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.

  • 8
    $\begingroup$ Okay. I drew the Cantor set on the circle. What next? $\endgroup$
    – Asaf Karagila
    Mar 16 '13 at 0:50
  • 1
    $\begingroup$ 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. $\endgroup$
    – Dylan Yott
    Mar 16 '13 at 0:55
  • 1
    $\begingroup$ @David : Why not make your comment an answer? $\endgroup$ Mar 16 '13 at 1:32
  • 1
    $\begingroup$ OK.${}{}{}{}{}$ $\endgroup$ 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 $\phi(x,y,\lambda)= \lambda x + (1-\lambda )y$.

  • $\begingroup$ Is this set convex? $\endgroup$
    – MathCosmo
    Dec 1 '18 at 5:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.