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Subsets in Real Analysis

  1. Aug 28, 2010 #1
    1. The problem statement, all variables and given/known data

    The problem is attached. Please help me out in understanding this problem. This is not a HW question, just for my own understanding....

    2. Relevant equations



    3. The attempt at a solution
     

    Attached Files:

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  2. jcsd
  3. Aug 28, 2010 #2
    So, first, I would like to show that E intersects with F in 0.
    Since E= -1>=x>=0 and F E= 0>=x>=1, these two intervals overlap only in 0.
     
  4. Aug 28, 2010 #3
    What is f(E) in this question?
     
  5. Aug 28, 2010 #4

    lanedance

    User Avatar
    Homework Helper

    do you have a question about the problem? Even if its not for homeowrk, you should attempt it & get lead through
     
  6. Aug 28, 2010 #5
    Here was I tried to solve. I found that f(E) and f(F) are the same. I don't get that f(E overlap F)=0.
     

    Attached Files:

  7. Aug 28, 2010 #6
    Well you showed that [tex] E \cap F = \{ 0 \}. [/tex] So then we simply have

    [tex] f(E \cap F) = f(\{ 0 \}) = f(0) = 0. [/tex]

    I don't get where you're getting lost. Anything in specific?
    You also already showed that f(E) = f(F) = {y : 0 <= y <= 1}, so that parts good.

    All that's left is for you to answer the following: "What would happen if 0 is deleted from the sets E and F?"
    What would E intersect F be? What would f(E intersect F) be? Would f(E) still equal f(F)? And probably most importantly: would we still have [tex] f(E \cap F) \subset ( f(E) \cap f(F) ) [/tex] ?
     
  8. Aug 28, 2010 #7
    Answering your questions,
    (E intersect F)=N/A,
    f(E intersect F)=N/A,
    f(E) will still be equal f(F),
    And the last question is tricky for me.
     
  9. Aug 28, 2010 #8
    you already know [tex]
    f(E \cap F)=\left\{\right\},\emptyset, empty set
    [/tex] and [tex]( f(E) \cap f(F) ) =\left\{y\inR:0<y\leq1\right\}
    [/tex]

    so the question are [tex]
    \emptyset \subset \left\{y\inR:0<y\leq1\right\}
    [/tex]??
     
    Last edited: Aug 28, 2010
  10. Aug 28, 2010 #9
    No, there is no empty set in 0=<y=<1.....
    Thank you...
     
    Last edited: Aug 28, 2010
  11. Aug 28, 2010 #10
    i don't know this is definition or theorem, because i didn't take rigorous set theory yet.

    Definition.
    Empty set is the set having no element, and it is a subset of every set

    the answer to [tex]

    \emptyset \subset \left\{y\inR:0<y\leq1\right\}

    [/tex] is true.

    even [tex]

    \emptyset \subset \emptyset

    [/tex] is also true for your information because empty set itself is a set
     
  12. Aug 28, 2010 #11
    This is very helpful for a dummie like me, thank you a lot.
     
  13. Aug 29, 2010 #12

    Mark44

    Staff: Mentor

    No, E = {x | -1 <= x <= 0} and F = {x | 0 <= x <= 1}

    As you wrote E, it must be true that -1 >= 0, which is not true. For F, you have 0 >= 1, which is also not true.
     
  14. Aug 29, 2010 #13
    Mark44, that was a typo on his part. Look at his work in the latest attached thumbnail and his posts since then. He's got it pretty much now I think.
     
  15. Aug 29, 2010 #14

    Mark44

    Staff: Mentor

    Your first two answers are incorrect. E [itex]\cup[/itex] F = {0}. This is not the empty set. As Raskolnikov already mentioned, f(E [itex]\cup[/itex] F) = f(0) = 0, which is also not the empty set.

    When the problem asks about f(E), it is asking about the interval along the y-axis that the set E is mapped to. IOW, f(E) = {y | y = f(x) for some x in E}.
     
  16. Aug 29, 2010 #15

    Mark44

    Staff: Mentor

    The work in the attached file in post 5 looks pretty good, but post 7, which came later, has some errors, so I'm not so sure the OP has it quite yet.
     
  17. Aug 29, 2010 #16
    hmm, mark44, i think
    that answer was referred to "What would happen if 0 is deleted from the sets E and F?" and its
    [tex]E\cap F[/tex]
    ;P
     
  18. Aug 29, 2010 #17

    Mark44

    Staff: Mentor

    I came late to the party, so I missed it that we were removing 0 from the intersection. Sorry.
     
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