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Measure theory

  1. Dec 7, 2009 #1
    I just started learning some basic measure theory.

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

    Thanks!
     
  2. jcsd
  3. Dec 7, 2009 #2
    What is F(A)? This is not standard notation; you need to explain it before we can help.
     
  4. Dec 7, 2009 #3
    I'm not sure, the book refers to F(A) as the power set.
     
  5. Dec 7, 2009 #4
    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 [tex]|A \times A|=n^2[/tex] and [tex]|F(A)|=2^n[/tex], so [tex]|F(A) \times F(A)|=(2^n)^2=4^n[/tex], but [tex]|F(A \times A)|=2^{ ( n^2 ) }[/tex], so these sets have different cardinality. If you let me know what the bar is, I can say more.
     
  6. Dec 7, 2009 #5
    The bar means the number of elements in that set.


    I'm trying to understand what the difference is between [tex] \overline{F(A \times A)} [/tex] and [tex] \overline{F(A) \times F(A)} [/tex] so I can determine which has the most elements or which is "bigger'."
     
  7. Dec 7, 2009 #6
    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.
     
  8. Dec 7, 2009 #7
    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.
     
  9. Dec 7, 2009 #8

    Hurkyl

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    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.
     
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