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Sum and intersection of anihalator spaces

  1. Jan 25, 2010 #1
    1. The problem statement, all variables and given/known data


    prove that (U[tex]\bigcap[/tex]W)[tex]^{\circ}[/tex]=W[tex]^{\circ}[/tex]+U[tex]^{\circ}[/tex]


    First prove That
    (U[tex]\bigcap[/tex]W)[tex]^{\circ}[/tex][tex]\supseteq[/tex]W[tex]^{\circ}[/tex]+U[tex]^{\circ}[/tex]

    Take any [tex]f\in (U\bigcap W)^{\circ}[/tex]
    Then it is easy to see that for any [tex] f\in (U\bigcap W) f(v) =0 [/tex]

    but since [tex] v\in U and v \in W [/tex] then [tex] f \in U^{\circ} and f \in W^{\circ}[/tex]



    So we have

    (U[tex]\bigcap[/tex]W)[tex]^{\circ}[/tex][tex]\supseteq[/tex]W[tex]^{\circ}[/tex]+U[tex]^{\circ}[/tex]

    Next prove
    (U[tex]\bigcap[/tex]W)[tex]^{\circ}[/tex][tex]\subseteq[/tex]W[tex]^{\circ}[/tex]+U[tex]^{\circ}[/tex]

    [tex] W^{\circ} + U ^{\circ} = span(W)^{\circ} + span(U)^{\circ} [/tex]

    because

    [tex] S^{\circ} =Span(S)^{\circ} [/tex]

    [tex] span(W)^{\circ} + span(U)^{\circ} = span (span(W)^{\circ} \cup span(u)^{\circ} [/tex]

    By definition of addition of subspaces

    [tex] span (span(W)^{\circ} \cup span(u)^{\circ}= span (W \cup U) ^{\circ} [/tex]

    Which I am not sure of

    And after all of that, we know that [tex] span (W \cap U) ^{\circ} \subseteq span (W \cup U) ^{\circ} [/tex]

    Which proves it if I did not make a mistake? Am I correct?

    Thanks
    Tal
     
  2. jcsd
  3. Jan 25, 2010 #2
    Need some more information here. What are [tex]U[/tex] and [tex]W[/tex]?
     
  4. Jan 25, 2010 #3
    Sorry Finite subspaces of the finite space V
     
  5. Jan 25, 2010 #4
    I'll assume you mean "finite-dimensional".

    In the first half, your argument is both faulty and goes in the wrong direction. You say you intend to prove [tex](U \cap W)^\circ \supset U^\circ + W^\circ[/tex], and then you give an argument that begins with "take [tex]f \in (U \cap W)^\circ[/tex]" and concludes that "[tex]f \in U^\circ + W^\circ[/tex]". This argument, if correct, would prove [tex](U \cap W)^\circ \subset U^\circ + W^\circ[/tex], not [tex]\supset[/tex] : [tex]A \subset B[/tex] means that [tex]\alpha \in A[/tex] implies [tex]\alpha \in B[/tex].

    However, the argument itself is not correct. If [tex]f \in (U \cap W)^\circ[/tex], and [tex]v \in U \cap W[/tex], then you are correct that [tex]f(v) = 0[/tex]. However, you cannot conclude from this that [tex]f \in U^\circ[/tex] on the grounds that [tex]f(v) = 0[/tex] and [tex]v \in U[/tex]. To conclude that [tex]f \in U^\circ[/tex] you would have to prove that [tex]f(v) = 0[/tex] for every [tex]v \in U[/tex], and this need not be true. What you actually want to do is give an equation [tex]f = g + h[/tex] where [tex]g \in U^\circ[/tex] and [tex]h \in W^\circ[/tex]; this proves that [tex]f \in U^\circ + W^\circ[/tex].

    In the second half, I can't understand at all what you've written. This direction, [tex]U^\circ + W^\circ \subset (U \cap W)^\circ[/tex], actually has a simpler, direct argument like the above: take [tex]f \in U^\circ + W^\circ[/tex], and prove that [tex]f \in (U \cap W)^\circ[/tex].
     
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