# Thread: A set in a set?

1. ## A set in a set?

Hi!

I've just taken on a course in discrete mathematics, i think im doing fairly well but there is one thing i really don't understand about sets and i thought maybe someone here could help me out with it.

A= {a,b,c}
B= {a,b,g}
C= {c,d,e,f}
G= {a,b,c,d,e,f,g}

{a,b,{g}} = B true/false?

What does this last line mean? (The left side of the =)

I can't find it in my litterature either. :/

2. (Well, sets may contain other sets, that's not very upsetting - you 'll be used to it in no time)

When writing {a,b,{g}}, you have formed a set with elements a,b and {g} - careful, not just g, but the one-element set {g} (called a "singleton set").

So, writing "a belongs to {a,b,{g}}" is right, but "g belongs to {a,b,{g}}" is wrong; Actually {g} belongs to {a,b,{g}}.

Try now to show that {a,b,{g}} is not equal to B, keeping the singleton in mind!

3. Uhhm, ok. Here goes nothing.

Since {g} is not equal to g, {a,b,g} is not equal to {a,b,{g}} and therefore, {a,b,{g}} is not equal to B?

4. Equality of sets means they share the same elements,
and as you said $\{g\}=g$ is false. The sets are not equal.

Another method to use is given two sets $S,S'$, if you can show $S\subseteq S' \mbox{ and }S'\subseteq S$ then $S=S'$.
Thus, show that each set is a subset of each other. Which is not true here.

5. B= {a,b,g}

{{b}} belongs to B
{{b}} is a subset of B
{{b}} is a subset of the powerset of B

I hope i understood everything correctly.
According to "what i know" i would say that all of the above statements are false, is that correct?
If it would be {b} instead of {{b}}, they would all be true?

6. B= {a,b,g}

{{b}} belongs to B
{{b}} is a subset of B
{{b}} is a subset of the powerset of B
For this set B,

- b belongs to B (drop curly brackets once to obtain elements)

- {b} is a subset of B (enclosing elements in brackets, forms a subset)

- {{b}} is a subset of the powerset
P(B)={ {a},{b},{g},{a,b},{a,g}{b,g}{a,b,g} }

and, what is the relation of {b} tp P(B)?

7. {b} is a proper subset of P(B) as far as i know.

So {{b}} is a subset of P(B) because the element of {{b}}, {b} that is, is included in P(B) and therefore, {{b}} is a subset of P(B)?

Just checking, P(B) stands for "the powerset of B" right?

8. Hi Rebesques:

I suspect this was just an oversight (you obviously know your stuff) but,
nP(S) = 2^(nP(S)) for any set, S, where n denotes "cardinality of". Thus
nP(B) = 2^nB = 2^3 = 8. You omitted {} from the power set.

Regards,

Rich B.

9. {b} is a proper subset of P(B) as far as i know.

So {{b}} is a subset of P(B) because the element of {{b}}, {b} that is, is included in P(B) and therefore, {{b}} is a subset of P(B)?

Just checking, P(B) stands for "the powerset of B" right?
Everything is ok, except from the first sentence. it is {{b}} that is a proper (containing less!) subset of P(B).

=========================================
You omitted {} from the power set.
Yeah you are right Rich. I forget stuff all the time...
(better it be the empty set than my wallet )

10. In relation to all this, just WHY can we include {g} in a set along with the elements a and b? I realize we CAN do this, but doesn't it create some sort of logical equivalence of the "nature" (for lack of a better word) of the element a to the nature of the set {g}? To say an element and a set have some feature in common bothers me. It would seem to me that all the elements of a set would need to have the same "nature." (And that way we don't run into things like Russell's paradox...we couldn't form the set of all sets and call it a set...it would need to be called something else.)

-Dan