Question: We want to show that every real sequence {xn} contained in [a, b] has a subsequence that converges to a point x in the interval. Because {xn} is bounded, the Bolzano-Weierstrass theorem ensures that it does indeed have a convergent subsequence. Assume that the limit of this subsequence lies outside [a, b] (e.g., x > b). Show that this leads to a contradiction. (First, draw a picture.)
The following two results tell us that taking the limits of convergent sequences "preserves" weak inequalities and algebraic operations.