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