deluks917
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Is the following Idea well-known/usefull? The points of T not "surrounded" by A
Let V be a metric space. Given a subset A\subseteqV we define a new set A\widehat{} to be equal to the set {x|for any open ball B1 with x \inB1 there exists B2 s.t. B2 \subseteq B1 and B2 \cap A = ∅}.
Examples: Let T be ℝ2. Let Δ = open unit ball. Then Δ\widehat{} = ΔC (its compliment). Let Y = the line (t,t) t\inℝ then Y\widehat{} = ℝ2.
I'm thinking of A\widehat{} as the points not "surrounded on all sides by A". The motivation for this definition is my attempt to solve the following problem: If f:ℝ→ℝ is smooth and given any x \in ℝ fn(x) eventually vanishes for all n>k then f is a polynomial. If I could prove that the property BC = B\widehat{} is preserved for some families of sets under chains of inclusions I can solve the above problem. However I'm interested in if this property is equivalent to something obvious I am missing.
Let V be a metric space. Given a subset A\subseteqV we define a new set A\widehat{} to be equal to the set {x|for any open ball B1 with x \inB1 there exists B2 s.t. B2 \subseteq B1 and B2 \cap A = ∅}.
Examples: Let T be ℝ2. Let Δ = open unit ball. Then Δ\widehat{} = ΔC (its compliment). Let Y = the line (t,t) t\inℝ then Y\widehat{} = ℝ2.
I'm thinking of A\widehat{} as the points not "surrounded on all sides by A". The motivation for this definition is my attempt to solve the following problem: If f:ℝ→ℝ is smooth and given any x \in ℝ fn(x) eventually vanishes for all n>k then f is a polynomial. If I could prove that the property BC = B\widehat{} is preserved for some families of sets under chains of inclusions I can solve the above problem. However I'm interested in if this property is equivalent to something obvious I am missing.