Assignment 8
Table of contents
url: https://ocw.mit.edu/courses/18-100a-real-analysis-fall-2020/resources/mit18_100af20_hw8/
Chapters and exercises given with a numbering are from Basic Analysis: Introduction to Real Analysis (Vol I) by J. Lebl.
Exercise 3.1.3
Prove Corollary 3.1.11.
Corollary 3.1.11. Let $S \subset \mathbb{R}$ and let $c$ be a cluster point of $S$. Suppose $f : S \to \mathbb{R}$, $g : S \to \mathbb{R}$, and $h : S \to \mathbb{R}$ are functions such that
\[f(x) \le g(x) \le h(x) \qquad \text{for all } x \in S \setminus \{c\}.\]Suppose the limits of $f(x)$ and $h(x)$ as $x \to c$ both exist, and
\[\lim_{x \to c} f(x) \;=\; \lim_{x \to c} h(x).\]Then the limit of $g(x)$ as $x \to c$ exists and
\[\lim_{x \to c} g(x) \;=\; \lim_{x \to c} f(x) \;=\; \lim_{x \to c} h(x).\]Take ${x_n}_{n=1}^{\infty}$ be a sequence of numbers in $S \setminus {c}$ that converges to $c$.
Let $L_1 := \lim_{x \to c} f(x)$, $L_2 := \lim_{x \to c} h(x)$, then $L := L_1 = L_2$.
Lemma 3.1.7 says that ${f(x_n)}{n=1}^{\infty}$ converges to $L_1$, and ${h(x_n)}{n=1}^{\infty}$ converges to $L_2$.
Since $\forall n \in \mathbb{N},\; f(x_n) \le g(x_n) \le h(x_n)$, taking limits we get
\(\lim_{n \to \infty} f(x_n) \le \lim_{n \to \infty} g(x_n) \le \lim_{n \to \infty} h(x_n).\)Given $\lim_{n \to \infty} f(x_n)=L_1$ and $\lim_{n \to \infty} h(x_n)=L_2$, we have $L = L_1 = L_2 \implies \lim_{n \to \infty} g(x_n) = \lim_{x \to c} g(x) = L$.
Problem 2
Let
\[f(x)= \begin{cases} 0 & \text{if } x\in\mathbb{Q},\\ 2x & \text{if } x\notin\mathbb{Q}. \end{cases}\]Prove that $f$ is continuous at $x=0$ and discontinuous at $x=1$.
(i) Let $\varepsilon>0$. Choose $\delta=\dfrac{\varepsilon}{2}>0$. If $\vert x-0\vert <\delta$ then $\vert x\vert <\dfrac{\varepsilon}{2}$.
- If $x\in\mathbb{Q}$, then $f(x)=0$ and $\vert f(x)-0\vert =0<\varepsilon$.
- If $x\notin\mathbb{Q}$, then $f(x)=2x$ and $\vert f(x)-0\vert =\vert 2x\vert <2\cdot\frac{\varepsilon}{2}=\varepsilon.$
In both cases $\vert f(x)-f(0)\vert <\varepsilon$, so $f$ is continuous at $x=0$.
(ii) Take a sequence ${x_n}_{n=1}^{\infty}\subset\mathbb{R}\setminus\mathbb{Q}$ with $x_n\to1$.
Since $x_n\notin\mathbb{Q}$, we have $f(x_n)=2x_n$ and therefore
\(\lim_{n\to\infty}f(x_n)=\lim_{n\to\infty}2\vert x_n\vert =2.\)But $1\in\mathbb{Q}$, so $f(1)=0$. Hence $\displaystyle\lim_{n\to\infty}f(x_n)=2\neq0=f(1)$; thus $f$ is not continuous at $x=1$.
Exercise 3.2.11
Let $f : \mathbb{R} \to \mathbb{R}$ be continuous. Suppose $f(c) > 0$. Show that there exists an $\alpha > 0$ such that for all $x \in (c - \alpha, c + \alpha)$, we have $f(x) > 0$.
Since $f$ is continuous at $c$, for every $\varepsilon > 0$, there exists $\delta > 0$ such that if $\vert x - c\vert < \delta$, then $\vert f(x) - f(c)\vert < \varepsilon$.
Let $\varepsilon = \frac{f(c)}{2}$ and set $\alpha = \delta$. Then for all $\vert x - c\vert < \alpha$, we have
\(f(x) > f(c) - \varepsilon = f(c) - \frac{f(c)}{2} = \frac{f(c)}{2} > 0.\)Thus, $f(x) > 0$ for all $x \in (c - \alpha, c + \alpha)$.
Exercise 3.2.14
Suppose $f : [-1, 0] \to \mathbb{R}$ and $g : [0, 1] \to \mathbb{R}$ are continuous and $f(0) = g(0)$. Define
$h : [-1, 1] \to \mathbb{R}$ by
Show that $h$ is continuous.
Only need to show $h(x)$ is continuous at $x = 0$, since $f$ and $g$ are already continuous on their domains.
We want $\lim_{x \to 0} h(x) = h(0)$. That is, $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} g(x)$.
Since $f$ and $g$ are continuous at $x = 0$, we have
\(\lim_{x \to 0^-} f(x) = f(0), \quad \lim_{x \to 0^+} g(x) = g(0).\)Given $f(0) = g(0)$, we conclude $\lim_{x \to 0} h(x) = h(0)$, so $h$ is continuous on $[-1, 1]$.
Problem 5
Let $f : \mathbb{R} \to \mathbb{R}$. Recall that if $U \subset \mathbb{R}$, the inverse image of $U$ is the set
\[f^{-1}(U) := \{x \in \mathbb{R} : f(x) \in U\}.\]Prove that $f$ is continuous if and only if for every open set $U \subset \mathbb{R}$, $f^{-1}(U)$ is open.
($\Rightarrow$) Suppose $f$ is continuous and let $U \subset \mathbb{R}$ be open.
Take an arbitrary $x \in f^{-1}(U)$, then $f(x) \in U$. Since $U$ is open, $\exists \varepsilon > 0$ such that $I := (f(x) - \varepsilon, f(x) + \varepsilon) \subset U$.
By continuity of $f$ at $x$, $\exists\delta > 0$ such that if $\vert y - x\vert < \delta$, then $\vert f(y) - f(x)\vert < \varepsilon$. Hence $f(y) \in I \subset U$, so $y \in f^{-1}(U)$.
Therefore, $\forall x \in f^{-1}(U)$, $(x - \delta, x + \delta) \subset f^{-1}(U)$, showing that $f^{-1}(U)$ is open.
($\Leftarrow$) Now assume that for every open $U \subset \mathbb{R}$, $f^{-1}(U)$ is open.
Let $x_0 \in \mathbb{R}$, and fix $\varepsilon > 0$. Define the open interval $V := (f(x_0) - \varepsilon, f(x_0) + \varepsilon)$, then $f^{-1}(V)$ is open and contains $x_0$. So $\exists\delta > 0$, s.t. $(x_0 - \delta, x_0 + \delta) \subset f^{-1}(V)$. This means whenever $\vert x - x_0\vert < \delta$, we have $f(x) \in V$, i.e. $\vert f(x) - f(x_0)\vert < \varepsilon$.
Thus $f$ is continuous at $x_0$, and since $x_0$ was arbitrary, $f$ is continuous.