REVIEW: Set Theory - Basics

Answer explanation

The symbol "⊃" is "superset"

"Superset" means the first set contains everything in the second set, but also other elements as well. For example:

{ 1, 2, 3 } ⊃ { 1, 2 }

For the symbol "⊂" means "subset".

"Subset" means that the first set contains some, but not all the elements of the second set, and that every element in the first set is also in the second set. For example:

{ 1, 2 } ⊂ { 1, 2, 3 }

Notice that, just like greater than, ">", and less than, "<", superset and subset are opposites. So if:

S ⊃ A

is true, then

A ⊂ S

is also true.

Tags

CCSS.HSS.CP.A.1

Answer explanation

"A ⊆ S" is read:

"A is a subset of or equal to S"

For the symbol "⊂" means "subset".

"Subset" means that the first set contains some, but not all the elements of the second set, and that every element in the first set is also in the second set. For example:

{ 1, 2 } ⊂ { 1, 2, 3 }

The symbol "⊆" means "subset or equal to".

If "A ⊂ S" is true,

then "A ⊆ S" is also true.

If "A = S" is true, then "A ⊆ S" is also true, but "A ⊂ S" is NOT true.

Tags

CCSS.HSS.CP.A.1

Answer explanation

"A ∩ S" is read:

"A intersection S"

or as

"the intersection of A and S"

The result of "A ∩ S" is a set that contains the elements that are in both A AND S.

In this case, since the numbers 1, 2, and 3 are in both sets, the resulting set is { 1, 2, 3 }. We can see this set is exactly set A, so:

A ∩ S = A

Tags

CCSS.HSS.CP.A.1

Answer explanation

"A ∪ S" is read:

"A union S"

or as

"the union of A and S"

The result of "A ∪ S" is a set that contains the elements that are in either A OR S.

In this case, since the numbers 1, 2, 3, 4, 5, and 6 are in set S, the resulting set will contain all these numbers. We should then include all the numbers from set A, but since every number in set A is already in set S, there are no other numbers to add! The resulting set is:

A ∪ S = { 1, 2, 3, 4, 5, 6}

which we can see is exactly equal to set S.

Tags

CCSS.HSS.CP.A.1

• { 5 } 

Answer explanation

"A ∪ B" is read:

"A union B"

or as

"the union of A and B"

The result of "A ∪ B" is a set that contains the elements that are in either A OR S.

In this case, since the numbers 1, 2, and 3 are in set A, the resulting set will contain all these numbers:

{ 1, 2, 3 }

We should then include all the numbers from set B. Since number 2 is already included, we just need to add numbers 4, and 6, so the final result is:

{ 1, 2, 3, 4, 6 }

Tags

CCSS.HSS.CP.A.1

• { 5 } 

Answer explanation

"A ∩ B" is read:

"A intersection B"

or as

"the intersection of A and B"

The result of "A ∩ B" is a set that contains the elements that are in both A AND B.

In this case, the only number that is in both A and B is the number 2, so the final answer is the set containing the number 2:

{ 2 }.

Note that the result of an intersection, "∩", is a set, so the answer "2" (not in a set) is incorrect!

Tags

CCSS.HSS.CP.A.1

Answer explanation

"x ∈ B" is read:

"x is an element of B"

and it means the set B will contain the element x.

If the left side of the expression were a set, that means the right set contains a set. For example, if set C is equal to:

{ {1, 2}, {1, 3} }

Then set C contains two sets, the set {1, 2} and the set {1, 3}. If this is true, then the following expression is also true:

{ 1, 2 } ∈ C

However, if set A = { 1, 2, 3}, then

{ 1, 2 } ∉ A

because although set A does contain the numbers 1 and 2, it does not contain a set { 1, 2 }.

Tags

CCSS.HSS.CP.A.1

Answer explanation

If A is a set, then "| A |" is read:

"the cardinality of set A"

and it means the number of elements in set A.

Since there are 3 elements in set A, the cardinality of set A is equal to 3, or:

| A | = 3

This 3 should be a number, not a set with the number 3 in it.

Tags

CCSS.HSS.CP.A.1

• { 1, 3 } 

Answer explanation

"A - B" is read:

"A set minus B",

"A take away B",

or

"the difference between A and B"

The result of "A - B" is a set that starts with the elements of set A, but then removes any elements that are also in set B.

In this case, set A is:

{ 1, 2, 3 }

But the number 2 is also an element of set B, so it is removed, leaving the set

{ 1, 3 }