Reading guide · Proof index

Sequences and limits

Jiří Lebl, Basic Analysis I–II, version 6.3. Free author edition of this section. Selection and attribution · Notation.

L2.1.10-inc: The fully written increasing half only; decreasing half is P6.2

Proof.

Consider a monotone increasing sequence {xn}n=1∞.\{ x_n \}_{n=1}^\infty\text{.} Suppose first the sequence is bounded, that is, the set {xn:n∈N}\{ x_n : n \in \N \} is bounded. Let
x≔sup⁡{xn:n∈N}.\begin{equation*} x \coloneqq \sup \{ x_n : n \in \N \} . \end{equation*}
Let ϵ>0\epsilon > 0 be arbitrary. As xx is the supremum, there must be at least one M∈NM \in \N such that xM>x−ϵ.x_{M} > x-\epsilon\text{.} As {xn}n=1∞\{ x_n \}_{n=1}^\infty is monotone increasing, it is easy to see (by induction) that xn≥xMx_n \geq x_{M} for all n≥M.n \geq M\text{.} Hence, for all n≥M,n \geq M\text{,}
∣xn−x∣=x−xn≤x−xM<ϵ.\begin{equation*} \sabs{x_n-x} = x-x_n \leq x-x_{M} < \epsilon . \end{equation*}
So {xn}n=1∞\{ x_n \}_{n=1}^\infty converges to x.x\text{.} Therefore, a bounded monotone increasing sequence converges. For the other direction, we already proved that a convergent sequence is bounded.
The proof for monotone decreasing sequences is left as an exercise.

L2.1.17: Subsequences preserve real limits; index induction supplied in P6.1

Proof.

Suppose lim⁡n→∞xn=x.\lim_{n\to \infty} x_n = x\text{.} So for every ϵ>0,\epsilon > 0\text{,} there is an M∈NM \in \N such that for all n≥M,n \geq M\text{,}
∣xn−x∣<ϵ.\begin{equation*} \sabs{x_n - x} < \epsilon . \end{equation*}
It is not hard to prove (do it!) by induction that ni≥in_i \geq i for all i∈N.i \in \N\text{.} Hence, i≥Mi \geq M implies ni≥M.n_i \geq M\text{.} Thus, for all i≥M,i \geq M\text{,}
∣xni−x∣<ϵ,\begin{equation*} \sabs{x_{n_i} - x} < \epsilon , \end{equation*}
and we are done.