If it is an open interval, it can be said that the function is extended to a closed interval, and the endpoint function value can take the corresponding unilateral limit.
In addition, if it is an unbounded interval, let it be [a, +∞), as long as it is proved that for any m >；; A, both f([a, M]) are intervals.
In fact, this problem boils down to proving the intermediate value theorem of continuous functions on a closed interval by using finite covering theorem, and only need to prove the zero point theorem, that is, if f∈C[a, b] and f (a) f (b).
If the conclusion is not true, then for any x_{0}, there exists f(x0)≠0. According to the local sign-preserving property of continuous function, there is a neighborhood U0 corresponding to x0, which makes f sign-preserving on U0. Therefore, when x0 traverses all the points in [a, b], we get the open cover of [a, b].
U0，U 1，…
(Unilateral interval at the endpoint) Therefore, there are a finite number of V 1, V2, ..., vk covers [a, b], and f is sign-preserving on every Vi.
Moreover, each Vi must intersect with another (the concept of Lebesgue number), so it is obvious that the signs of F (a) and F (b) are the same, which contradicts the hypothesis.