📂Topology

Separation Properties in Topology

Definition ¹

Let $X$ be a topological space. For $a,b \in X$, suppose $a \ne b$ and $U, V \subset X$ are open sets in $X$.

0. $T_{0}$: If for any $a$ and $b$ there exists $U$ that contains exactly one of them, then $X$ is a Kolmogorov space.
1. $T_{1}$: If for any $a,b$ there exists $U,V$ satisfying $$ a \in U, b \notin U \\ a \notin V, b \in V $$ then $X$ is a Fréchet space.
2. $T_{2}$: If for any $a,b$ there exists $U,V$ satisfying $$ a \in U, b \in V \\ U \cap V = \emptyset $$ then $X$ is a Hausdorff space.
3. $T_{3}$: If $X$ is a $T_{1}$-space and for every closed set $C \subset X$ not containing $a$ there exists $U,V$ satisfying $$ a \in U , C \subset V \\ U \cap V = \emptyset $$ then $X$ is a regular space.
4. $T_{4}$: If $X$ is a $T_{1}$-space and for any two disjoint closed sets $A, B \subset X$ that are $A \cap B = \emptyset$ there exists $U,V$ satisfying $$ A \subset U , B \subset V \\ U \cap V = \emptyset $$ then $X$ is a normal space.

Explanation

These properties, also called the separation axioms, literally focus on separating a space into parts. The classification presented above as $T_{i}$ is called the Kolmogorov classification. At first glance one may get the impression $$ T_{4} \implies T_{3} \implies T_{2} \implies T_{1} \implies T_{0} $$, and indeed it is a convenient classification in that sense.

Of particular interest is the $T_{2}$ Hausdorff space: its condition is neither too strong nor too weak, making it practically useful. Many pathological spaces commonly used as counterexamples often fail to satisfy $T_{2}$. Examples of non-Hausdorff spaces include the Sierpiński space and the particular point topology.

Below are some useful properties of Hausdorff spaces. Since every metric space is Hausdorff, there are many immediate applications.

Results

• [2-1]: $T_{2}$ is a topological property.
• [2-2]: $T_{2}$ is a hereditary property.
• [2-3]: In a $T_{2}$-space, a sequence $\left\{ x_{n} \right\}$, if it converges, converges to a unique point.

Proof

[2-1]

Let $f : X \to Y$ be a homeomorphism and suppose $X$ is a Hausdorff space. It suffices to show that $Y$ is a Hausdorff space.

Since $f$ is a bijection, for any two distinct $y_{1}, y_{2} \in Y$ there exist two distinct $x_{1}, x_{2} \in X$ satisfying $$ a = f(x_{1}) \\ b = f(x_{2}) $$. By assumption $X$ is Hausdorff, so there exist open sets $U, V \subset X$ satisfying $$ x_{1} \in U \\ x_{2} \in V \\ U \cap V = \emptyset $$. Because $f$ is an open map by continuity, $f(U)$ and $f(V)$ are open in $Y$, and by $$ a \in f(U) \\ b \in f(V) \\ f(U) \cap f(V) = \emptyset $$, therefore $Y$ is a Hausdorff space.

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