Calculus TA

Problem Call a mapping of $X$ into $Y$ open if $f(V)$ is an open set in $Y$ whenver $V$ is an open set in $X$. Prove that every continuous open mapping of $\mathbb{R}^1$ into $\mathbb{R}^1$ is monotonic. Answer Suppose $f:\mathbb{R} \rightarrow \mathbb{R}$ is an open mapping but not a monotonic function. Then, without loss of generality, there exists $a,b,c \in \mathbb{R}$ such that $a<b<c$, $f(a) < f(b)$, and $f(c) < f(b)$. Note that the interval $[a,c] \subset \mathbb{R}$ is compact, so there exists $p \in [a,c]$ such that $f(p) = \sup f([a,c])$. $p$ is an interior point of $[a,c]$ since $f$ does not attain a maximum at $a$ or $c$. Choose $\epsilon > 0$ such that $N_\epsilon(p) \subset (a,c)$. Then $f(N_\epsilon(p))$ is open because $f$ is an open mapping. For $f(p) \in f(N_\epsilon(p))$, we can choose $\epsilon_1>0$ such that $N_{\epsilon_1}(f(p)) \subset f(N_\epsilon(p))$. This means $f(q) > f(p)$ for some $q \in N_\epsilon(p) \subset [a,c]$, which is contradic...

Problem Show that the requirement in the definition of uniform continuity can rephrased as follows, in terms of diameters of sets: To every $\epsilon > 0$ there exists a $\delta > 0$ such that $\text{diam}f(E)<\epsilon$ for all $E \subset X$ with $\text{diam}E < \delta$. Definition 4.18 Let $f$ be a mapping of a metric space $X$ into a metric space $Y$. We say that $f$ is uniformly continuous on $X$ if for every $\epsilon >0$ there exists $\delta >0$ such that $d_Y(f(p),f(q))<\epsilon$ for all $p$ and $q$ in $X$ for which $d_X(p,q) < \delta$. Answer Let us denote Definition 4.18 and the new definition as Definition A and Definition B. We will show that Definition A implies Definition B and vice versa. A $\Rightarrow$ B: For $\epsilon>0$, there exists $\delta_A>0$ such that $d_Y(f(p),f(q))<\epsilon/2$ for all $p$ and $q$ in $X$ for which $d_X(p,q) < \delta_A$. Let $\delta:=\delta_A$ and consider $E \subset X$ satisfying $\text{diam}E < \de...

Problem Let $f:(0,1) \rightarrow \mathbb{R}$ be uniformly continuous. Must $f$ be bounded? Answer Since $f$ is a uniformly continuous, there exists a $\delta >0$ such that $|x-y|<\delta$  $\Rightarrow$  $|f(x)-f(y)|<1$. Suppose $f$ is not bounded. Then, we can choose a sequence $x_n \in (0,1)$ such that $f(x_{n+1}) > f(x_n) + 1$. Since $x_n$ is a sequence in the compact set $[0,1]$, there exists a convergent subsequence $x_{n_k}$. Since $x_{n_k}$ is a Cauchy sequence, there exists $K \in \mathbb{N}$ such that $|x_{n_k} - x_{n_l}|<\delta$ for all $k,l \geq K$. However, $|f(x_{n_K}) - f(x_{n_{K+1}})|>1$, which is contradiction.

Problem Let $I=[0,1]$ be the closed unit interval. Suppose $f$ is a continuous mapping of $I$ into $I$. Prove that $f(x)=x$ for at least one $x \in I$. Answer In Rudin's book, the intermediate value theorem states that Let $f$ be a continuous real function on the interval $[a,b]$. If $f(a) < f(b)$ and if $c$ is a number such that $f(a) < c < f(b)$, then there exists a point $x \in (a,b)$ such that $f(x)=c$. Suppose $f(x) \neq x$ for all $x \in I$. Define a function $g(x) = f(x) -x$. Then $g$ is continuous and $g(0) = f(0) >0$, $g(1)=f(1) - 1 <0$. By the intermediate value theorem, there exists a point $c \in (0,1)$ such that $g(c)=0$  $\Rightarrow$  $g(c) = c$, which is a contradiction.

Problem If $f$ is defined on $E$, the graph of $f$ is the set of points $(x,f(x))$, for $x \in E$. In particular, if $E$ is a set of real numbers, and $f$ is real-valued, the graph of $f$ is a subset of the plane. Suppose $E$ is compact, and prove that $f$ is continuous on $E$ if and only if its graph is compact. Answer $\Rightarrow$: Consider a map $\phi(x):=(x,f(x))$. Since $\phi$ is continuous and $E$ is compact, the graph $\phi(E)$ is compact. $\Leftarrow$: Suppose $f$ is not continuous on $E$. Then for any sequence $x_n \rightarrow x$, $f(x_n) \not\rightarrow f(x)$. Since $\phi(E)$ is compact, the sequence $(x_n,f(x_n)) \in \phi(E)$ has a convergent subsequence $(x_{n_k},f(x_{n_k})) \rightarrow (p,q) \in \phi(E)$. Since $x_n \rightarrow x$, $x_{n_k} \rightarrow x = p$. Furthermore, $q=f(p)$ because $(p,q)$ is on the graph. Hence, $f(x_{n_k}) \rightarrow f(x)$, which is a contradiction.