# The Cantor Set

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• Apr 22nd 2008, 02:25 AM
squarerootof2
The Cantor Set
The hint given for the problem is that the set of isolated pts of a countable complete metric space X forms a dense subset of X, and using this i need to prove that the cantor set is uncountable. so i was thinking that we need to use the fact that since cantor set has no isolated points (which was proven in previous exercise), but i keep getting stuck. can someone help me?
• Apr 22nd 2008, 09:24 AM
ThePerfectHacker
Quote:

Originally Posted by squarerootof2
The hint given for the problem is that the set of isolated pts of a countable complete metric space X forms a dense subset of X, and using this i need to prove that the cantor set is uncountable. so i was thinking that we need to use the fact that since cantor set has no isolated points (which was proven in previous exercise), but i keep getting stuck. can someone help me?

Suppose that $C$, the Cantor set, is countable. Since $(\mathbb{R}, | ~ |)$ is a complete metric space it would mean the isolated points of $C$, call this subset $S$, would be dense in $C$. Topologically it means $\bar S = C$. But $S = \emptyset$ by your previous exercise, so the closure of $S$ is the empty set, but that is a contradiction because it has to be $C$. This means the Cantor set is uncountable.
• Apr 22nd 2008, 09:29 PM
squarerootof2
thanks so much for the input, but one small question, when you say "Since http://www.mathhelpforum.com/math-he...4a7f9cef-1.gif is a complete metric space " what is the metric you are using? just the ordinary euclidean metric? thanks.
• Apr 22nd 2008, 09:49 PM
ThePerfectHacker
Quote:

Originally Posted by squarerootof2
thanks so much for the input, but one small question, when you say "Since http://www.mathhelpforum.com/math-he...4a7f9cef-1.gif is a complete metric space " what is the metric you are using? just the ordinary euclidean metric? thanks.

Yes. Here it is the same thing as taking the absolute value of the number.