closure and interior of Cantor set

The Cantor set is closed and its interior is empty.

To prove the first assertion, note that each of the sets $C_0,C_1,C_2,\mathrm{\dots }$, being the union of a finite number of closed intervalsis closed. Since the Cantor set is the intersection of all these setsand intersections of closed sets are closed, it follows that theCantor set is closed.

To prove the second assertion, it suffices to show that given any openinterval $I$, no matter how small, at least one point of that intervalwill not belong to the Cantor set. To accomplish this, the ternarycharacterization of the Cantor set is useful. Because rationalnumbers whose denominators are powers of 3 are dense, there exists arational number $n/3^m$ contained in $I$. Expressed in base 3, thisrational number has a finite expansion. If this expansion containsthe digit “1”, then our number does not belong to Cantor set, and weare done. If not, since $I$ is open, there must exist a number $k>m$ such that $n/3^m+1/3^k\in I$. Now, the last digit of the ternaryexpansion of this number is “1” by construction, so we also find anumber not belonging to the interval in this case.