Exercise 4.2.3

Question

Review the definition of Thomae's function $t(x)$ from Section 4.1.
  $$
  t(x)= \begin{cases}1 & 	ext { if } x=0 \ 1 / n & 	ext { if } x=m / n \in \mathbf{Q} \backslash\{0\} 	ext { is in lowest terms with } n>0 \ 0 & 	ext { if } x 
otin \mathbf{Q} .\end{cases}
  $$

\begin{enumerate}[label=(\alph*)] 
  \item Construct three different sequences $\left(x_{n}ight),\left(y_{n}ight)$, and $\left(z_{n}ight)$, each of which converges to 1 without using the number 1 as a term in the sequence.
  \item Now, compute $\lim t\left(x_{n}ight), \lim t\left(y_{n}ight)$, and $\lim t\left(z_{n}ight)$.
  \item Make an educated conjecture for $\lim _{x ightarrow 1} t(x)$, and use Definition $4.2 .1 \mathrm{~B}$ to verify the claim. (Given $\epsilon>0$, consider the set of points $\{x \in \mathbf{R}: t(x) \geq \epsilon\}$ Argue that all the points in this set are isolated.)
   \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] 
  \item $x_n = (1+n)/n$, $y_n = 1 - 1/n^2$ and $z_n = 1 + 1/2^n$.
  \item $\lim t(x_n) = 0$ since the size of the denominator becomes arbitrarily large. Same for the others
  \item I claim $\lim_{x 	o 1} t(x) = 0$. Let $\epsilon > 0$ be arbitrary; we must show there exists a $\delta$ where every $|x - 1| < \delta$ has $t(x) < \epsilon$. For $x 
otin \mathbf{Q}$ we have $t(x) = 0 < \epsilon$, and we can easily set $\delta$ small enough that $t(0)=1$ is excluded. That leaves us with the case $x \in \mathbf{Q}$ in which case we can write $x - 1 = m/n$ in lowest terms.

To get $t(x) = 1/n < \epsilon$ we observe that $|m/n| < \delta$ implies $t(x) = 1/n \le |m/n| < \delta$ so setting $\delta = \epsilon$ gives $t(x) < \epsilon$. To complete the proof set $\delta = \min\{\epsilon, 1\}$.
   \end{enumerate}

Exercise 4.2.3

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Exercise 4.2.3

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