Prove property of classes (introductory set theory)

[demonelite123]

this problem is from Apostol's Calculus Vol.1, i just started doing proofs so I'm still getting used to it.

B - U (A) = ∩ (B-A)

the U is the union of sets in a class F and ∩ is the intersection of sets in a class F. written under both the U and the ∩ are A∈F.

so i let an element x ∈ B - U (A) which means that its in the set B but in none of the sets A1, A2,... in the class F. i think what I'm having trouble with is the right side of the equation. i know what it means when i have U (A) or ∩ (A) but I'm confused about ∩ (B-A). does that mean B - (A1∩A2∩A3∩...)? also is my logic correct so far?

 

[Landau]

but I'm confused about ∩ (B-A). does that mean B - (A1∩A2∩A3∩...)?

No, this is (somewhat) what you want to prove, not what it means. (Btw, you are assuming that F is countable by saying writing A1,A2,A3... I don't think this is given, and in any case you don't need it).
Go back to the definition! x is in the intersection [tex]\bigcap_{i\in F} (B-A_i)[/tex] iff x is in [tex]B-A_i[/tex] for all [tex]i\in F[/tex]. Hence x is in B, but x is none of the A_i's. Hence...

 

[demonelite123]

thanks for clarifying! so the left side and the right side are subsets of each other and that means they are equal.

 

[Martin Rattigan]

Btw, you are assuming that F is countable by saying writing A1,A2,A3...

That's an interesting point. Doesn't it depend on how long what the dots are intended represent goes on for? Could it not mean A1, A2, ... A[itex]\omega[/itex], A([itex]\omega+1[/itex]), ... , A[itex]c[/itex] with [itex]c[/itex] being the cardinal (first ordinal with the cardinality) of the continuum, for example?

 

[HallsofIvy]

No, it is the fact that you are assuming they can be written in a particular order with one coming right after the other that makes them "countable".

 

[Martin Rattigan]

But have you ever seen a definition of the ellipsis that states that assumption is involved? I thought the ellipsis could be used more or less as Humpy Dumpty would have used it.

In any case if you remove A[itex]c[/itex] from the end of sequence I suggested, the sequence is in an order with one coming right after any other (though not necessarily before) and there are still an uncountable number of terms.

 

[HallsofIvy]

The definitions of the ellipsis I have always seen say that it means something continues in the same way. Before the ellipsis, I see "A1, A2, A3" three things (here sets) coming one right after the other. That's "countable" no matter how long it continues.

 

[Landau]

[offtopic]

No, Martin does make a point. The ordening is really not what "makes them countable", otherwise every totally ordered set would be countable! The ellipsis is the essence.

Let L be any well-ordered set. If L is non-empty, it (as a subset of itself) has a least element, which we call 0_L. If 0_L is the only element, we stop. If not we go on: if x is not the greatest element of L, the set [tex]\{y\in L|x<y\}[/tex] is non-empty and hence has a least element which we call x+1. Therefore, if L is infinite, L contains the set
[tex]F=\{0_L,0_L+1=1_L,1_L+1=2_L,3_L,...\}[/tex].
This is a countably infinite subset. If F is not the whole of L, its complement L\F has a least element, which we call [tex]\omega_L[/tex]. We can continue this game en get [tex]\omega_l,\omega_L+1,\omega_L+2,...[/tex]. This can be the whole of L, or not, etc. This may go on forever :)

Some terminology: a succesor element (of L) is an element of the form x+1 for some x in L. In other words, an element of L is a succesor element if it is the smallest element in L strictly greater than some x in L. A limit element is an element of L which is not a succesor element. In the above, 0_L and [tex]\omega_L[/tex] are limit elements.

I guess when HallsofIvy says "they can be written in a particular order with one coming right after the other", he means there is just one limit element. 'In F, we can never reach [tex]\omega_L[/tex]' :)

Of course, you may wonder why F should be countable. It's all in the dots. I think if you want to give a precise meaning of the ellipsis, you're going to need [tex]\mathbb{N}[/tex] somewhere. (E.g. a function with domain [tex]\mathbb{N}[/tex]).

 

[Martin Rattigan]

What you see before the dots is "A" followed by 1, 2 and 3. You can take 1, 2 and 3 as the first few natural numbers or the first few ordinals or cardinals. If you understand the natural numbers to be the ordinals less than [itex]\omega[/itex], it's the same thing.

There seems to me to be nothing in the ellipsis to tell you whether it represents "A" followed by numbers less than [itex]\omega[/itex], or "A" followed by numbers (ordinal or cardinal it doesn't matter in this case) less than any other ordinal [itex]o>3[/itex] (some of which are uncountable).

In each case the series is carrying on in the same way, if you take that way to be a well order.

 

[Martin Rattigan]

What demonelite123 actually said was

A1∩A2∩A3∩...

using the binary intersection operation, ∩, between consecutive sets. This extends only to a finite sequence of sets, so I think Landau was actually correct in saying that demonelite123 had assumed a countable (countably finite) number of sets.

Regarding dots in general; HallsofIvy's suggestion that they should indicate a sequence in which each term has a sucessor and Landau's suggestions that they indicate a well order having just one term with no immediate predecessor would, taken together, constitute a reasonable reproduction of Peano's postulates and, if adopted generally, ensure that dots always referred to a countably infinite sequence.

But I would say no such assumptions are in general circulation. I think dots are generally used just as a means of waving one's hands without being physically present to do it, and best avoided if lack of ambiguity is required (though looking back through the thread I have to admit that my own posts are among the dottiest.)