Re: [微積] 中間值定理
我沒用中間值定理
用Weierstrass extreme value theorem
pf：
Assueme there exists a continuous function f defined on [0,1] with range (0,1)
by Weierstrass extreme value theorem
there exsists a,b€[0,1] , s.t. f(a) <= f(x) <= f(b) , for all x€[0,1]
i.e. range of f <= [f(a),f(b)] (這邊的<=代表被包含)
(實際上，如果用高微的連續函數會把連通打到連通，可以證range of f
恰恰好就是等於[f(a),f(b)]，不過這邊我們用被包含即可)
from the assumption, we know
range of f = (0,1) <= [f(a),f(b)] (這邊的<=代表被包含) ----- (1)
But f(a),f(b)€range of f , too .
so we get [f(a),f(b)] <= (0,1) (這邊的<=代表被包含) ------(2)
(1) contradict (2)
so it's impossible that
there exists a continuous function f defined on [0,1] with range (0,1)
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