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Proof about linear space, product equal to zero

  1. Mar 25, 2013 #1
    Hello! I have a problem with one proof. The task is:

    Suppose that X is linear space, x belongs to X and λ is real number. Proof if λx=0 so λ=0 or x=0. And there are conditions. Can use only this properties μ also is real number:

    2vl88xw.jpg

    I tryed to prove that but completely fails.
    Let λ=0. According to d, λx=0
    Let x=0. From c and g => λ(0+0)=λ0+λ0. And then i dont know what to do next. And also i think need to prove that if λ and x is not equal to zero when the product of them also not equal to zero.

    Thanks for helping!
     
  2. jcsd
  3. Mar 25, 2013 #2

    jbunniii

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    Hint: if ##\lambda \neq 0##, then ##\lambda## has a multiplicative inverse.
     
  4. Mar 25, 2013 #3
    Understand :) Thank you a lot
     
  5. Mar 25, 2013 #4
    sorry but after few hours i have little doubt about my proof :blushing: so if i missunderstand just say.

    if λ=0 then λx=0 according to (d).
    if λ isnt equal to zero, then there is a multiplicative inverse. λx=0, λ not equal to zero and μ=1/λ . when μ(λx)=(μλ)x=1x=x. And there is one way, x=0. So if λx=0 then λ=0 or x=0.

    its good or not? thanks for answer :)
     
  6. Mar 25, 2013 #5

    jbunniii

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    Yes, I think the logic is right. You have shown that if ##\lambda x = 0## and ##\lambda \neq 0##, then ##x = 0##. But you should try to state it more clearly, and specify which vector space properties you are using. Something like the following:

    Suppose ##\lambda x = 0## and ##\lambda \neq 0##. Then ##\lambda## has a multiplicative inverse, call it ##\mu = \lambda^{-1}##. By property (e), ##\mu(\lambda x) = (\mu \lambda) x = 1x = x##, where the last equality is true because of property (d). But ##\lambda x = 0##, so this means...
     
  7. Mar 25, 2013 #6
    thanks dude very much. Its difficult to me all that write in english because my english isnt very good. in my native language i will try do the best. :)))
     
  8. Mar 26, 2013 #7
    Hello. I dont want to create another thread so i ask here.

    There is task:

    "A" is set of Euclidean space and x is "A" point of contact (x belong to Ā, dont know how is calling the set of all points of contact in english.). Need to prove that there is sequence of A elements which converges to x.

    I dont have the idea how that prove. I found one theorem but it say that there is sequence of set elements which converges to x when x is borderline point.

    If something not understand i'll try to say more cleary if it will be possible. Thanks.
     
  9. Mar 26, 2013 #8

    jbunniii

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    It's better to create a new thread if you have a new question, unless it is very closely related to the existing thread.

    In English, ##\bar{A}## is usually called the closure of ##A##, and the elements of ##\bar{A}## are sometimes called the points of closure of ##A##. The closure ##\bar{A}## consists of ##A## and all limit points of ##A##.

    Here is a hint to get you started: if ##x \in \bar{A}## then either ##x \in A## or ##x## is a limit point of ##A##.

    If ##x \in A## then it's trivial to create a sequence in ##A## which converges to ##x##.

    If ##x## is a limit point of ##A##, then every neighborhood of ##x## contains an element of ##A## that is different from ##x##. Use that fact to construct a sequence converging to ##x##.
     
  10. Mar 26, 2013 #9
    omg :D thank you very much :))) you are awesome!!!!
     
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