test2

I want to display diverse kinds of mathematics formular in this blog, so I will try to work out some exercises of baby rudin as samples.

Sample 1(exercise 4.2):

If $f$ is a continuous mapping of a metric space $X$ into a metric space $Y$, prove that

$$ \begin{equation} f(\overline{E})\subset \overline{f(E)} \end{equation} $$

for every set $E\subset X$. ($\overline{E}$ denotes the closure of $E$.) Show, by an example, that $f(\overline{E})$ can be a proper subset of $\overline{f(E)}$.

Solution:
Since $\overline{f(E)}$ is closed, $f^{-1}[\overline{f(E)}]$ is closed. Since $E\subset f^{-1}[\overline{f(E)}]$, we have $\overline{E}\subset f^{-1}[\overline{f(E)}]$, thus $f(\overline{E})\subset \overline{f(E)}$.
Next, let $X$ and $Y$ be $R^1$, $E=\{x|x\in N^{*}\}$, and $f(x)=\frac{1}{x}$. Then $f$ is continuous, and $f(\overline{E})=\{\frac{1}{x}|x\in N^{*}\}$, while $\overline{f(E)}=\{\frac{1}{x}|x\in N^{*}\}\cup \{0\}$. Hence $f(\overline{E})$ is a proper subset of $\overline{f(E)}$.

Sample 2(Exercise 3.14 (a)):

If $s_n$ is a complex sequence, define its arithmetic means $\sigma_n$ by

$$ \begin{equation} \sigma _n=\frac{s_0+s_1+\cdot\cdot\cdot+s_n}{n+1}\hspace{1cm} (n=0,1,2,...). \end{equation} $$

If $\lim{s_n}=s$, prove that $\lim{\sigma_n}=s$.

Solution:

Given an $\epsilon_0>0$.

$\lim{s_n}=s$, that is to say, $\forall \epsilon_1>0$, $\exists N_1>0$, s.t. $\forall n\geq N_1$, $\mid s_n-s\mid <\epsilon_1(1)$.

We noticed that

$$ \begin{equation} |\sigma_n-s|=|\frac{\sum_{i=0}^{n}s_i-s}{n+1}|\leq \frac{\sum_{i=0}^{n}|s_i-s|}{n+1} \end{equation} $$

We now let $\epsilon_1=\frac{\epsilon_0}{2}$, and then apply (1). Now we have

$$ \begin{equation} \begin{aligned} |\sigma_n-s|&\leq \frac{\sum_{i=0}^{N_1-1}|s_i-s|+\sum_{i=N_1}^{n}|s_i-s|}{n+1}\\ &\leq \frac{\sum_{i=0}^{N_1-1}|s_i-s|+(n-N_1+1)\epsilon_1}{n+1}\\ &\leq \frac{\sum_{i=0}^{N_1-1}|s_i-s|}{n+1}+\frac{\epsilon_0}{2} \end{aligned} \end{equation} $$

if $n\geq N_1$.

Put $A_n=\frac{\sum_{i=0}^{N_1-1}\mid s_i-s \mid}{n+1}$. We can see $\lim A_n=0$, i.e., $\forall \epsilon_2>0$, $\exists N_2>0$, s.t. $\forall n\geq N_2$, $A_n<\epsilon_2(2)$. Then we let $\epsilon_2=\frac{\epsilon_0}{2}$, and $N_0=max\{N_1,N_2\}$. Hence, $\mid\sigma_n-s\mid \leq A_n+\frac{\epsilon_0}{2}\leq \frac{\epsilon_0}{2}+\frac{\epsilon_0}{2}=\epsilon_0$, if $n\geq N_0$.

Thus we obtain the conclusion that $\forall \epsilon_0>0$, $\exists N_0>0$, s.t. $\forall n\geq N_0$, $\mid\sigma_n-s\mid<\epsilon_0$, which means that $\lim{\sigma_n}=s$.