By Mischa Cotlar
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Extra info for An Introduction to Functional Analysis (North-Holland Texts in Advanced Mathematics)
33. [0, 1] is uncountable. Proof. Assume, to the contrary, that there is a surjective function f : Z+ → [0, 1]. Let I1 be any closed interval in [0, 1] which does not contain f (1), I2 be any closed subinterval of I1 which does not contain f (2), and so on. In this way, we obtain a nested sequence of closed intervals In such / In . 32, there exists x ∈ [0, 1] that, for each n ∈ Z+ , f (n) ∈ such that x ∈ In for all n ∈ Z+ . Hence, x = f (n) for all n ∈ Z+ , that is, x is not in the range of f .
Can we prove this? 8. Claim: the sequence an = (−1)n does not converge (to any limit). Proof. Assume, to the contrary, that an → L. Then, given any positive number ε, there exists N ∈ Z+ such that |an − L| < ε for all n ≥ N . In particular, this must be true in the case ε = 1: there exists N ∈ Z+ such that, for all n ≥ N , |an − L| < 1. But then N and N + 1 are consecutive integers, so one is odd and the other is even, and hence |aN +1 − aN | = 2. But, by the Triangle Inequality, |aN +1 − aN | = |(aN +1 − L) − (aN − L)| ≤ |aN +1 − L| + |aN − L| < 1 + 1 = 2 by the deﬁnition of N .
Our ﬁnal limit theorem in this section is an extremely useful and versatile result: the Squeeze Rule. This says that if a sequence is trapped (“squeezed”) between two other sequences which are known to converge to a common limit, then the trapped sequence also converges (to the same limit). 10 (The Squeeze Rule). Let (an ), (bn ) and (cn ) be sequences such that (an ) and (cn ) converge to the same limit L. If an ≤ bn ≤ cn for all n, then (bn ) converges to L also. Proof. Let ε > 0 be given. Since an → L, there exists N1 ∈ Z+ such that, for all n ≥ N1 , |an − L| < ε, and hence, an > L − ε.