MATH 447: Real Variables - Homework #6

Proof. Our goal is to show that $f=x$ and $g=e^{x}$ are both continuous, then invoke Theorem 17.4 part iii): Let $f$ and $g$ be real-valued functions that are continuous at $x_0$ in $\mathbb{R} $. Then: $$\begin{align} fg \text{ is continuous at } x_0 \end{align}$$ To show that $f=x$ is continuous, we can show the following: $$\begin{align} \left\vert x_n -x\right\vert <\varepsilon \end{align}$$ Choosing $\delta{} =\epsilon{} $, we get: $$\begin{align} \left\vert x_{n} -x \right\vert <\delta \implies \left\vert f(x_n)-f(x) \right\vert <\epsilon \end{align}$$ We wish to prove that the function $ f(x) = e^x $ is continuous. Let $ \epsilon{} > 0 $. We need to show that for every $ \epsilon{} > 0 $, there exists $ \delta{} > 0 $ such that if $ |x - x_0| < \delta{} $, then $ |e^x - e^{x_0}| < \epsilon{} $. Starting with the expression: $$\begin{align} |e^x - e^{x_0}| &= e^{x_0} |e^{x - x_0} - 1| \end{align}$$ Now, we use the elementary inequality: $$\begin{align} e^y &\geq 1 + y \end{align}$$ For $ y = -y $, we get: $$\begin{align} e^{-y} &\geq 1 - y \end{align}$$ which implies: $$\begin{align} \frac{1}{1 - y} &\geq e^y \quad \text{for } y < 1 \end{align}$$ Hence: $$\begin{align} |e^y - 1| &\leq \max \left\{ |y|, \left| \frac{y}{1 - y} \right| \right\} \quad \text{for } y < 1 \end{align}$$ Thus, taking $ y = x - x_0 $, we have: $$\begin{align} |e^y - 1| &\leq \max \left\{ |y|, \left| \frac{y}{1 - y} \right| \right\} \end{align}$$ Now, choose $ \delta{} $ small enough such that $ |y| = |x - x_0| < \delta{} $ satisfies: $$\begin{align} \max \left\{ |y|, \left| \frac{y}{1 - y} \right| \right\} &< e^{-x_0} \epsilon \end{align}$$ and $ |y| < 1 $. Therefore, we can conclude that $ e^x $ is continuous at $ x_0 $. \vspace{.5cm} Invoking Theorem 17.4 iii), we show that $xe^{x}$ is a continuous function. To show that there exists $x$ such that $h=xe^{x}=2$, we can show that pick two points in $(0,1)$ : $x_1=0.01$, and $x_2=0.99$, and substitute these into $h$, getting $h(x_1)=0.01$ and $f(x_2)=2.66$. $h(x_1)<2<h(x_2)$, and because $h$ is continuous on $(0,1)$, we can invoke the IVT, which proves that there exists $x$ such that $xe^{x}=2$.

Proof. $\implies{}:$ The goal is to show that every point in $V$ has a neighborhood, i.e., is open. Because $U$ is open, we know that there exists $r=\varepsilon{}$ for each point $y\in{} U$. Because $f$ is continuous, we also know that there is a corresponding $\delta{}>0$ such that $s\in{} B_{\delta{}}(x)=V$, so this implies that there exists an open ball $V$ around each point $s$ in $S$. Because $d(s,s_0)<\delta{} \implies{} d(f(s),f(s))<\varepsilon{}$, and $\delta{} = V$, $\varepsilon{} \subseteq{} U$, as there may be other points not in $\delta{}$ that map into $U$, this implies $f(V)\subseteq{} U$. \vspace{.5cm} $\impliedby{}:$ We know that $V$ is open for every $U$ in $S^{*}$. Choose $\varepsilon{} \in{} U \ s.t. \ \varepsilon{} >0$. Then choose $\delta{} \in{} V \ s.t. \ \delta{} >0$. By implication, we know that $\delta{}$ is open. Then, we can state that $d(s, s_0)<\delta{} \implies{} d(f(s),f(s_0))<\varepsilon{}$, which is the definition of continuity at a point.

Proof. We to show the definition of continuity: $$\begin{align} \forall \epsilon > 0, \exists \delta > 0 \text{ such that } d_Y(f(p), f(q)) &< \epsilon \\ \forall p, q \in X \text{ for which } d_X(p, q) &< \delta. \end{align}$$ We know that: $$\begin{align} f(s_0) = d(s_0, s_0) = 0 \\[10pt] f(p)= d(p, s_0) \\[10pt] f(q)= d(q, s_0) \\[10pt] d_Y(f(p),f(q))\leq d(f(p)) + d(f(q))\\[10pt] \end{align}$$ We want to show using the triangle inequality: $$\begin{align} d(f(p),f(q))\leq d(f(p),f(s_0))+d(f(q),f(s_0))< \varepsilon \end{align}$$ From Eqn 13, 14, and 15, we can say: $$\begin{align} d(f(p),f(q))\leq d(p,s_0)+d(q,s_0). \end{align}$$ Choosing $p,q$ such that $d(p,s_0)<\frac{\varepsilon}{2}$ and $d(q,s_0)<\varepsilon{}$ , we can pick $\delta{} =\varepsilon{}$ and use Eqn 14 and 15 to show: $$\begin{align} d(f(p),f(q))\leq d(f(p))+d(f(q))\\[10pt] \leq d(p,s_0)+d(q,s_0)<\varepsilon \end{align}$$

Proof. $1\implies{} 2$: We know that any open interval $(a,b)$ in $R^{1}$ is complete, i.e., every subsequence converges to a limit point contained in $\mathbb{R}$. Therefore, $(a,b)$ is an open set in $\mathbb{R}$. We know from Problem 2 that if $f$ is continuous, then every open set in the range corresponds to an open set in the domain. $(a,b)$ is open, so $f^{-1}(a,b)$ is also open.

Proof. $1\implies{} 3$: By $1\implies{} 2$, we know that if $a,b\in{} \mathbb{Q} $, then $(a,b)$ is open by the denseness of the rationals, i.e., between every real there exists a rational.

Proof. $3\implies{} 1$: We achieve this by taking the converse of $1\implies{} 3$, which exists by the bijection of $1\implies{} 2$.

Proof. $2\implies{} 1$: This exists by the bijection of $1\implies{} 2$.

Proof. $3\implies{} 2$: This is always true by the denseness of the rationals.

Proof. $2\implies{} 3$: This is always true, as $\mathbb{Q} \subset{} \mathbb{R}$