- #1
Edwin
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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
(K+V) \cap (C+V) = \emptyset"
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
U + U \subset W."
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
V_{x}
such that
x + V_{x} + V_{x} + V_{x}
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
-x+C^{*} is a neighborhood of 0 in X. Thus by the preceding proposition, there exists a symmetric neighborhood
V_{x}
of 0 in X such that
V_{x} + V_{x} + V_{x} \subset -x + C^{*}.
Since the topology of X is translation invariant, then
V_{x} + V_{x} + V_{x} \subset -x + C^{*} iff
x+V_{x} + V_{x} + V_{x} \subset x+(-x) + C^{*} = C^{*},
so that x+V_{x} + V_{x} + V_{x} does not intersect C...?
Is this line of reasoning correct?
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
(K+V) \cap (C+V) = \emptyset"
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
U + U \subset W."
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
V_{x}
such that
x + V_{x} + V_{x} + V_{x}
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
-x+C^{*} is a neighborhood of 0 in X. Thus by the preceding proposition, there exists a symmetric neighborhood
V_{x}
of 0 in X such that
V_{x} + V_{x} + V_{x} \subset -x + C^{*}.
Since the topology of X is translation invariant, then
V_{x} + V_{x} + V_{x} \subset -x + C^{*} iff
x+V_{x} + V_{x} + V_{x} \subset x+(-x) + C^{*} = C^{*},
so that x+V_{x} + V_{x} + V_{x} does not intersect C...?
Is this line of reasoning correct?
