Exercise 4.4.4

Question

Decide whether each of the following statements is true or false, justifying each conclusion.
  \begin{enumerate}[label=(\alph*)] 
  \item If $f$ is continuous on $[a, b]$ with $f(x)>0$ for all $a \leq x \leq b$, then $1 / f$ is bounded on $[a, b]$ (meaning $1 / f$ has bounded range).
  \item If $f$ is uniformly continuous on a bounded set $A$, then $f(A)$ is bounded.
  \item If $f$ is defined on $\mathbf{R}$ and $f(K)$ is compact whenever $K$ is compact, then $f$ is continuous on $\mathbf{R}$.
   \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] \item True, the Algebraic Limit Theorem implies $1/f$ is continuous (well defined since $f > 0$) and the Extreme Value Theorem implies $1/f$ attains a maximum and minimum and so is bounded. \item Let $\epsilon = 1$ and choose $\delta > 0$ so that $|x - y| < \delta$ implies $|f(x)-f(y)|<\epsilon$. Define the set $X = \{x_0, \dots, x_n\}$ consisting of evenly spaced values $x_i$, ranging from $x_0 = \inf A$ to $x_n =\sup A$, with the spacing between each value less than $\delta/2$ (i.e. $\forall k,\; x_k - x_{k-1} < \delta/2$). Now define the set $P = \{p_0, \dots, p_m\}$ where for each $x_i \in X$, we add one element $p_i \in A \cap V_{\delta/2}(x)$, if $A \cap V_{\delta/2}(x) \neq \emptyset$ (and do not add anything if $A \cap V_{\delta/2}(x) = \emptyset$). Now, every element $a \in A$ is at most $\delta$ from an element $p \in P$ (i.e. $|a - x| < \delta$). To see this, for any $a \in A$, there must be some $x_i \in X$ so that $|a - x_i| < \delta / 2$, and since $A \cap V_{\delta/2}(x) \neq \emptyset$ (it at least contains $a$), there must also be an element $p_i\in P$ so that $|x_i - p| < \delta /2$. By the Triangle Inequality, $|a - p| < \delta$ for some $p \in P$. Noting that $P \subseteq A$ is finite, we can consider $M = \max (f(P))$. Let $a \in A$ be arbitrary, and identify the nearest $p \in P$. We have $|a - p| < \delta$ so $|f(y) - f(p)| < \epsilon$ and since $f(p) \leq M$, $f(a) < \epsilon + M$, completing the proof. Alternative proof approach: $\bar{A}$ is closed, bounded, and thus compact. Extend the definition of $f$ to cover limit points of $A$ via a limit on $f$; these limits exist since $f$ is uniformly continuous (\textbf{TODO } write explicit proof). The extended $f$ is continuous, and thus preserves the compactness of $\bar{A}$; therefore $f(A)$ is bounded. \item Any function with finite range preserves compact sets, since all finite sets are compact. Meaning Dirichlet's function $$ g(x) = \begin{cases} 1 &\text{ if } x \in \mathbf{Q} \ 0 &\text{ if } x \in \mathbf{Q} \end{cases} $$ ``preserves'' compact sets, but is nowhere-continuous. \end{enumerate}

Exercise 4.4.4

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Exercise 4.4.4

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