# Thread: Set Theory basics: an element of an element

1. ## Set Theory basics: an element of an element

Hi everyone,

In a proof I'm writing I inferred from, x is an element of D, and D is an element of A, that x is a subset of A.

There is something wrong here, right? What can I infer from knowing the first two statements?

Thank you!

2. ## Re: Set Theory basics: an element of an element

You musn't confuse elements and subsets.

For example if D = {1,2,3}, and A is { {1,2,3}, {2,4}} (A consists of 2 elements {1,2,3} and {2,4} and D has 3 elements 1,2,3)
2 is an element of D, and D is an element of A, but 2 is not a subset of A.

3. ## Re: Set Theory basics: an element of an element

It is not always true that $x\in D$ and $D\in A$ imply $x\subseteq A$, but for some x, D and A this may happen.

4. ## Re: Set Theory basics: an element of an element

Originally Posted by kripkey
Hi everyone,

In a proof I'm writing I inferred from, x is an element of D, and D is an element of A, that x is a subset of A.

There is something wrong here, right? What can I infer from knowing the first two statements?

Thank you!
Hi kripkey!

Suppose x={1}.
And suppose D={{1}, y}. It means that x is an element of D.
Then let A={{{1}, y}, 1}. We'd have that D is an element of A and also that x is a subset of A.

Erm... what's wrong? ;P

5. ## Re: Set Theory basics: an element of an element

Originally Posted by kripkey
In a proof I'm writing I inferred from, x is an element of D, and D is an element of A, that x is a subset of A.

Are you working with an ultra-filter on a set?

6. ## Re: Set Theory basics: an element of an element

Thanks for the responses guys!

Not sure what an ultra-filter on a set is, so probably not! Now I'm curious

I was actually trying to somehow justify that if D = arbitrary union of G, then D is a subset of G.
So I guess I got a little lost on the way.

7. ## Re: Set Theory basics: an element of an element

Originally Posted by kripkey
Thanks for the responses guys!

Not sure what an ultra-filter on a set is, so probably not! Now I'm curious

I was actually trying to somehow justify that if D = arbitrary union of G, then D is a subset of G.
So I guess I got a little lost on the way.
The definition of a subset from D is that it contains only elements that are also in D.
If you take the union of G with 1 or more other sets, the elements of G will also be in that union, so G will be a subset of that union.
If D is that union, it means that G is a subset of D and not the other way around.

8. ## Re: Set Theory basics: an element of an element

That makes sense, thanks ILikeSerena.

9. ## Re: Set Theory basics: an element of an element

Originally Posted by kripkey
I was actually trying to somehow justify that if D = arbitrary union of G, then D is a subset of G.

If $G$ is a collection of sets then $x \in \bigcup G \; \Leftrightarrow \,\left( {\exists b \in G} \right)\left[ {x \in b} \right]$.
Suppose $G=\{\{1,2,8\},\{2,8\},\{4,8\}\}$ then
$D=\bigcup G=\{1,2,4,8\}$.
Now it is perfectly clear that $D\notin G\text{ and }D\not\subset G.$
Moreover, $G\notin D\text{ and }G\not\subset D.$