# Basic sets question

1. Apr 10, 2008

### DavidWhitbeck

[SOLVED] Basic sets question

It's been years since I've taken analysis, and so I thought I would have a refresher by studying Abbott's Understanding Analysis.

Anyway to the point-- there is a simple exercise in the beginning that stumps me (don't laugh I'm a physicist).

First of all I am fine with, and have proven that $$(\cup_{n=1}^{N}A_n)^{c} = \cap_{n=1}^{N}A_n^c$$ using induction, but I don't see why induction can't be used to say that $$(\cup_{n=1}^{\infty}A_n)^{c} = \cap_{n=1}^{\infty}A_n^c$$?

Abbott then wants me, the reader, to prove that set equality if it's valid using another method. There was a hint given, and that was to use the fact that if $$B_1 \supset B_2 \supset \cdots$$ and each $$B_n$$ is countably infinite, their intersection $$\cap_{n=1}^{\infty}B_n$$ does not have to be.

The only natural construction of sets that I could think of that would fit with the hint would be something like $$B_m = \cup_{n=m}^{\infty}A_n$$ or perhaps $$B_m = \cap_{n=1}^{m}A_n^c$$ so that $$B_1 \supset B_2 \supset \cdots$$ is satisfied and I have an expression either way that appears in the conjecture.

It's probably a standard result, but I can't figure it out, can someone help me with this?

2. Apr 10, 2008

### tiny-tim

Hi David!

Hint: try B_n is all the real numbers between 0 and 1/n.

3. Apr 10, 2008

### DavidWhitbeck

Oh well I already have an example for when the intersection is not countably infinite, if that's what you meant.

I am supposed to use that fact to help solve the problem that I'm actually concerned with-- is the compliment of the union of a countably infinite family of sets the intersection of the compliment?

I think Abbott is trying to get at a round about way that the order of the compliment of the union is not the same as the intersection of the compliments. Thanks btw for the reply.

4. Apr 10, 2008

### lurflurf

Don't stop at countable do uncounntable too muhahaha
let ' denote complement (^c not confusing enough)
is x' a devivative a complement or alternate value i'll never tell
we desire to show
Union(A_i)'=Intersection(A_i')
Blech another homomorphism blasted things are every where
usual argument A=B iff x є A(resp B)->x є B(resp A)
supose U=union A=A_i ~=not
xє(UA)'->~xєUA->~xєA (all i)->stuff->xєIntersection(A')
do similar stuff to show
xєIntersection(A')->xє(UA)'
similarly show
Intersection(A_i)'=Union(A')

note: above valid for finite,countable, and uncountable cases

5. Apr 11, 2008

### LukeD

What does induction do? You prove it for a base case and then show that given that it is true for any integers k, k-1, k-2, ... down to your base case, that it is true for k+1. Therefore, you conclude that it's true for every integer greater than your base value.

So why doesn't this show the infinite case? (whatever that may mean, there's a lot of different infinities) Well, so let's say that we can prove our statement P(1), and this implies that we can prove P(2), which lets us prove P(3), and so on. We can keep counting as far as we want. When do we ever reach infinity?

The problem is that the term "infinity" can be ambiguous. Sometimes when someone says infinity, they mean "large, arbitrary value" (e.g. when a limit is taken to infinity), but in other cases, such as the problem that you are looking at, it actually means a set with an infinite members such as the set of integers or the set of real numbers. I.e., the set is not just arbitrarily large; it actually has too many to count with ANY finite number (some call this the distinction between potential and actual infinity)

---

David: I don't understand the "hint" because it seems to be hinting that the statement is false, but I'm pretty sure I came up with a proof. (By the way tiny-tim's counter example doesn't work. De Morgan's Law holds for every interpretation of his statement I could think of. I think tiny-tim was thinking about a counter example to a similar claim in Topology). Although lurflurf's post is a little confusing and misleading, he essentially gives a sketch of the proof.

Last edited: Apr 11, 2008
6. Apr 11, 2008

### DavidWhitbeck

Thanks Luke for explaining why induction doesn't work, and thanks lurflurf for providing a proof that doesn't rely on induction. It was so similar to how I proved the n=2 case that I could kick myself for not seeing that it can be easily generalized.

Luke your post especially made it clear to me why I was having trouble with infinite collections of sets (this isn't an isolated case) I was thinking of infinity in terms of large, arbitrary value which is was driving me in circles.