two numbers of the given sequence that are in the open interval (a, b). Denote the smaller of these two numbers by a1 and the larger by b1. Similarly, find the first two numbers of the given sequence that are in (a1, b1). Denote the smaller by a2 and the larger by b2. Continuing this procedure generates a sequence of intervals (a1, b1), (a2, b2), (a3, b3), ... such that each interval in the sequence contains all succeeding intervals—that is, it generates a sequence of nested intervals. This implies that the sequence a1, a2, a3, ... is increasing and the sequence b1, b2, b3, ... is decreasing. Either the number of intervals generated is finite or infinite. If finite, let (aL, bL) be the last interval. If infinite, take the limits a∞ = limn → ∞ an and b∞ = limn → ∞ bn. Since an < bn for all n, either a∞ = b∞ or a∞ < b∞. Thus, there are three cases to consider: Case 1: There is a last interval (aL, bL). Since at most one xn can be in this interval, every y in this interval except xn (if it exists) is not in the given sequence. Case 2: a∞ = b∞. Then a∞ is not in the sequence since for all n: a∞ is in the interval (an, bn) but xn does not belong to (an, bn). In symbols: a∞ ∈ (an, bn) but xn ∉ (an, bn). {| class="wikitable collapsible collapsed" ! style="background: f5f5f5;" |Proof that for all n: |- style="text-align: left; vertical-align: top; background: white" | style="padding-left: 1em; padding-right: 1em" | This lemma is used by cases 2 and 3. It is implied by the stronger lemma: For all n, (an, bn) excludes x1, ..., x2n. This is proved by induction. Basis step: Since the endpoints of (a1, b1) are x1 and x2 and an open interval excludes its endpoints, (a1, b1) excludes x1, x2. Inductive step: Assume that (an, bn) excludes x1, ..., x2n. Since (an+1, bn+1) is a subset of (an, bn) and its endpoints are x2n+1 and x2n+2, (an+1, bn+1) excludes x1, ..., x2n and x2n+1, x2n+2. Hence, for all n, (an, bn) excludes x1, ..., x2n. Therefore, for all n, xn ∉ (an, bn). |} Case 3: a∞ < b∞. Then every y in [a∞, b∞] is not contained in the given sequence since for all n: y belongs