# Cantor set is compact?

Cantor set See the link, I am referring to cantor set on the real line. I wish to show that it is compact. I am doing this by pointing following arguments. I am not sure if this is enough.

1. Cantor set is bounded by definition in the region $$[0,1]$$
2. Cantor set is the union of closed intervals, and hence it is a closed set.
3. Since the Cantor set is both bounded and closed it is compact by Heine-Borel Theorem.
• 2. Complement of union of open intervals.
– user545497
Mar 27 '18 at 3:50
• Yes true. However is there something wrong with taking it as union of closed set. Both facts imply one another. Mar 27 '18 at 3:55
• A union of closed sets need not be closed, so you should amend statement 2 accordingly. Mar 27 '18 at 3:56
• You could fix that by saying, each partial level of the construction is a finite union of closed intervals, and hence is closed, and the full set is the intersection of closed sets, and hence closed. Mar 27 '18 at 3:58
• But the Cantor set is not a union of intervals of any kind: it’s totally disconnected, even. Mar 27 '18 at 4:26

Cantor set is defined as $C=\cap_n C_n$ where $C_{n+1}$ is obtained from $C_n$ by dropping 'middle third' of each closed interval in $C_n$
Since each $C_n$ is closed and $C$ is an intersection of such sets, $C$ is closed (arbitrary intersection of closed sets is a closed set).
As $C$ is closed and bounded, it is compact by Heine-Borel theorem.