IMC 1997 · Day 2 · P12

a) $f(x) = x \sin \dfrac{1}{x}$.

b) Yes. The Cantor set is given by $$C = \left\{ x \in [0,1) : x = \sum_{j=1}^{\infty} b_j 3^{-j},\ b_j \in \{0, 2\} \right\}.$$ There is an one-to-one mapping $f : [0,1) \to C$. Indeed, for $x = \sum\limits_{j=1}^{\infty} a_j 2^{-j}$, $a_j \in \{0,1\}$ we set $f(x) = \sum\limits_{j=1}^{\infty} (2 a_j) 3^{-j}$. Hence $C$ is uncountable.

For $k = 1, 2, \dots$ and $i = 0, 1, 2, \dots, 2^{k-1} - 1$ we set $$a_{k,i} = 3^{-k} \left( 6 \sum_{j=0}^{k-2} a_j 3^j + 1 \right), \qquad b_{k,i} = 3^{-k} \left( 6 \sum_{j=0}^{k-2} a_j 3^j + 2 \right),$$ where $i = \sum\limits_{j=0}^{k-2} a_j 2^j$, $a_j \in \{0,1\}$. Then $$[0,1) \setminus C = \bigcup_{k=1}^{\infty} \bigcup_{i=0}^{2^{k-1}-1} (a_{k,i}, b_{k,i}),$$ i.e. the Cantor set consists of all points which have a trinary representation with 0 and 2 as digits and the points of its compliment

have some 1's in their trinary representation. Thus, $\bigcup\limits_{i=0}^{2^{k-1}-1} (a_{k,i}, b_{k,i})$ are all points (exept $a_{k,i}$) which have 1 on $k$-th place and 0 or 2 on the $j$-th ($j < k$) places.

Noticing that the points with at least one digit equals to 1 are everywhere dence in $[0,1]$ we set $$f(x) = \sum_{k=1}^{\infty} (-1)^k g_k(x).$$ where $g_k$ is a piece-wise linear continuous functions with values at the knots $$g_k \left( \frac{a_{k,i} + b_{k,i}}{2} \right) = 2^{-k}, \qquad g_k(0) = g_k(1) = g_k(a_{k,i}) = g_k(b_{k,i}) = 0,$$ $i = 0, 1, \dots, 2^{k-1} - 1$.

Then $f$ is continuous and $f$ “crosses the axis” at every point of the Cantor set.