Who can help me prove the boundedness of function and the maximum and minimum theorem The property theorem of continuous function on closed interval: the continuous function on closed interval is bounded on the interval, and it must be able to obtain its maximum and minimum value. This theorem is easy to understand with graph, I hope someone can prove it with mathematical language,

Who can help me prove the boundedness of function and the maximum and minimum theorem The property theorem of continuous function on closed interval: the continuous function on closed interval is bounded on the interval, and it must be able to obtain its maximum and minimum value. This theorem is easy to understand with graph, I hope someone can prove it with mathematical language,

The basic steps to prove the extremum theorem are as follows
1. Prove the boundedness theorem
2. Find a sequence whose image converges to the minimum upper bound of F
3. Prove that there is a subsequence which converges to a point in the domain
4. Using continuity to prove that the image of subsequence converges to the minimum upper bound
The proof of boundedness theorem
Suppose that the function f has no upper bound in the interval [a, b]. Then, according to the Archimedes principle of real numbers, for every natural number n, there exists an xn in [a, b], such that f (xn) > n. this defines a sequence {xn}. Since [a, b] is bounded, according to the polcharno Weierstrass theorem, we can deduce a convergent subsequence {x} of {xn}_ {n_ k} Since [a, b] is a closed interval, it must contain X. because f is continuous at x, we know {f (x)_ {n_ k} But for all K, there is f (x)_ {n_ k} ) > NK ≥ K, which means {f (x_ {n_ k} Therefore, f has an upper bound in [a, b]
Proof of extreme value theorem
According to the boundedness theorem, f has an upper bound. Therefore, according to the dedkin completeness of real numbers, the minimum upper bound m of F exists. We need to find a D in [a, b] such that M = f (d). Let n be a natural number. Since m is the minimum upper bound, m – 1 / N is not the minimum upper bound of F. therefore, there exists DN in [a, b], Let m – 1 / N < f (DN). This defines a sequence {DN}. Since m is an upper bound of F, we have m – 1 / N < f (DN) ≤ m for all n. therefore, the sequence {f (DN)} converges to M
According to the polchano Weierstrass theorem, there is a subsequence {D}_ {n_ k} Since [a, b] is a closed interval and D is in [a, b], the sequence {f (d) is continuous at D_ {n_ k} But {f (d)} converges to f (d)_ {n_ k} )} is a subsequence of {f (DN)} and converges to m, so m = f (d). Therefore, f obtains the minimum upper bound m at D