# Prove property of classes (introductory set theory)

1. Apr 12, 2010

### 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?

2. Apr 13, 2010

### Landau

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 $$\bigcap_{i\in F} (B-A_i)$$ iff x is in $$B-A_i$$ for all $$i\in F$$. Hence x is in B, but x is none of the A_i's. Hence....

3. Apr 13, 2010

### demonelite123

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

4. Apr 14, 2010

### Martin Rattigan

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$\omega$, A($\omega+1$), ... , A$c$ with $c$ being the cardinal (first ordinal with the cardinality) of the continuum, for example?

5. Apr 15, 2010

### 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".

6. Apr 15, 2010

### 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$c$ 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.

7. Apr 15, 2010

### 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.

8. Apr 15, 2010

### 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 $$\{y\in L|x<y\}$$ is non-empty and hence has a least element which we call x+1. Therefore, if L is infinite, L contains the set
$$F=\{0_L,0_L+1=1_L,1_L+1=2_L,3_L,...\}$$.
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 $$\omega_L$$. We can continue this game en get $$\omega_l,\omega_L+1,\omega_L+2,...$$. 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 $$\omega_L$$ 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 $$\omega_L$$' :)

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 $$\mathbb{N}$$ somewhere. (E.g. a function with domain $$\mathbb{N}$$).

Last edited: Apr 15, 2010
9. Apr 15, 2010

### 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 $\omega$, 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 $\omega$, or "A" followed by numbers (ordinal or cardinal it doesn't matter in this case) less than any other ordinal $o>3$ (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.

Last edited: Apr 15, 2010
10. Apr 16, 2010

### Martin Rattigan

What demonelite123 actually said was
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.)