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Prove sup AC = supA supC.

  1. Feb 25, 2010 #1
    1. The problem statement, all variables and given/known data
    If A and C are subsets of R, let AC={ac: a E A, c E C}. If A and C are bounded, prove that sup AC = supA supC.

    2. Relevant equations


    3. The attempt at a solution
    I acutally think it's not always true, for example when A and C each consists of only negative numbers.
    If A consists of only negative numbers and C consists of only positive numbers, will it be true then?
    In general, when will sup AC = supA supC be true? And how can we "prove" the claim in this case?

    Any help is appreciated!
     
    Last edited: Feb 25, 2010
  2. jcsd
  3. Feb 25, 2010 #2
    Please help...I still can't figure out how to prove...
     
  4. Feb 25, 2010 #3

    LCKurtz

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    Well, you seem to have doubts about its truth in all cases. So have you looked for counterexamples? Have you tried looking at any of the things you asked about? What did you find out?
     
  5. Feb 25, 2010 #4
    I have no idea what I had in mind before. When A and C each consists of only negative numbers. Then sup AC >=0, supA<=0 supC<=0, so it's not contradicatory.

    So I guess the claim is ok as it stands, and it's always true.
    But now how to prove it? (yes, I know the definitions, but I have no idea how to prove)
     
  6. Feb 25, 2010 #5

    LCKurtz

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    Until you actually try some specific examples with various choices for A and C (even simple examples like intervals), positive, negative, or both, you aren't going to have a "feel" for why it is true or whether there are cases where it isn't true. Get your hands dirty with some numbers.
     
  7. Feb 25, 2010 #6

    jgens

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    As the problem is stated, it doesn't make any sense. For example let [itex]A = \{-5, 0, 1\}[/itex] and [itex]C = \{-6, 0 , 2\}[/itex]. Clearly [itex]\sup\{AC\} = 30[/itex] but [itex]\sup\{A\}\sup\{C\} = 2[/itex].

    Edit: Sorry for jumping in LCKurtz, I didn't see your posts.
     
  8. Feb 25, 2010 #7
    Then what should be the correct statment of the problem?
    I tried positive numbers, and the claim seems to be true. But how can we prove it (in the case when A and C consist of only positive numbers)? This is my main question. I want to see how to prove statements like this.

    Thank you!
     
  9. Feb 25, 2010 #8

    Dick

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    If everything is positive, you seem to be asking how to prove ac<=(sup A)(sup C) given that a<=(sup A) and c<=(sup C). That's the same thing as asking how to prove ac<=bd, given that a<=b and c<=d. Is that really what you are asking? Then use that multiplication by a positive number preserves order. I.e. a<=b and c>=0 -> ac<=bc.
     
  10. Feb 25, 2010 #9

    jgens

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    Dick, that only proves that [itex]\sup\{A\}\sup\{C\}[/itex] is an upper bound for [itex]AC[/itex]. The OP needs to prove that this is the least upper bound.
     
  11. Feb 25, 2010 #10

    Dick

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    Hah. Good point. We both know how to do this, right? Does kingwinner?
     
  12. Feb 25, 2010 #11

    jgens

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    Yes, the claim is true if the sets consist of strictly positive elements. To prove something like this I would show that [itex]\sup\{A\}\sup\{C\}[/itex] is an upper bound for [itex]AC[/itex]. Then, I might say let [itex]\varepsilon > 0[/itex] be some real number such that [itex]\sup\{A\} - \varepsilon > 0[/itex] and [itex]\sup\{C\} - \varepsilon > 0[/itex]. Clearly we can always find [itex]a \in A[/itex] and [itex]c \in C[/itex] such that [itex](\sup\{A\} - \varepsilon/n)(\sup\{C\} - \varepsilon/n) <[/itex][itex]ac \leq \sup\{AC\} \leq \sup\{A\}\sup\{C\}[/itex]. Using this inequality and choosing large enough [itex]n[/itex], it's possible to show that [itex]\sup\{A\}\sup\{C\} = \sup\{AC\}[/itex]. There's probably a simpler way to do it but this is just what I can think of off the top of my head.
     
  13. Feb 25, 2010 #12

    Dick

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    I think you've got the right idea in your head. Pick a_n in A that lies in the interval [sup(A)-1/n,sup(A)]. Ditto for c_n. Now a_n*c_n must lie in the interval [(sup(A)-1/n)*(sup(C)-1/n),sup(A)*sup(C)]. Let n approach infinity.
     
  14. Feb 25, 2010 #13

    jgens

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    Ah! That's so much simpler and cleaner!
     
  15. Feb 25, 2010 #14

    Dick

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    Yeah, but I thought I was responding to a post by kingwinner. Didn't check the byline.
     
  16. Feb 25, 2010 #15
    "a_n*c_n must lie in the interval [(sup(A)-1/n)*(sup(C)-1/n),sup(A)*sup(C)]" <----why? What if (sup(A)-1/n) is negative?

    And I'm sorry, but I don't understand the proof. I don't get the general idea of the proof. Maybe someone can explain a little more?

    So FIRST we must prove that supA supC is an upper bound for AC. THEN, we begin picking sequences?? Why does this prove that sup AC = supA supC ??

    thanks.
     
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