the real numbers are indecomposable as a topological space
Let be the set of real numbers with standard topology. We wish to show that if is homeomorphic to for some topological spaces and , then either is one point space or is one point space. First let us prove a lemma:
Proof. Let and such that and (we assumed that such points exist). It is sufficient to show that for any point from there exists a continous map such that , and .
Let . Therefore either or . Assume that (the other case is analogous). Choose paths from to and from to . Then we have induced paths:
Then the path defined by the formula
is a desired path.
Proposition. If there exist topological spaces and such that is homeomorphic to , then either has exactly one point or has exactly one point.
Proof. Assume that neither nor has exactly one point. Now is path connected since it is homeomorphic to , so it is well known that both and have to be path connected (please see this entry (http://planetmath.org/ProductOfPathConnectedSpacesIsPathConnected) for more details). Therefore for any point the space is also path connected (due to lemma), but there exists a real number such that is homeomorphic to . Contradiction, since is not path connected.
|Title||the real numbers are indecomposable as a topological space|
|Date of creation||2013-03-22 18:30:59|
|Last modified on||2013-03-22 18:30:59|
|Last modified by||joking (16130)|