Principles of Mathematical Analysis, Rudin, 3th ed, Chapter 4, Problem 1

Problem

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

$$f(\overline{E}) \subset \overline{f(E)}$$

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 $\overline{f(E)}$.

Answer

For $y \in f(\overline{E})$, there exists $x \in \overline{E}$ such that $y=f(x)$. For $\epsilon > 0$, there exists $\delta > 0$ such that $d_X(x,x')<\delta$  $\Rightarrow$  $d_Y(f(x),f(x'))<\epsilon$. Since $x \in \overline{E}$, there exists $z \in E$ such that $d_X(x,z) < \delta$, which implies $f(z) \in N_\epsilon(f(x))$. This means $y=f(x) \in \overline{f(E)}$.

Comments

Post a Comment

Popular posts from this blog

Elementary Classical Analysis, Marsden, 2nd ed, Chapter 2, Problem 1

Problem Discuss whether the following sets are open or closed: (a) (1,2) in $\mathbb{R}$ (b) [2,3] in $\mathbb{R}$ (c) $\bigcap_{n=1}^\infty [-1,1/n)$ in $\mathbb{R}$ (d) $\mathbb{R}^n$ in $\mathbb{R}^n$ (e) A hyperplane in $\mathbb{R}^n$ (f) $\{r \in (0,1) \;|\; \text{$r$ is rational} \}$ in $\mathbb{R}$ (g) $\{(x,y) \in \mathbb{R}^2 \;|\; 0<x \leq 1\}$ in $\mathbb{R}^2$ (h) $\{x \in \mathbb{R}^n \;|\; \|x\|=1\}$ in $\mathbb{R}^n$. Answer (a) Open: For $x \in (1,2)$, we can choose $\epsilon = \min(x-1,2-x)$ so that $D(x,\epsilon) \subset (1,2)$. (b) Closed: For $x \in [2,3]^c$, we can choose $\epsilon = \min(2-x,x-3)$ so that $D(x,\epsilon) \subset [2,3]^c$. (c) Closed: We first prove that $[-1,0] = A:=\bigcap_{n=1}^\infty [-1,1/n)$. Since $[-1,0] \subset [-1,1/n)$ for all $n \in \mathbb{N}$, $[-1,0] \subset A$. To prove $A \subset [-1,0]$, we assume that $x \in A$. Then obviously, $x \geq -1$. If $x > 0$, then there exists $n \in \mathbb{N}$ such that $x > 1/n$, which means
Read more

Principles of Mathematical Analysis, Rudin, 3th ed, Chapter 3, Problem 13

Problem Prove that the Cauchy product of two absolutely convergent series converges absolutely. Answer Let $\sum a_n, \sum b_n$ be two absolutely convergent series. There Cauchy product is defined as $c_n:=\sum_{k=0}^n a_kb_{n-k}$. $$ \sum_{n=0}^N |c_n| \leq \sum_{n=0}^N \Bigg|\sum_{k=0}^n a_kb_{n-k}\Bigg| \leq \sum_{n=0}^N \sum_{k=0}^n |a_k||b_{n-k}| = \sum_{n=0}^N |a_n|\sum_{k=0}^{N-n} |b_k| \leq \sum_{n=0}^\infty |a_n| \sum_{k=0}^\infty |b_k| < \infty.$$ Since $C_N:= \sum_{n=0}^N |c_n|$ is bounded and increasing, $C_N$ convergence, and therefore, $\sum c_n$ converges absolutely.
Read more

Principles of Mathematical Analysis, Rudin, 3th ed, Chapter 2, Problem 14

Problem Give an example of an open cover of the segment $(0,1)$ which has no finite subcover. Answer Consider a family of sets $\{(1/n,1)\}_{n=2}^\infty$. For any $x \in (0,1)$, there exists $N \in \mathbb{N}$ such that $1/N < x$, which means $x \in (1/N,1)$. This means that $\{(1/n,1)\}_{n=2}^\infty$ covers $(0,1)$. Consider an arbitrary finite subcover $\{(1/n_1,1),\cdots,(1/n_K)\}$ of $\{(1/n,1)\}_{n=2}^\infty$. Then $\bigcup_{k=1}^K(1/n_k,1)=(1/n_\ast,1)$ where $n_\ast=\max(n_1,\cdots,n_K)$, which cannot cover $(0,1)$.
Read more