Problem Set 04

Apr 2, 2023

Problem. 1

Lemma.   If $x_{n+1} \leq \lambda x_n$, where $0 \leq \lambda < 1$, Then the sequence ${x_n}$ gets artbitrarily small
Clearly $x_{1+k} \leq \lambda^k x_1$, by substituting successive terms in the inequality. Given $\epsilon > 0$ we can reach $\lambda^k x \leq \epsilon$ by setting $k \geq \log_\lambda y/x$.

Fix any x in the metric space, Then construct the following sequence: ${f^n(x)} = f^0(x), f^1(x), f^2(x), \dots$. We prove it is cauchy. Consider $d(f^n(x),f^m(x))$ of some tail where $n < m$. By the triangular inequality, We know the distance is upper-bounded by $d(f^n(x),f^{n+1}(x)) + d(f^{n+1}(x),f^{n+2}(x)) + \dots + d(f^{m-1}(x),f^m(x)) \leq (m-n+1) \lambda^{n-1} d(f^1(x),f^2(x))$. By Lemma 1 and substituting distances by a sequence ${x_n}$ our intended result is concluded.

Given $X$ is complete we know our sequence ${f^n(x)}$ converges. Call it $q$. We show it converges also to $f(q)$, and by the uniqueness of limits, The main theorem of $f(x) = x$ for some $x$ is concluded. Observe $d(f^{n+1}(x),f(q)) \leq d(f^n(x),q)$, but the right hand side of the inequality is arbitrarily small. QED.

Note. This problem was solved with assistance by wonderful friends. The main key idea of using the uniqueness of limits was given by them. See the following chat:

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Problem. 2

Suppose $(x_k)$ converges to $q$. Let $\epsilon > 0$ be arbitrary. We already have $N_0$ where for any $k \geq N_0$ $x_k - q < \epsilon$. For a given permuted sequence $(x_{g(k)})$, We now show there’s $N_1$ where for any $n \geq N_1$, $x’_{n} - q < \epsilon$.

Observe $x_1, \dots, x_{N_0-1}$ are finite. Consider indices $g(1), \dots, g(N_0-1)$ and take the maximum. Call it $g_{max}(N_0-1)$. Clearly for any index $i$ greater than it, we know $x_i’$ is not equal to any one of $x_1, \dots, x_{N_0-1}$. So it is contained in the trail $x_{N_0}, x_{N_0+1}, \dots$. Thus, $x_i’ - q < \epsilon$ for any $i > g_{max}(N_0-1)$.

It is not true if we dropped the assumption that $g$ is one-to-one. A counter example is a permutation function whose range is exactly one element of $\mathcal{N}$.

Problem. 3

The is exactly the same as theorem 3.4 in Rudin’s page 50.

Problem. 4

Postponed