Let C be a Cantor set and let U be a subset of C which is both open and closed. Then U is a Cantor set.
I don't know how to prove that...
Any hints would be appreciated.
Let C be a Cantor set and let U be a subset of C which is both open and closed. Then U is a Cantor set.
I don't know how to prove that...
Any hints would be appreciated.
On
A 1910 Brouwer theorem states that any (non-empty) compact zero-dimensional metrisable space without isolated points is homeomorphic to the Cantor set.
Now, if $U \subset C$ and $U \neq \emptyset$ is clopen, it is a compact (as $U$ is closed in $C$) metrisable zero-dimensional space (metrisable and zero-dimensional are hereditary) without isolated points (an isolated point of $U$ would also be one of $C$ (as $U$ is open in $C$), but $C$ doesn't have any isolated points, so neither has $U$). So the theorem implies your result almost immediately.
It is a theorem of Brouwer that non-empty compact Hausdorff spaces without isolated points and having countable bases consisting of clopen sets are homeomorphic to each other.
See Theorem 1 here, proven in section 4.
Brouwer's original paper.
The condition that $U$ is non-empty needs to be added to your question for the statement that you want to prove to be correct. With this condition, note that $U$ satisfies the hypotheses of Brouwer's theorem. Therefore, it is homeomorphic to $C$.