

I don't think this is a correct writing, unless you can prove (A\B)\C=A\(B\C) :p
But surely if the set is A not B not C then there would just be A regardless of where you put brackets?

Quote:

$\displaystyle A \backslash B=A \cap B^c$
Where did this come from? I'm trying to think of it as a Venn diagram but I can't see what it would look like.

Quote:

$\displaystyle A\backslash \left( {B\backslash C} \right) = A \cap \left( {B \cap C^c } \right)^c = A \cap \left( {B^c \cup C} \right)$
ohhh, so A \ B \ C is not the same as (A \ B) \ C. That seems weird! does "\" mean "not"?

My foundations lecturer introduced "¬" that apparently also means not. It seems strange that he would introduce two different symbols that apparently mean the same thing! (he also did the same for ^ and $\displaystyle \cap$ and it's counterpart with the union. Is there a difference between all this notation???)
• Nov 2nd 2008, 04:29 AM
Moo
Quote:

Originally Posted by Showcase_22
But surely if the set is A not B not C then there would just be A regardless of where you put brackets?

Why ?

Quote:

Where did this come from? I'm trying to think of it as a Venn diagram but I can't see what it would look like.
Hmm well, it's the definition oO
How do you define $\displaystyle A \backslash B$ ?
It's the set of elements that are in A but not in B, that is to say they are in A and in $\displaystyle B^c$
• Nov 2nd 2008, 04:49 AM
Showcase_22
Hang on, I think i'm missing something.

"\" means "not" doesn't it?

For example, "A \ B" means "A not B".

I'm pretty sure that's what our lecturer told us. =S
• Nov 2nd 2008, 04:55 AM
Moo
Quote:

Originally Posted by Showcase_22
Hang on, I think i'm missing something.

"\" means "not" doesn't it?

For example, "A \ B" means "A not B".

I'm pretty sure that's what our lecturer told us. =S

Hmmm "not A" usually stands for $\displaystyle A^c$, it affects only one set.
Here, \ is a binary operator, that is to say it deals with two sets.

If you prefer, it's "A and not B" and we can refer to
Quote:

elements that are in A but not in B
• Nov 2nd 2008, 05:12 AM
Showcase_22
Okay then. Going back to the first question:

A \ (B \ C)=A \ $\displaystyle (B \cap C^d) \neq$ (A \ B) $\displaystyle \cup C$

(I used d instead of c since C was already a set).

The second question would become:

Need to show: A \ $\displaystyle (B \cup C)$=(A \ B) \ C
(A \ B) \ C= $\displaystyle A \cap B^d$ \ C

Would I need another identity to show that it's true? (if it's true, I think it is). Truthfully, I would much rather remember identities and try to get one side to equal the other side. Venn diagrams are rather limited.
• Nov 2nd 2008, 05:50 AM
Plato
#3 is correct.
$\displaystyle \begin{array}{rcl} {A\backslash \left( {B \cap C} \right)} & = & {A \cap \left( {B \cap C} \right)^c } \\ {} & = & {A \cap \left( {B^c \cup C^c } \right)} \\ {} & = & {\left( {A \cap B^c } \right) \cup \left( {A \cap C^c } \right)} \\ {} & = & {\left( {A\backslash B} \right) \cup \left( {A\backslash C} \right)} \\ {} & {} & {} \\ \end{array}$

This set operation is known as setminus.
The are two notations in use: $\displaystyle \left( {A\backslash C} \right) = A-C$