Prove that every continuous function on a closed interval [a, b] attains its maximum and minimum values.
To prove that every continuous function on a closed interval [a, b] attains its maximum and minimum values, we will use the Extreme Value Theorem.

The Extreme Value Theorem states that if a function f is continuous on a closed interval [a, b], then there exist points c and d in [a, b] such that f(c) is the minimum value of f on [a, b] and f(d) is the maximum value of f on [a, b].

Proof:

Let f be a continuous function on the closed interval [a, b]. By the definition of continuity, for any sequence (x_n) in [a, b] that converges to a point x in [a, b], the sequence (f(x_n)) also converges to f(x).

Now consider the set S = {f(x) : x ∈ [a, b]}. Since f is continuous on [a, b], S is non-empty and bounded. By the Completeness Axiom, S has a supremum and an infimum, which we denote by M and m, respectively.

We claim that there exist points c and d in [a, b] such that f(c) = m and f(d) = M.

Since m is the infimum of S, for each natural number n, there exists a point x_n in [a, b] such that f(x_n) > m - 1/n. As [a, b] is closed and bounded, it is compact. Therefore, there exists a subsequence (x_{n_k}) of (x_n) that converges to some point c in [a, b]. Since f is continuous, the sequence (f(x_{n_k})) converges to f(c). However, we have f(x_{n_k}) > m - 1/{n_k} for all k, so f(c) ≥ m. On the other hand, since m is the infimum of S, we have f(c) ≤ m. Combining these inequalities, we get f(c) = m.

Similarly, we can show that there exists a point d in [a, b] such that f(d) = M.

Thus, every continuous function on a closed interval [a, b] attains its maximum and minimum values at points c and d, respectively.