By G. N. Watson
An Unabridged, Digitally Enlarged Printing, to incorporate: The Tabulation Of Bessel services - Bibliography - Index Of Symbols - checklist Of Authors Quoted, And A finished Index
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Extra info for A Treatise on the Theory of Bessel Functions
Then the internal sets x ∈ ∗R x ∈ / ∗S ∧ |x − r| < 1/n , n ∈ N, satisfy the finite intersection property. e. µ(r) ⊂ ∗S. A subset is closed iff the complement is open, so the following dual is an easy consequence of the above proposition. 1. S ⊂ R is closed iff ◦ r ∈ S for all r ∈ ∗S ∩ Fin( ∗R). 28 Nonstandard Methods in Functional Analysis Intuitively one thinks of compactness as being tight enough so that no true expansion is possible. 2. Let C ⊂ R. Then C is compact iff ∀r ∈ ∗C ◦ r∈C . Proof.
Intuitively, a set is open if every point is cushioned inside the set—in fact by its monad. 10. S ⊂ R is open iff µ(r) ⊂ ∗S for all r ∈ S. Proof. e for some ∈ R+ , the interval (r − , r + ) ⊂ S. Then by transfer, ∀u ∈ ∗R |u − r| < ⇒ u ∈ ∗S . In particular ∀u ≈ r u ∈ ∗S . That is, µ(r) ⊂ ∗S. (⇐) : If S is not open, for some r ∈ S, there are n ∈ R+ , n decreasing to 0, so that r + n ∈ / S. Then the internal sets x ∈ ∗R x ∈ / ∗S ∧ |x − r| < 1/n , n ∈ N, satisfy the finite intersection property. e.
Hence (iv): Let A ∈ L(B). So there are Cn , Dn ∈ B such that Cn ⊂ A ⊂ Dn with µ(Dn \ Cn ) ≈ 0. Let C = n∈N Cn and D = n∈N Dn . By saturation, there is B ∈ B such that C ⊂ B ⊂ D. e. A is approximated by some B ∈ B under L(µ). (ii): Clearly L(µ) extends µ. So it suffices to show that L(µ) is σadditive. First let An ∈ L(B) with L(µ)(An ) = 0, n ∈ N. Let ∈ R+ and Dn ∈ B so that An ⊂ Dn and µ(Dn ) ≤ 2−n . Extend the Dn to an internal sequence by Prop. 1, then for small N ∈ ∗N \N, we have An ⊂ Therefore L(µ) Dn n