Prove if and only if for each , .
If then for every , (1)
If then for every , (2)
So for every for each , (3)
R. H. S.
For each for each , (4)
Are (3) & (4) equivalent (they look like theyare to me but the indices are throwing me)?
That's correct. Let's call these statements (1) and (2).
This is obviously false and in no way follows from the previous statements. If either the right- or the left-hand side of the original claim is true, then for all . However, x can be arbitrary; it can belong to B or not.
Combine (1) and (2) and write without using or . Then do the same with the right-hand side, i.e., "for each , ." Compare the results.
Prove if and only if for each , .
If then for every , (1)
If then for every , (2)
So for every for each , (3)
R. H. S.
For each for each , (4)
Are (3) & (4) equivalent (they look like theyare to me but the indices are throwing me)?
i, too, am confused by the indices. i think it should read:
if and only for each
or:
if and only for each
which i find preferable.
in plain english, what is being said is:
"B is contained in the mutual intersection of all the A's, if and only if B is contained in each A".
now one direction of this equivalence is obvious:
if B is in the intersection of all of the A's, B is obviously contained by each and every single A, since the intersection of a family of sets is a subset of each set (this follows directly from the definition of intersection).
the other direction is a little more subtle. basically, you are generalizing to an infinite family the statement:
[(B ⊂ A_{1}) & (B ⊂ A_{2})] → (B ⊂ (A_{1}∩A_{2}))
the "&" is most important, here.
Thanks yes I transcribed the original question wrong and have propagated the error for the last 2 days (I would say that'll teach me but it probably wont )
OK so lets see if I'm getting the hang of this,
Let be an indexed family of subsets of a set S and let
Prove that iff for each ,
iff for every , (1)
iff for a least one , (2)
So iff for each , (as we dont know which x is in)?
Doesnt look right to me, is it?
(1) and (2) are correct.
This is what you need to prove, so it is right. However, you don't get it immediately. Statements (1) and (2) are combined into
For every x, if there exists a such that , then .
If we write for "for all" and for "there exists," then this is
(3)
What you need to prove is (after expanding )
. (4)
It is important that in (3), the whole is in the premise of the implication. Now, one of logical laws says that is equivalent to , where P(x) is any property of x and Q is any proposition that does not depend on x. (Make sure it is obvious to you,) Apply it to (3) and then swap the consecutive to get (4).
i don't think that reducing the statement to the equivalence of logical forms involving universal and existential quantifiers is particularly enlightening, here (although, admittedly, that is *one* way to proceed).
suppose that , and choose an arbitrary .
now pick any . since .
since . this shows that for any (thus all) .
in the other direction, we suppose that for all .
now we pick an (arbitrary) .
by the *definition* of union, for some .
thus , since for any , which shows that .
OK I'm really struggling here...
1. and are 2 indexed families of subsets in S.
a) Prove for each ,
My attempt...
Let for any
If then for some (at least one) ,
............lost it at this point.......
2. For each ,
Prove
Attempt one...
If then for all , (1)
If then for some , and (2)
............lost it at this point.......
Attempt two...
Let and then
As then so
....err yeah lost it at this point......
I'm not having big issues with the other proofs I'm trying (besides have to think about them for a long time) but for some reason I cant get my head around these changes of indicies.
It is not recommended to ask more than two questions in one thread. Otherwise, threads become difficult to navigate. You should create a new thread and copy your attempts there.
What you must mean is, "To prove it is sufficient to show that for some (at least one) , ." So, is it really difficult to show this when you assumed that for some ?
It's useless to say "If ." Suppose the claim that starts with this "If" is true. How are you going to use it? To use "If A, then B," one needs to provide a proof of A; then he/she will obtain a proof of B. But you can't provide a proof of the premise because the premise in your claim is exactly what needs to be ultimately proven!
This reminds me the following joke.Here, the goal is to get the can opened. What's the use to assume it? One needs to achieve/produce it!As an experiment, an engineer, a physicist and a mathematician are placed in separate rooms and left with a can of food but no can-opener. A day later, the rooms are opened one by one.
In the first room, the engineer is snoring, with a battered, opened and emptied can. When asked, he explains that when he got hungry, he beat the can to its failure point.
In the second room, the physicist is seen mouthing equations, with a can popped open beside him. When asked, he explains that when he got hungry, he examined the stress points of the can, applied pressure, and 'pop'!
In the third room, the mathematician is found sweating, and mumbling to himself, 'Assume the can is open, assume the can is open...'
For another example, a proven implication "If A, then B" is similar to a store that sells B's for the price A. If you have an A in your pocket, you can come to the store, pay A and get a B. This is different from the situation where you need to prove "If A, then B." This situation is similar to a workshop that produces B's from A's. If somebody gives you an A, you need to know how to manufacture a B from it.
I recommend that you review how to prove claims of the form "For all x, P(x)," "There exists an x such that P(x)," "If A, then B," "A and B," "A or B" and "not A." Also review how to use such claims when they are already proven.