Thomae's Function

Consider the following function:

t (x) = {\begin{cases} 1 & x = 0 \\ \frac{1}{n} & x = \frac{m}{n} \in Q - {0} is in lowest terms with n > 0 \\ 0 & x \notin Q \end{cases}

If $c \in Q$ then $t (c) > 0$ . Because the $I$ is dense in $R$ then we can find a sequence $y_{n}$ in $I$ converging to $c$ . The result is that:

lim_{n \to \infty} t (y_{n}) = 0 \neq t (c)

and Thomae's Function fails to be continuous at any rational point.

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However, what if we consider a specific point $c = \sqrt{2}$ that's irrational. Because all the values are now irrational, then via $t$ they all get mapped to 0. Consider such a sequence $(x_{n}) \subseteq Q$ where $x_{n} \to \sqrt{2}$ . The idea is that $x_{n}$ are approximations to $\sqrt{2}$ . For example it might be:

x_{n} \in {1, \frac{14}{10}, \frac{1414}{1000}, \frac{14142}{10000}, \frac{141421}{100000}, \dots}

But notice that the denominators of these fractions are getting larger. In this case, the sequence of $t (x_{n})$ begins,

t (x_{n}) \in {1, \frac{1}{5}, \frac{1}{100}, \frac{1}{500}, \frac{1}{5000}, \frac{1}{100000}, \dots}

and is fast approaching $0 = t (\sqrt{2})$ . We will see that this always happens, and as a consequence it has the bizarre property of being continuous at every irrational point on $R$ and discontinuous at every rational point.