1. ## Sets

Consider the sets of values.

A = {x: - infinity < x < 3}

B = {x:3<x< infinity}

C = {x:x=-3,3}

Find = ( A intersect b ) union C.

Would this be somthing like A = numbers between -3 and negative infinity

b = numbers between 3 and positive infinity

and c = numbers between -3 and 3?

So a intersect b is any number between -3 and 3 and c union ( a intersect b ) is again between -3 and 3?

2. Originally Posted by el123
Consider the sets of values.

A = {x: - infinity < x < 3}

B = {x:3<x< infinity}

C = {x:x=-3,3}

Find = ( A intersect b ) union C.

Would this be somthing like A = numbers between -3 and negative infinity

b = numbers between 3 and positive infinity

and c = numbers between -3 and 3?

So a intersect b is any number between -3 and 3 and c union ( a intersect b ) is again between -3 and 3?
$A \cap B = \emptyset$ ... they have no elements in common

$\emptyset \cup C = C$

as you have it written, C is a set with only two elements, 3 and -3.

3. Thanks skeeter!

4. ## Re: Sets

Originally Posted by el123
Consider the sets of values.

A = {x: - infinity < x < 3}

B = {x:3<x< infinity}

C = {x:x=-3,3}

Find = ( A intersect b ) union C.

Would this be somthing like A = numbers between -3 and negative infinity

b = numbers between 3 and positive infinity

and c = numbers between -3 and 3?

So a intersect b is any number between -3 and 3 and c union ( a intersect b ) is again between -3 and 3?

Sorry this is supposed to be
a = {negative infinity < x < 3}

b = {-3 < x < infinity.}

c ={ x = -3 , 3}

So as i see it , a intersect b = {-3<x<3}

a intersect b includes all numbers between -3 and 3 but not -3 and 3 themselves.

And since c includes only the values -3 and 3 , the union on (a intersect b) union c = nothing.???

Is this correct?

5. Originally Posted by el123
Sorry this is supposed to be
a = {negative infinity < x < 3}

b = {-3 < x < infinity.}

c ={ x = -3 , 3}

So as i see it , a intersect b = {-3<x<3}

a intersect b includes all numbers between -3 and 3 but not -3 and 3 themselves.

And since c includes only the values -3 and 3 , the union on (a intersect b) union c = nothing.???

Is this correct?
Considering your revisions, then, no, you are not correct. It is true that

$A\cap B = \{x: -3 < x < 3\}$,

but now, since $C=\{-3, 3\}$

we have

$(A\cap B)\cup C = \{x: -3\leq x \leq 3\}.$

6. ok i get it.

So im right with the other values but since c contains -3, 3 and it is now a union of the values in a and b then all the values together are the values in (a intersect b) union c.

Did i just make sense?

7. Originally Posted by el123
ok i get it.

So im right with the other values but since c contains -3, 3 and it is now a union of the values in a and b then all the values together are the values in (a intersect b) union c.

Did i just make sense?
I think so...

Here's a tip: Think of intersection as "and" and union as "or." So for $A\cap B$ we want all values that are in both $A$ and in $B$. $(A\cap B)\cup C$ is then all values that are in the first set or in $C$.