This blog post discusses the elementary separation axioms , , and . There are also the more sophisticated separation axioms , , and , which are explained (but not numbered) in Section 4-2 of Munkres’s Topology.
The separation axioms , , and are properties that a specific topological space may or may not have. By giving examples, we will see that the condition that a space be is strictly stronger than the condition that a space be . Finally, we will discuss how the properties are inhertied by the constructions on topological spaces that we know, namely taking subspaces and product spaces.
The Axioms. Here are the axioms. We say a space is when:
- : For any two distinct points , either there is an open set around not containing (or both).
- : For any two distinct points , there is an open set around not containing and there is an open set around not containing .
- : For any two distinct points , there are disjoint open sets and with and .
The axiom is especially important and is also called the Hausdorff axiom. We say a space is Hausdorf if it is satisfies . We know some examples of Hausdorff spaces. For example a discrete space is Hausdorff. More generally, metric spaces are Hausdorff.
Theorem. If is a metric space, then is Hausdorff.
Proof. Let be distinct points. Let . The balls and are disjoint (by the triangle inequality) open sets containing and respectively.
Clearly . Also, clearly not all spaces satisfy even the weakest axiom : an indiscrete space with more than one point does not satisfy . Let us give some examples to show that does not imply , and does not imply .
Example 1. A space that is but not . Consider the space whose open sets are given by the “dot-diagram” below.
There is an open set around the point on the left not containing the point on the right, but not vice-versa.
Example 2. A space that is but not . Let be any infinite set with the cofinite topology. Thus, a subset is open in if and only if either is empty or is finite. For distinct points , and are open sets containing but not contaning , and containing but not containing respectively, so is . On the other hand, let and be any open sets containing but not containing , and containing but not containing respectively. Then and are finite sets and so
is a finite set. Since is infinite, we must have that is not empty. Since and were arbitrary, is not .
Example 3. Another example of a space that is but not is with the Zariski topology. Recall that the Zariski topology has as a basis the set
Given and , distinct points in , the open sets
contain but not and but not respectively. This shows that this topology is . To see that the topology is not , it suffices to show that the intersection of nonempty open sets is nonempty. For arbitrary open sets and , write
Then , so it suffices to show that the intersection of two nonempty basic open sets is nonempty, i.e., that given nonzero polynomials and , there is a point such that and are both nonzero. This is clear if (since and each ahve ony finitely many zeros) and can be shown for general by induction: for all but at most finitely many values of , and are nonzero polynomials in .
Note that both examples of spaces that are not are infinite. It turns out that a finite space must be discrete. Let be a topological space whose underlying set is a finite set. For any the property implies that the intersection of all open sets containing is just the set . There are only a finite number of open sets containing (since there are only finitely many subsets of ), so eveyr one point set is open; this implies that is discrete. This observation is closely related to the following fact.
Proposition. A topological space is if and only if every one point subset is closed.
Proof. If is and , then the property implies that for each not equal to , we can find an open set containing and not containing . Then is an open set, and so is closed. Suppose on the other hand that eveyr one point set of is closed. Then for distinct points , is an open set containing but containing and is an open set containing but not containing . Hence is .
Finally, let us talk about subspaces and product spaces.
Theorem. Let be a topological space, and let be a subspace. If is , , or then so is .
Proof. Let and be distinct points in . Then and are also distinct points in , and if is , we can find an open subset of around one not containing the other. Then is an open subset of containing (the same) one of or and not the other, and so is . If is , we can find open subset and of such that , , , and . Then and are open subset of containing but not and containing but not , respectively, and so is . If is , then we can choose and to be disjoint. Then and are disjoint, and so is .
Theorem. Let and be topological spaces. If and are both , , or then so is .
Proof. Let and be disjoint points in . Then either or . We will treat the case ; the other case is entirely similar. If is , then we can find an open subset of containing one of or but not the other. Then is an open subset of containing one of or but not the other, and so is . If is , we can find open subset and of such that , , , and . Then and are open subset of containing but not and containing but not , respectively, and so is . If is , then we can choose and to be disjoint. Then and are disjoint, and so is .