# Subsets of sets

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#1
Hi,

If I have a the following sets, what will be the subsets?

S1= {{B}, B}

S2= {{B}}

I'm not sure is S1 is both and how to write it and if S2 will keep both brackets?

0
4 years ago
#2
(Original post by loryn95)
Hi,

If I have a the following sets, what will be the subsets?

S1= {{B}, B}

S2= {{B}}

I'm not sure is S1 is both and how to write it and if S2 will keep both brackets?

Lets take a slightly different case:

Suppose S={a,b}

Then the subsets are sets consisting of different combinations of elements of S.
i.e. {a} {b} {a,b} and {}.............................. ..(1)

Note that the whole set and the empty set are always subsets of our given set.

Now suppose instead of the element "a", be have a set "{c}", I.e. S={{c},b}
As far as the set S is concerned "{c}" is just an element of S. Note, "c" is NOT and element of S.
We can just replace "a" with "{c}" in (1) and we have all the subsets of our new S.

Can you now do your original question?
1
4 years ago
#3
(Original post by loryn95)
Hi,

If I have a the following sets, what will be the subsets?

S1= {{B}, B}

S2= {{B}}

I'm not sure is S1 is both and how to write it and if S2 will keep both brackets?

In set notation, elements in a set are separated by commas.

So S1 is a set containing the elements B and {B} (which is itself a set containing B).

A subset of a set A is a set containing elements that are in A.

So {B} is a subset of S1 because {B} is a set containing only the element B. And B is an element of S1.

{{B}} is also a subset of S1. This is because {{B}} is a set containing only the element {B} and {B} is an element of S1.

Does that help? So what are all the subsets of S1 and S2? Remeber that the empty set {} is a subset of any set so you'll need to include the empty set in your list of subsets.
0
#4
(Original post by notnek)
In set notation, elements in a set are separated by commas.

So S1 is a set containing the elements B and {B} (which is itself a set containing B).

A subset of a set A is a set containing elements that are in A.

So {B} is a subset of S1 because {B} is a set containing only the element B. And B is an element of S1.

{{B}} is also a subset of S1. This is because {{B}} is a set containing only the element {B} and {B} is an element of S1.

Does that help? So what are all the subsets of S1 and S2? Remeber that the empty set {} is a subset of any set so you'll need to include the empty set in your list of subsets.
Yes that's really helped thanks! So the subsets of S1 are {B} and {{B}} and S2 {{B}} I believe?
0
#5
I just thought of another quick question actually. For S1={{B}, B} what will the cardinality be if we have another set which is S3= B= {e,f,g}?

Would it be 4 or would it be 2?
0
4 years ago
#6
(Original post by loryn95)
Yes that's really helped thanks! So the subsets of S1 are {B} and {{B}} and S2 {{B}} I believe?
Those are correct subsets but you've missed a few.

A subset can have more than 1 element.

S1 contains B and {B} so you can make a subset with both of these elements:

{B, {B}}

For any set, one of the susbets of it will be the set itself.

Also, as I mentioned above, the empty set { } is a subset of any set.
0
4 years ago
#7
(Original post by loryn95)
I just thought of another quick question actually. For S1={{B}, B} what will the cardinality be if we have another set which is S3= B= {e,f,g}?

Would it be 4 or would it be 2?
Can you clarify your question - which set do you want to know the cardinality of?
0
#8
(Original post by notnek)
Can you clarify your question - which set do you want to know the cardinality of?
Yeah sorry, I meant what is the cardinality of S1?
Is it |S1| 1 or |S1|4?

We know B has 3 members due to S3 which is why I was wondering whether we take this into account if B is within another set?
0
4 years ago
#9
(Original post by loryn95)
Yeah sorry, I meant what is the cardinality of S1?
Is it |S1| 1 or |S1|4?

We know B has 3 members due to S3 which is why I was wondering whether we take this into account if B is within another set?
The cardinality of a set is how many elements are in the set.

S1 = {{B}, B}

As I explained before, this is a set containing only the elements {B} and B (since these are separated by commas).

So the cardinality is 2 because there are only 2 elements. What these elements are is irrelevant.
0
#10
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