# Cantor set on the circle

Draw a Cantor set C on the circle and consider the set A of all the chords between points of C. Prove that A is compact.

• Okay. I drew the Cantor set on the circle. What next? Mar 16 '13 at 0:50
• If $A \subset \mathbb R^{2}$, then we have to show $A$ is closed and bounded. Clearly $A$ is bounded, so we just have to show $A$ is closed. Mar 16 '13 at 0:55
• @David : Why not make your comment an answer? Mar 16 '13 at 1:32
• OK.${}{}{}{}{}$ Mar 16 '13 at 1:51

$C$ is compact as it's closed and bounded. Then, $A$ is compact as it's the image of the compact set $C\times C\times [0,1]$ under the continuous map $\phi: {\Bbb R}^2\times {\Bbb R}^2\times [0,1]\to {\Bbb R}^2$ given by $\phi(x,y,\lambda)= \lambda x + (1-\lambda )y$.