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Tricky subspace & intersection Problem

  1. Apr 15, 2009 #1
    1. The problem statement, all variables and given/known data

    I am trying to solve this problem:
    Let W_1, W_2, W_3 be subspaces of a vector space, V.
    Prove that W_1 ∩ (W_2 + ( W_1 ∩ W_3)) = (W_1 ∩ W_2) + (W_1 ∩ W_3).
    Can someone help me show this? I have tried using Dedekind's law, but not sure it that is the way to go.


    The attempt at a solution

    I tried with in my mind very trivial case...can somebody please show me a more detailed solution with more steps?

    This is what I did....Since a subspace is a set, the laws of set operations apply. I assume (not sure if this is a valid assumption) that + here is the same as "union".

    Now intersection is distributive over union,
    i.e. a∩(b+c) = a∩b + a∩c
    so in this case,
    W_1 ∩ (W_2 + ( W_1 ∩ W_3))
    = (W_1 ∩ W_2) + (W_1 ∩ (W_1 ∩ W_3))
    In the second term, I use the properties that intersection is associative, and W_1 ∩ W_1 = W_1, and that term becomes W_1 ∩ W_3 which proves the required result.


    Now can anyone answer if this always holds and please show me a more detailed solution with more steps that would make more sense? Thanks...
     
  2. jcsd
  3. Apr 15, 2009 #2

    George Jones

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    I haven't thought about how to do the question, but
    is not a valid assumption. Have you looked in your notes or text for a definition?
     
  4. Apr 15, 2009 #3
    Yes I did...but unfortunately, I am totally lost and confused now...
     
  5. Apr 15, 2009 #4

    George Jones

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    Write down Dedekind's law (in a post).
     
  6. Apr 15, 2009 #5
    V ∩ (U + W)= U + (V ∩ W) this is what I got....
     
  7. Apr 15, 2009 #6

    George Jones

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    Is there a condition that U and V are required to satisfy?
     
  8. Apr 15, 2009 #7
    U, V and W, are just subsets of a set S, right?
     
  9. Apr 15, 2009 #8

    George Jones

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  10. Apr 15, 2009 #9
    In my question those are all the conditions specified. Do you mean that U <= V? How does, this help me understand whether the LHS is equal to the RHS? The real question is thus, when is this valid, and how I can get the RHS from LHS?
     
  11. Apr 15, 2009 #10

    George Jones

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    The LHS of your question and the LHS of Dedekind's law look similar.

    [edit}

    To make them the same, take [itex]V = W_1[/itex], and either [itex]U = W_2[/itex] and [itex]W = W_1 \cap W_3[/itex], or [itex]W = W_2[/itex] and [itex]U = W_1 \cap W_3[/itex].

    [/edit]

    With respect to Dedekind's law, does either of these choices allow you to show the desired result?
     
  12. Apr 15, 2009 #11
    I dont think so:
    The first one results in:
    RHS = W_1 + (W_2 + W_1 ∩ W_3)

    The second one results in:
    W_1 + (W_1 ∩ W_3 ∩ W_2)

    What should I do now?? What conclusion can I take away from this?
     
  13. Apr 15, 2009 #12

    George Jones

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    Take a look at the second choice more carefully.
     
  14. Apr 15, 2009 #13
    I realize that for the equality to hold then:

    W_1 + (W_1 ∩ W_3 ∩ W_2) = W_1 ∩ (W_2 + ( W_1 ∩ W_3))

    However I can not see how... Could you please guide?
     
  15. Apr 15, 2009 #14

    George Jones

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    I'm a little confused.

    With the second of my choices in post #10 (note the edit):

    what are V, U , and W;

    what is the LHS of Dedekind's laws;

    what is the RHS of Dedekind's laws?
     
  16. Apr 15, 2009 #15
    I just noticed the edit:
    Using the edited U, V, W from post 10 I get that:
    W_1 ∩ (W_1 ∩ W_3 +W_2) = (W_1 ∩ W_3) + (W_1 ∩ W_2)

    Hence, I have shown the equality in the original problem statement (Thank you).

    One question remains and that is... Is this equality ALWAYS valid? Are there any restrictions? I thought the condition in the original Dedekind's law with U <= V could be one. Are there others?
     
  17. Apr 16, 2009 #16
    Dear George,
    Could you please help me out regarding my previous questions (In Post #15)? Thank you....
     
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