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:
Lemma. Let and be path connected topological spaces such that cardinality of both and is at least . Then for any point the space with subspace topology is path connected.
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 |
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Canonical name | TheRealNumbersAreIndecomposableAsATopologicalSpace |
Date of creation | 2013-03-22 18:30:59 |
Last modified on | 2013-03-22 18:30:59 |
Owner | joking (16130) |
Last modified by | joking (16130) |
Numerical id | 12 |
Author | joking (16130) |
Entry type | Theorem |
Classification | msc 54F99 |