induced partial order on an Alexandroff space
Let be a , Alexandroff space. For denote by the intersection of all open neighbourhoods of . Define a relation on as follows: for any we have if and only if . This relation will be called the induced partial order on .
Proposition 1. is a poset.
Proof. Of course for any . Thus is reflexive.
Assume now that and for some . Assume that . Then, since is a space, there is an open set such that and or there is an open set such that and . Both cases lead to contradiction, because we assumed that and . Thus every open neighbourhood of one element must also contain the other. Thus is antisymmetric.
Proof. ,,” Assume that is continuous and suppose that are such that . We wish to show that , so assume this is not the case. Let . Then . But is open, so is also open (because we assumed that is continuous). Furthermore and because , then , but this implies that . Contradiction.
,,” Assume that preserves the induced partial order and let be an open subset. Let . Then for any we have (because preserves the induced partial order) and since (because is open and is the smallest open neighbourhood of ) we have that . Thus
which implies that is open because contains a small neighbourhood of each point. This completes the proof.
|Title||induced partial order on an Alexandroff space|
|Date of creation||2013-03-22 18:45:55|
|Last modified on||2013-03-22 18:45:55|
|Last modified by||joking (16130)|