Exercise 4.3.8

Question

Decide if the following claims are true or false, providing either a short proof or counterexample to justify each conclusion. Assume throughout that $g$ is defined and continuous on all of $\mathbf{R}$.
  \begin{enumerate}[label=(\alph*)] 
  \item If $g(x) \geq 0$ for all $x<1$, then $g(1) \geq 0$ as well.
  \item If $g(r)=0$ for all $r \in \mathbf{Q}$, then $g(x)=0$ for all $x \in \mathbf{R}$.
  \item If $g\left(x_{0}\right)>0$ for a single point $x_{0} \in \mathbf{R}$, then $g(x)$ is in fact strictly positive for uncountably many points.
   \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] 
  \item True, using the sequential definition for functional limits letting $(x_n) 	o 1$ we have $g(x_n) \ge 0$ and $g(x_n) 	o g(1)$ so by the Order Limit Theorem $g(1) \ge 0$
  \item True, since if there was some $x \in \mathbf R$ with $g(x) 
e 0$ then $g$ would not be continuous at $x$ because we could never make $\epsilon$ smaller then $|g(x) - g(r)| = |g(x)|$ as we can always find rational numbers satisfying $g(r) = 0$ inside any $\delta$-neighborhood.
  \item True, let $\epsilon < g(x_0)$ and pick $\delta$ so that every $x \in V_\delta(x_0)$ satisfies $g(x) \in (g(x_0)-\epsilon, g(x_0)+\epsilon)$ and thus $g(x) > 0$ since $g(x_0)-\epsilon > 0$.
   \end{enumerate}

Exercise 4.3.8

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Exercise 4.3.8

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