我没用中间值定理 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)] (这边的<=代表被包含) (实际上，如果用高微的连续函数会把连通打到连通，可以证range of f 恰恰好就是等於[f(a),f(b)]，不过这边我们用被包含即可) from the assumption, we know range of f = (0,1) <= [f(a),f(b)] (这边的<=代表被包含) ----- (1) But f(a),f(b)€range of f , too . so we get [f(a),f(b)] <= (0,1) (这边的<=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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