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Unformatted text preview: s is well–deﬁned. For example, if {xn }n∈IN ∼ {xn }n∈IN (so that
{xn }n∈IN = {xn }n∈IN ) and {yn }n∈IN ∼ {yn }n∈IN , then
lim xn , yn n→∞ V = lim xn , yn
n→∞ V Finally, we deﬁne U : V → H by
U x = {x, x, x, · · ·}
The conclusions of Theorem 1 are now proven as a series of Lemmata.
Lemma 6 H with the operations of Lemma 5 is an inner product space.
Lemma 7 H is complete
Lemma 8 U is linear and obeys U x, U y H = x, y V for all x, y ∈ V . Lemma 9 U is one–to–one.
Lemma 10 U (V ) is dense in H.
Lemma 11 If V is complete, then U (V ) = H. (1) A binary relation ∼ on a set S is an equivalence relation if and only if, for all s, t, u ∈ S , (1) s ∼ s (reﬂexivity)
(2) if s ∼ t, then t ∼ s (symmetry) and (3) if s ∼ t and t ∼ u, then s ∼ u (transitivity). September 7, 2011 Completion 2 So now we just have to prove all of the Lemmata.
Lemma 3 ∼ is an equivalence relation. In particular, if {xn }n∈IN , {yn }n∈IN ∈ V then
either {xn }n∈IN = {yn }n∈IN {xn }n∈IN ∩ {yn }n∈IN = ∅ or Proof: The three equivalence relation axioms are trivially veriﬁed, so we only prove the last
claim. Suppose that {xn }n∈IN ∩ {yn }n∈IN = ∅ but {xn }n∈IN = {yn }n∈IN . As {xn }n∈IN
a...
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This note was uploaded on 02/13/2014 for the course MATH 511 taught by Professor Joelfeldman during the Spring '13 term at The University of British Columbia.
 Spring '13
 JoelFeldman

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