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作者  b218h (Gordon (In Canada)) 站內  ALGEBRA
標題  Limsup
時間  2008/07/06 Sun 04:16:47


Definition :

Let {x_n} C |R be a sequence
           ̄
For n=1,2,3,...
                          y_n = sup {x_k}   // decreasing
                                k≧n

Then,
         ___                                               ┌          ┐
         lim x_n = limsup x_n = lim y_n = inf y_n = inf│ sup {x_k}│
                     n             n          n          n └ k≧n     ┘

Claim
┌─────────────────────────────────────┐
│If {x_n} is bounded above and has at least one cluster point. Then        │
│                                                                          │
│   ___            ┌          ┐      ┌                       ┐        │
│   lim x_n = inf │ sup {x_k}│= sup│ cluster points of {x_n}│        │
│               n  └ k≧n     ┘      └                       ┘        │
└─────────────────────────────────────┘
[pf]
               ┌          ┐         ┌                        ┐
    Let α=inf│ sup {x_k}│, β=sup│ cluster points of {x_n}│
             n └ k≧n     ┘         └                        ┘

Then,                   -∞<α,β<∞ (proof omitted)


"α≦β"

    There exists a subsequence {x_nj} that converges to α, and therefore α is

also a cluster point of {x_n}. This implies

                                   α≦β

"α≧β" 

    Suppose α<β.

From the definition of β, there exists an x, a cluster point of {x_n} satisf-

ying
                                 α<x<β

              x-α
Now, let ε=───>0, there exists an N in |N s.t.
               2
                                     α+x
                        x_n<α+ε=─── for n≧N
                                       2

Since x is a cluster point of {x_n}, there exists a subsequence {x_nj} where

                         n_1≧N, and  lim x_n =x
                                     j→∞   j

Note that
                               α+x
                         x_n <───, for all j
                            j     2

                            α+x
=>                       x≦───  =>  x<α, a contradiction 
                              2



--
想說兩個定義是互通的

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  •FROM [b218h 從 d154-20-178-83.bchsia.telus.net 發表]
□ Modify: 2008/07/06 Sun 04:23:37  d154-20-178-83.bchsia.telus.net 修改

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