# Couple of questions about sets

1. Mar 8, 2013

### setvectorgroup

1. The problem statement, all variables and given/known data

I am confused with sets- just wanted some clarification.

Say, I have a set A={b, {1,a},{3}, {{1,3}}, 3}

What are the elements of set A?

What are the subsets of set A?

Are the subsets also the elements of the set A?

3. The attempt at a solution

I think the elements of the set A are all the objects within it like b, {1,a},{3}, {{1,3}}, 3. Is that right?

The subsets of the set A are the elements(?) {1,a},{3}, {{1,3}}? Here's my confusion: I've seen a textbook show the element(?) {1,a} is a subset of set A like {{1,a}}⊂ A, rather than {1,a} ⊂ A. If that's the right notation, what's the difference between two different ways of parenthesizing?

Thanks.

2. Mar 8, 2013

### tiny-tim

welcome to pf!

hi setvectorgroup! welcome to pf!
no, for the reason you've referred to …
a subset is a set, so it must start and end with {}

inside the {}, you put all the elements of the subset

eg if you look carefully, you'll see that the two-element subset {b,3} is a subset of A

So is {b} so is {3} so is {3, {3}}, and so on …

3. Mar 8, 2013

### setvectorgroup

Thank You, tiny-tim.

So, if {3, {3}} is a subset...

Say, I make up a set S={1,2,3...}. Can I scoop up any collection of numbers from the set S and call it a subset? For example: T={1, 1000000000, 8} ⊂ S. Would that be right?

edit: Also, if I were to modify the set A={b, {1,a},{3}, {{1,3}}, 3} to make {1,a} ⊂ A true would I modify the set A like A={1, {1,a},{3}, {{1,3}}, a} ?

Last edited: Mar 8, 2013
4. Mar 8, 2013

### tiny-tim

yes, exactly
yes, for {1,a} ⊂ A to be true, both 1 and a must be elements of A

({1,a} ⊂ A is the same as saying that 1 ε A and a ε A)

5. Mar 8, 2013

6. Mar 8, 2013

### Joffan

$$\{1,a\} \in A$$
The set {1,a} is a member (element) of A, and

$$\{\{1,a\}\} \subset A$$
The set containing only the set {1,a} is a subset of A

7. Mar 9, 2013

### setvectorgroup

Thanks, Joffan, for the additional insight.

What I wrote below might sound redundant, but I am trying to internalize all this set business:

Is the notation above just a statement that two random sets happen to share the same elements while the set A also contains more of other elements which makes the lefthand side of this relation to be the subset?

Thanks.

edit:

Am I right( at least conceptually) to visualize all this like :

{1,a} ∈ A

also

{1,a} ∈ {{1,a}}

while ( if you count the elements of each set)

{{1,a}} < A

which is

{{1,a}} ⊂ A

???

Last edited: Mar 9, 2013
8. Mar 9, 2013

### vela

Staff Emeritus
A subset of A doesn't have to have fewer elements than A. The set A is a subset of itself.

What's called a proper subset of A does have to contain fewer elements than A.

9. Mar 9, 2013

### setvectorgroup

Thank You, vela.

Here's yet another attempt to differentiate between 'element of' and 'subset of' (also, by subset I'll mean proper subset)

Say, I got the following relations below and asked to see which ones are true, which ones- false:

2 ∈ {1,2,3}

{2} ∈ {1,2,3}

{2} ∈ {{1}, {2}}

Anything to the left of ∈ is an object. All I care about is if that object can be found in the set on the right side of ∈ in exactly the same form it's on the left side i. e.

2 ∈ {1,2,3} is true because the object- number 2- can be found in the set {1,2,3}

{2} ∈ {1,2,3} is false because there's no object- a set containing number 2- in the set {1,2,3}

{2} ∈ {{1}, {2}} is true because an object- a set containing number 2- is also in {{1}, {2}}

Also, say, I have these below and need to know if any of them are true:

2 ⊆ {1,2,3}

{2} ⊆ {1,2,3}

{2} ⊆ {{1}}, {2}}

The statements above, sort of, implicitly imply (to me) comparison between sets. Here we are concerned if all the elements in the subset are also in the superset i. e.

2 ⊆ {1,2,3} is false because the number two is not even a set to begin with.

{2} ⊆ {1,2,3} is true because the number 2 in the set {2} also happens in the set {1,2,3}

{2} ⊆ {{1}}, {2}} is false because the set {{1}}, {2}} does not contain the number 2

Am I looking at this right?

I am just paranoid something might slip by me unnoticed hence this (redundant) dissection.

10. Mar 9, 2013

### ArcanaNoir

You've got it :)

11. Mar 9, 2013

### setvectorgroup

Well, I guess this does it.

Thank You, ArcanaNoir, for (hopefully) closing this thread.