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!

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- Feb 14th 2013, 02:20 PMkripkeySet 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! - Feb 14th 2013, 02:38 PMjakncokeRe: 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. - Feb 14th 2013, 02:41 PMemakarovRe: Set Theory basics: an element of an element
It is not always true that $\displaystyle x\in D$ and $\displaystyle D\in A$ imply $\displaystyle x\subseteq A$, but for some x, D and A this may happen.

- Feb 14th 2013, 02:42 PMILikeSerenaRe: Set Theory basics: an element of an element
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 - Feb 14th 2013, 02:57 PMPlatoRe: Set Theory basics: an element of an element
- Feb 14th 2013, 03:16 PMkripkeyRe: 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. - Feb 14th 2013, 03:56 PMILikeSerenaRe: Set Theory basics: an element of an element
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. - Feb 14th 2013, 05:24 PMkripkeyRe: Set Theory basics: an element of an element
That makes sense, thanks ILikeSerena.

- Feb 14th 2013, 08:33 PMPlatoRe: Set Theory basics: an element of an element

It seems as if you are asking about generalized unions.

This is an example of the use of the*Union Axiom*.

If $\displaystyle G$ is a collection of sets then $\displaystyle x \in \bigcup G \; \Leftrightarrow \,\left( {\exists b \in G} \right)\left[ {x \in b} \right]$.

Here is an example given by Herbert Enderton.

Suppose $\displaystyle G=\{\{1,2,8\},\{2,8\},\{4,8\}\}$ then

$\displaystyle D=\bigcup G=\{1,2,4,8\}$.

Now it is perfectly clear that $\displaystyle D\notin G\text{ and }D\not\subset G.$

Moreover, $\displaystyle G\notin D\text{ and }G\not\subset D.$