推 arthurduh1 :推! 10/15 14:29
推 suhorng :謝謝你!!!! 10/15 19:05
※ 引述《suhorng ( )》之銘言：
: f 為定義在 [0,1] 上的函數. 若對任意 x∈[0,1] 都是 f 的極小點,
: [即 ∀x∈[0,1], ∃δ( ∀x'∈[0,1]∩(x-δ,x+δ), f(x)≦f(x') ) ]
: 則 f 的值域 at most countable.
: 題目的提示是, 若集合 S 具有以下性質：
: ∀y∈S, ∃δ>0 s.t. (y,y+δ）∩ S = empty set
: 則 S at most countable
: -
: 想了很久, 完全不知道怎麼下手, 希望能幫忙解答(或提示)
: 謝謝各位先進 m(_ _)m
Let A(n) = {x| x belongs to [0,1] ; for each z belonging to
[0,1]∩(x-1/n,x+1/n) ,f(x)<=f(z)}. where n is a natural number,
then [0,1] = union of all A(n).
Now fixed n, A(n) = union of S(k), where:
S(k) = A(n)∩[k/n, (k+1)/n], k = 0,1,2,....,n-1.
On the other hand, f(S(k)) only contains one value ( if f(S(k)) contains
more than one value, there exist x_1 and x_2, x_1 is not equal to x_2 and
x_1 and x_2 belong to S(k), such that f(x_1) is not equal to x_2. however,
by the definition of S(k), it is easy to show that this assumption would
lead to a contradiction ). Therefore, f(A(n)) is at most countable. In the
end, f([0,1]) is at most countable because countable union of countable sets
are countable
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