Question

Show that the subset of cantor set which are used in making of the cantor set are NOT homoeomorphic and homogenous.

Answer 1

Answer:

10

Hint: Look at the complement of a point—the Cantor space is not discrete.

Later: Since the Cantor set is homeomorphic to a countable product ∏∞n=1Z/2Z of the cyclic group of order two (use the identification of the cantor set with the points in [0,1] whose infinite ternary expansion contains no 1), it is homogeneous. This means in particular that the Cantor set has no isolated points and hence it has no open points.

Now note that a basic open set is of the form ∏∞n=1Xn with all but finitely many Xn=Z/2Z. But this means that a basic open set contains a space homeomorphic to the entire Cantor space, hence non-empty open sets are uncountable (in fact of cardinality c). In particular we see that a convergent sequence (which is of course closed) can't be open. Passing to complements we get an open set that's not closed, as required.

Step-by-step explanation: