[分析] 一題關於finite intersection prop.
其實這題是明天考試的內容
和同學討論之後 大家都沒什麼頭緒
期待版上的眾多強者解惑阿!
感激不盡(拜)
*題目 (finite intersection property)
let T be any metric space, and define
X:={ f ∈ R^T : { x∈T: |f(x)| >= ε} is compact for any ε>0 }
show that X ⊆ B(T).
R 表 實數
R^T 表 定義於T上
B(T) 表 定義於T上的bounded function
*目前想到的證法
令Ai :={x ∈T: |f(x)|>= i } for all interger
Supposed Ai ≠ ∮ (空集合)
∵ Ai is compact
∴ use finite intersection property
∩ Ai ≠ ∮ where i=1,2,3 ....∞
∴ ∃ x ∈ ∩ Ai
|f(x)| >= i ,for all integer
然後就卡住了= =
該怎麼證bounded呢?
課本有個hint: if f(x) is unbounded, ∩Ai=∮ 違反finite intersection property
但該怎麼進行呢@@?
另外還有個問題想請教
該怎麼證 C[0,1] is complete?
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