# Measure theory

1. Dec 7, 2009

### zeebo17

I just started learning some basic measure theory.

Could someone explain the difference between $$\overline{F(A \times A)}$$ and $$\overline{F(A) \times F(A)}$$ where A is a finite set. Also, how would this be different in A was an countably infinite set?

Thanks!

2. Dec 7, 2009

### rochfor1

What is F(A)? This is not standard notation; you need to explain it before we can help.

3. Dec 7, 2009

### zeebo17

I'm not sure, the book refers to F(A) as the power set.

4. Dec 7, 2009

### rochfor1

Ok, then what's the bar over them? I figured there would be some sort of topological thing involved. Without the bars, say |A|=n. Then $$|A \times A|=n^2$$ and $$|F(A)|=2^n$$, so $$|F(A) \times F(A)|=(2^n)^2=4^n$$, but $$|F(A \times A)|=2^{ ( n^2 ) }$$, so these sets have different cardinality. If you let me know what the bar is, I can say more.

5. Dec 7, 2009

### zeebo17

The bar means the number of elements in that set.

I'm trying to understand what the difference is between $$\overline{F(A \times A)}$$ and $$\overline{F(A) \times F(A)}$$ so I can determine which has the most elements or which is "bigger'."

6. Dec 7, 2009

### rochfor1

Ah well I used |A| for the number of elements of A. Note that |F(A)|=2^|A|. Try proving this; it shouldn't be very hard.

7. Dec 7, 2009

### zeebo17

Ok, great- Thanks! I think I can get the rest from there.

The other thing I was wondering about was how to deal with that when A it is instead a countably infinite set. My book says that they would be equal in this case, but I'm not sure I see how.

8. Dec 7, 2009

### Hurkyl

Staff Emeritus
Again, it's just arithmetic of cardinal numbers. The set structure doesn't matter -- replace the sets with cardinal numbers, and the set arithmetic operations with the appropriate cardinal arithmetic operators.