Question

Math Proofs

Answer 1

Suppose F and G are families of sets.
\[(\cup F) /(\cup G)\subseteq \cup (F/G)\]

Answer 2

Is the slash set subtraction?

Answer 3

yeah, i couldnt do the backslash

Answer 4

I haven't actually worked with families of sets very much. Could you enlighten me to what \(\cup F\) means?

Answer 5

Hrmm.... I don't know how I could explain.. hold on.

Answer 6

Its pretty much the union of a bunch of sets into F. If you're thinking about \[\cup F\]

Answer 7

Well, we would need to prove that that any element in \[(\cup F) \backslash (\cup G)\]is also in\[\cup (F\;\backslash G)\]

Answer 8

Right, so here's what I have

Answer 9

Let \(a\in (\cup F)\backslash(\cup G)\). Then \(a\in \cup F\) and \(a \notin \cup G\).

Answer 10

Let \[x \in(\cup F) / (cupG)\]
so \[x \in \cup F\] and \[x \notin \cup G\]
then \[\exists S \in F\] such that \[x \in S\] Also, \[\forall T \in G\] where \[x \notin T\]

Answer 11

That's all i have..

Answer 12

That at least gives me a good idea of what \(\cup F\) means. let's see where we get from there.

Answer 13

Also, you're using S as a set correct?

Answer 14

Correct

Answer 15

This means that no set in G contains x, and at least one set in F contains x. Now look at \(\cup (F\;\backslash G)\). This is equal to the collection of sets that are in F, but not in G.

By what we just said, there must be at least one set in F that contains x, while no set in G contains x. We can now see that x must be in both \((\cup F)\backslash(\cup G)\) and \(\cup(F\;\backslash G)\).

Therefore we are done.