Homework Help: 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.