intervals are connected
Let be a metric space. Recall that for and we have
Lemma. Let be a metric space and such that is nonempty and bounded. Then for any we have
Proof. Assume that . Then there is such that and thus , so .
Proposition. The space of real numbers is connected.
Proof. Assume that are open subsets of such that and . Furthermore assume that and take any . Then (since is open) there is such that the open ball
is contained in . Consider the set
Thus is nonempty.
Assume that is bounded. Denote by . We can apply the lemma:
Thus (due to the definition of ) is a maximal open ball (with the center in ) which is contained in . Now
Without loss of generality we can assume that . Then , because . But then (since is open) there is such that and . Thus . Contradiction. Therefore is unbounded.
Take any unbounded sequence from . Then we have
and thus , so . This completes the proof.
Corollary. For any such that intervals , , and are connected.
Proof. One can easily show that intervals are continous image of and therefore intervals are connected.
|Title||intervals are connected|
|Date of creation||2013-03-22 18:32:49|
|Last modified on||2013-03-22 18:32:49|
|Last modified by||joking (16130)|