topological condition for a set to be uncountable
Let be a nonempty compact Hausdorff space with no isolated points. To each finite -sequence associate a point and an open neighbourhood as follows. First, since is nonempty, let be a point of . Second, since is not isolated, let be another point of . The fact that is Hausdorff implies that and can be separated by open sets. So let and be disjoint open neighborhoods of and respectively.
Now suppose for induction that and a neighbourhood of have been constructed for all of length less than . A -sequence of length has the form or for some of length . Define . Since is not isolated, there is a point in besides ; call that point . Now apply the Hausdorff property to find disjoint open neighbourhoods and of and respectively. The neighbourhoods and can be chosen to be proper subsets of . Proceed by induction to find an for each finite -sequence .
Now define a function as follows. If is eventually zero, put . Otherwise, consider the sequence of points in . Since is compact and Hausdorff, it is closed and limit point compact, so the sequence has a limit point in . Let be such a limit point. Observe that for each finite prefix of , the point is in .
Suppose and are distinct sequences in . Let be the first position where . Then and , and by construction and are disjoint. Hence , implying that is an injective function. Since the set is uncountable and is an injective function from into , is also uncountable. ∎
The set is uncountable.
|Title||topological condition for a set to be uncountable|
|Date of creation||2013-03-22 16:15:15|
|Last modified on||2013-03-22 16:15:15|
|Last modified by||mps (409)|