Exercise 4.5

By Exercise 2.29, the open complement of $E$ is a countable collection of disjoint open intervals ${\text{∪}}_i(a_i,b_i)$. If $b_i=\infty$ define $g_i$ on $[a_i,\infty )$ to take the constant value $f(a_i)$. Similarly, if $a_i=-\infty$, define $g_i$ on $(-\infty ,b_i]$ to take the constant value $f(b_i)$. Otherwise, on $[a_i,b_i]$, let $g_i$ be the linear function

Note that $g_i(a_i)=f(a_i)$, $g_i(b_i)=f(b_i)$, and the values of $g_i$ lie between $f(a_i)$ and $f(b_i)$ on $(a_i,b_i)$.

Let $g(x)=f(x)$ for $x\in E$ and $g(x)=g_i(x)$ for $x\in (a_i,b_i)$. It is clear that $g$ is continuous at any $x\in (a_i,b_i)$, so suppose $x\in E$ and let $𝜀>0$. Then there is a $\delta >0$ such that $|f(x)-f(y)|<𝜀$ for $y\in (x-\delta ,x+\delta )\cap E$. If $x-\delta \notin E$, then $x\in (a_i,b_i)$ for some $i$, and we can replace $x-\delta$ with $x-{\delta }_1=b_i\in E$. Similarly, if $x+\delta \notin E$, we can replace $x+\delta$ with some $x+{\delta }_2=a_j\in E$. Hence, if any of the open intervals $(a_i,b_i)$ intersect $(x-{\delta }_1,x+{\delta }_2)$, then both $a_i$ and $b_i$ must be in $(x-{\delta }_1,x+{\delta }_2)$. By the construction of the $g_i$, we must have $|g(x)-g(y)|<𝜀$ for $y\in (x-{\delta }_1,x+{\delta }_2)$.

If ${\delta }_1=0$, so that $x=b_i$ for some $i$, then by the linearity of $g_i$ we can increase ${\delta }_1$ by an amount small enough so that $|g(x)-g(y)|=|g_i(x)-g_i(y)|<𝜀$ for $y\in (x-{\delta }_1,x)$. Similarly, if ${\delta }_2=0$ so that $x=a_j$ for some $j$, we can increase ${\delta }_2$ by an amount small enough so that $|g(x)-g(y)|<𝜀$ for $y\in (x,x+{\delta }_2)$. Hence both ${\delta }_1>0$ and ${\delta }_2>0$, so we can conclude that $g$ is continuous at $x$.

Let $f(x)=x^{-1}$ for $x\in (0,1)$. There can be no extension of $f$ to all of $R$ since $\underset{x\to 0+}{\lim <!--\; FUNCTION\; APPLICATION\; -->}f(x)=\infty$.

If $f(x)=\left(f_1(x),\dots ,f_k(x)\right)$ is a continuous map from a closed set $E$ in $R$ into $R^k$, then each of the component functions $f_n$ are continuous functions on $E$ by Theorem 4.10(a). Extend each of the $f_n$ to a continuous function $g_n$ defined on all of $R$. Then, also by Theorem 4.10(a), the vector-valued function $g(x)=\left(g_1(x),\dots ,g_k(x)\right)$ is a continuous extension of $f$ to all of $R$.