Topic: Let f(x) be continuous on A and B, and prove that f(x) must be bounded on A and B. Proof: Let f(x) be unbounded on A and B, and b= [a, (a+b)/2]+[(a+b)/2, b] The above two subintervals are A/. A 1，B 1 = [A 1，(a 1+b 1)/2]+[(a 1+b 1)/2，B 1]。 By dividing A2 a2 and b2 into two equal intervals, at least one A3 a3 and b3 makes f(x) unbounded on it. In this way, a series of closed intervals an, bnn = 1, 2, 3, 4 ... so f(x) is unbounded on it. It is easy to see that: ... is included in an, bn is included in ... is included in a3, b3 is included in a2, b2 is included in a 1, b 1 is included in A, and B exists with convergence criterion II: lim an(n→∞) and lim bn(n→∞). And bn-an = (b-a)/2 n n, so lim (bn-an) = 0 (where n→∞), thus it is deduced that lim bn = lim an = (an≤ ≤bn, ∈ a, b) is continuously derived from f(x) on a and b.6? 9σ& gt； 0 is-σ when ∣ x-∣ exists.