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The question seemed simple enough, but something feels funny about my proof. I would appreciate if someone could please check it.

Question: Prove that if f(x) is monotonic on [a,b] and satisfies the intermediate value property, then f(x) is continuous.

Proof: Let e denote epsilon and d denote delta. For a point p in [a,b], we wish to prove |f(x) - f(p)| < e whenever |x - p| < d, for e > 0 and d > 0. Suppose we have a neighborhood about p such that p - d < p < p + d. Also suppose there is an x in the interval, so we have p - d < x < p + d. By the intermediate value property, and since the function is monotonic, we have:

f(p-d) < f(x) < f(p+d). Since the function satisfies the intermediate value property and is monotonic, then we can find an e such that f(p-d) < f(p) - e < f(x) < f(p) + e < f(p+d) for some neighborhood about f(p). The inequality f(p) - e < f(x) < f(p) + e implies |f(x) - f(p)| < e. We also have the inequality p - d < x < p + d, so that implies |x-p| < d. Therefore the function is continuous in the closed interval [a,b]. QED.

Question: Prove that if f(x) is monotonic on [a,b] and satisfies the intermediate value property, then f(x) is continuous.

Proof: Let e denote epsilon and d denote delta. For a point p in [a,b], we wish to prove |f(x) - f(p)| < e whenever |x - p| < d, for e > 0 and d > 0. Suppose we have a neighborhood about p such that p - d < p < p + d. Also suppose there is an x in the interval, so we have p - d < x < p + d. By the intermediate value property, and since the function is monotonic, we have:

f(p-d) < f(x) < f(p+d). Since the function satisfies the intermediate value property and is monotonic, then we can find an e such that f(p-d) < f(p) - e < f(x) < f(p) + e < f(p+d) for some neighborhood about f(p). The inequality f(p) - e < f(x) < f(p) + e implies |f(x) - f(p)| < e. We also have the inequality p - d < x < p + d, so that implies |x-p| < d. Therefore the function is continuous in the closed interval [a,b]. QED.

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