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Showing a set is null

  1. Mar 22, 2012 #1
    1. The problem statement, all variables and given/known data

    I'm trying to get to grips with how to show a set is null.

    For example, how would I show that the set [tex]\{ (x,y) : x+y=0 \} \subseteq \mathbb{R}^2[/tex] is null using the definition below?

    2. Relevant equations

    [itex]A \subseteq \mathbb{R}^k[/itex] is null if, given any [itex]\varepsilon >0[/itex] there exists a countable collection of intervals of [itex]\mathbb{R}^k[/itex], [itex](I_n)_{n\in\mathbb{N}}[/itex] such that [itex]A \subseteq \bigcup_{n\in\mathbb{N}} I_n[/itex] and [itex]\sum_{n\in\mathbb{N}} |I_n| \leq \varepsilon[/itex].

    3. The attempt at a solution

    How do I start?
     
    Last edited: Mar 22, 2012
  2. jcsd
  3. Mar 22, 2012 #2

    Dick

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    Start by drawing a picture of your set. It's an infinite line in R^2. The I_n will be rectangles in R^2. Now do a warm-up exercise. Take a finite piece of your line. Say 0<=x<=1 and -1<=y<=0 with x+y=0. Can you figure out how to cover that with a finite set of rectangles whose area sums to less than ε?
     
  4. Mar 22, 2012 #3
    I can do it with a picture but how do I transform this into maths? Since A is just a line, I can draw a picture showing that you can 'cover' the line with arbitrarily small squares with the top left and bottom right corners touching.
     
  5. Mar 22, 2012 #4

    Dick

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    You start formalizing it by saying how small the squares have to be and how many of them there have to be to have a total area less than ε. Pick ε=1/n for n an integer.
     
    Last edited: Mar 22, 2012
  6. Mar 22, 2012 #5
    What do you mean 'Pick ε=1/n for n an integer'? Isn't ε arbitrary?
     
  7. Mar 22, 2012 #6

    Dick

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    You can find an integer such that 1/n<ε for any ε. I was just thinking it might be easier to think about integer numbers for the line segment.
     
  8. Mar 22, 2012 #7
    If the [itex]I_n[/itex] have side length [itex]\frac{\delta}{n}[/itex] then the sum of all their areas will be [itex]\frac{\delta \pi^2}{6}[/itex]. So for any epsilon if I can choose an appropriate delta. In terms of a picture the squares are getting smaller the further they are from 0.
     
  9. Mar 22, 2012 #8

    Dick

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    You've left some parts of the argument out. That looks like something you found online. But if you understand it, that's fine with me.
     
  10. Mar 23, 2012 #9
    How could I present it similar to this:

    21o8pwo.jpg
     
  11. Mar 23, 2012 #10

    Dick

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    I was trying to give you hints to do that. You want to describe how to cover the line segment (t,-t) where 0<=t<=1 with squares in such a way that the sum of the areas of the squares is less than 1/N for an integer N.
     
  12. Mar 23, 2012 #11
    Why does [itex]\{(x,y) : x+y=0\} = \{ (t,-t) : 0 \leq t \leq 1 \}[/itex] ?
     
  13. Mar 23, 2012 #12

    Dick

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    They aren't equal. The first set is an infinite line, the second set is a finite segment of that line. Once you show the segment is null, it's pretty easy to show the whole line is null.
     
  14. Mar 23, 2012 #13
    Well can't I just put [itex]x=1[/itex] and [itex]y=-1[/itex] into the above and then write [tex]\{(t,-t) : t\in \mathbb{R} \} = \bigcup_{n\in\mathbb{N}} A_n[/tex] where [itex]A_n = \{(t,-t) : t\in [-n,n] \}[/itex].

    Each set [itex]A_n[/itex] is null so [itex]\{(t,-t) : t\in \mathbb{R} \}[/itex] is null, being a countable union of null sets.
     
  15. Mar 23, 2012 #14

    Dick

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    Well, yes. That's pretty much how it goes.
     
  16. Mar 23, 2012 #15
    But if I prove that [itex]\{ (-x,x) : 0 \leq x \leq n \}[/itex] is null then the set [itex]\{ (-x,x) : x \geq 0 \}[/itex] can be written as [itex]\bigcup_{n\in\mathbb{N}} \{ (-x,x) : x \in [0,n] \}[/itex] and each set [itex]\{ (-x,x) : x \in [0,n] \}[/itex] we've proved to be null but what about the other half of the line: I need the set [itex]\{ (-x,x) : x \in \mathbb{R} \}[/itex] to be null.

    Or can I just say by symmetry [itex]\{ (-x,x) : -n \leq x \leq 0 \}[/itex] is also null?
     
    Last edited: Mar 23, 2012
  17. Mar 23, 2012 #16

    Dick

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    The proof for -n<=x<=0 would go pretty much exactly the same way as for 0<=x<=n. If you don't want to write it out then saying 'by symmetry' should be fine. It is pretty obvious.
     
  18. Mar 25, 2012 #17
    In the argument I posted to show that the set [itex]\{(tx,ty) : 0\leq t \leq n \}[/itex] is null: 21o8pwo.jpg

    why is the measure of each interval [tex]I_n = \left [\frac{(j-1)nx}{N} , \frac{jnx}{N} \right] \times \left [\frac{(j-1)ny}{N} , \frac{jny}{N} \right][/tex] given by [tex]|I_n| = \frac{x^2y^2}{N^2} \;?[/tex]
    The measure of intervals in [itex]\mathbb{R}^2[/itex] is the product of the measure of the 2 intervals in [itex]\mathbb{R}[/itex] so [itex]|I| = |I_1|\cdot |I_2|[/itex].

    The measure of an interval [itex]I=[a,b][/itex] is [itex]|I|=b-a[/itex] so the measure of [tex]I_1= \left [\frac{(j-1)nx}{N} , \frac{jnx}{N} \right][/tex] is [tex]|I_1| = \frac{jnx}{N} - \left(\frac{jnx}{N} -\frac{nx}{N} \right) = \frac{nx}{N}[/tex] and the measure of [tex]I_2= \left [\frac{(j-1)ny}{N} , \frac{jny}{N} \right][/tex] is [tex]|I_2| = \frac{jny}{N} - \left(\frac{jny}{N} -\frac{ny}{N} \right) = \frac{ny}{N}[/tex] so the measure of each [itex]I_n[/itex] is [tex]\frac{nx}{N}\frac{ny}{N} = \frac{n^2xy}{N^2}[/tex] but they have it as [tex]\frac{x^2y^2}{N^2}[/tex]
     
    Last edited: Mar 25, 2012
  19. Mar 25, 2012 #18

    Dick

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    They were just being sloppy. Yes, it should be n^2*xy/N^2. It doesn't change the conclusion.
     
  20. Mar 25, 2012 #19
    Trouble is when I see something like that it makes me think I'm doing something wrong.

    Is there a way of showing the set [itex]\{ (t,-t) : t\in\mathbb{R} \}[/itex] is a null set directly from the definition (i.e. without using the result that a countable union of null sets is itself null)? In some questions I've seen it states the bit in italics which makes me think they don't want me to use any theorems for those questions.
     
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