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Functional Analysis (Big Rudin)

  1. Mar 7, 2009 #1
    I had a quick question on a part of a proof in chapter 1 of Functional Analysis, by Professor Rudin.

    Theorem 1.10 states

    "Suppose K and C are subsets of a topological vector space X. K is compact, and C is closed, and the intersection of K and C is the empty set. Then 0 has a neighborhood V such that

    [tex] (K+V) \cap (C+V) = \emptyset [/tex]"

    In the proof of this theorem, Professor Rudin starts out by proving the following proposition

    "If W is a neighborhood of 0 in X, then there is neighborhood U of 0 which is symmetric (in the sense that U = -U) and which satisfies

    [tex] U + U \subset W [/tex]."

    The question I have is about the next part of Rudin's proof

    "Suppose K is not empty, and consider x in K, since C is closed, and since x is not in C, and since the topology of X is invariant under translations, the preceding proposition shows that 0 has a symmetric neighborhood

    [tex] V_{x} [/tex]

    such that

    [tex] x + V_{x} + V_{x} + V_{x} [/tex]

    does not intersect C..."

    Is Professor Rudin's reasoning as follows:

    Since C is closed, the the complement C* of C is open in X by definition. Since x is not contained in C, then x is contained in the complement of C, C*. Since C* is open, and contains x, then C* is a neighborhood of x. Since C* is a neighborhood of x, then the set

    [tex]-x+C^{*}[/tex] is a neighborhood of 0 in X. Thus by the preceding proposition, there exists a symmetric neighborhood

    [tex] V_{x}[/tex]

    of 0 in X such that

    [tex] V_{x} + V_{x} + V_{x} \subset -x + C^{*} [/tex].

    Since the topology of X is translation invariant, then

    [tex] V_{x} + V_{x} + V_{x} \subset -x + C^{*} [/tex] iff

    [tex] x+V_{x} + V_{x} + V_{x} \subset x+(-x) + C^{*} = C^{*}[/tex],

    so that [tex] x+V_{x} + V_{x} + V_{x}[/tex] does not intersect C...?

    Is this line of reasoning correct?
  2. jcsd
  3. Mar 8, 2009 #2


    User Avatar
    Gold Member

    It looks correct, but I'm not Rudin. (-:
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