local dimension of a locally Euclidean space
Let be a locally Euclidean space. Recall that the local dimension of in is a natural number such that there is an open neighbourhood of homeomorphic to . This number is well defined (please, see parent object for more details) and we will denote it by .
Proposition. Function defined by is continuous (where on we have discrete topology).
Proof. It is enough to show that preimage of a point is open. Assume that and is such that . Then there is an open neighbourhood of such that is homeomorphic to . Obviously for any we have that is an open neighbourhood of homeomorphic to . Therefore , so . Thus (since was arbitrary) we’ve shown that around every point in there is an open neighbourhood of that point contained in . This shows that is open, which completes the proof.
Corollary. Assume that is a connected, locally Euclidean space. Then local dimension is constant, i.e. there exists natural number such that for any we have
Proof. Consider the mapping such that . Proposition shows that is continuous. Therefore is connected, because is. But has discrete topology, so there are no other connected subsets then points. Thus there is such that , which completes the proof.
Remark. Generally, local dimension need not be constant. For example consider such that
One can easily show that (with topology inherited from ) is locally Euclidean, but and .
|Title||local dimension of a locally Euclidean space|
|Date of creation||2013-03-22 18:55:34|
|Last modified on||2013-03-22 18:55:34|
|Last modified by||joking (16130)|