Answer:
To prove that every bounded sequence contains a convergent subsequence, we will use the Bolzano-Weierstrass theorem.
Bolzano-Weierstrass theorem: Every bounded sequence in Euclidean space has a convergent subsequence.
Proof: Let (a_n) be a bounded sequence in Euclidean space, which means that there exists some M > 0 such that |a_n| <= M for all n. We will use the bisection method to construct a subsequence that converges to some limit L.
Step 1: Divide the interval [-M, M] into two subintervals, [-M, 0] and [0, M]. Since (a_n) is bounded, at least one of these intervals contains infinitely many terms of the sequence. Let's call the interval that contains infinitely many terms I_1.
Step 2: Divide I_1 into two subintervals of equal length, and again, at least one of these subintervals must contain infinitely many terms of the sequence. Let's call the interval that contains infinitely many terms I_2.
Step 3: Repeat this process for each subinterval, dividing it into two subintervals of equal length and choosing the one that contains infinitely many terms of the sequence. We obtain a sequence of nested intervals I_1 ⊃ I_2 ⊃ I_3 ⊃ ... that contain infinitely many terms of the sequence.
Step 4: Since the length of each interval is decreasing and they are nested, the intersection of all the intervals is a single point L. By construction, every term of the sequence (a_n) belongs to one of the intervals I_k, which means that there exists a subsequence (a_nk) that converges to L.
Therefore, every bounded sequence in Euclidean space has a convergent subsequence, as required.
Explanation: