Topology/Separation Axioms

Separation Axioms for Topological Spaces

A topology on a space is a collection of subsets called open. We can then ask questions such as "can we separate any two distinct points in the space by enclosing them in two disjoint open sets?" For the real line with its usual topology, the answer is obviously yes, but there are spaces for which this is not so. It turns out that many properties of continous maps one could take for granted depend, in fact, on one of the conditions stated below holding. Topological spaces are classified according to which such conditions, called separation axioms, happen to hold.

Let $(X,𝒯)$ be a topological space, and let x,y be any two distinct points in that space. The following conditions, ordered from least to most restrictive, are ones we may wish to place on $(X,𝒯)$:

T₀

For every x, y, there exists an open set O that contains one point of the pair, but not the other.

T₁

For every x, y, there exist open sets $O_1$ and $O_2$, such that $O_1$ contains x but not y and $O_2$ contains y but not x.

T₂

For every x, y, there exist disjoint open sets $O_1$ and $O_2$, such that $O_1$ contains x and $O_2$ contains y. $T_2$ spaces are also called Hausdorff spaces.

T_2½

For every x, y, there exist disjoint closed neighborhoods $O_1$ and $O_2$ of x and y respectively.

regular

If C is a closed set, and z is a point not in C, there exist disjoint open sets $O_1$ and $O_2$, such that $O_1$ contains C and $O_2$ contains z. A topological space that is both regular and $T_1$ is called T₃.

completely regular

If C is a closed set, and z is a point not in C, there exists a continuous function $f:X\to [0,1]$ such that f(z)=0 and for any $w\in C$, we have f(w)=1 (i.e. f(C)={1}). A topological space that is both completely regular and $T_1$ is called T_3½.

normal

If $C_1$ and $C_2$ are disjoint closed sets, there exist disjoint open sets $O_1$ and $O_2$, such that $O_1$ contains $C_1$ and $O_2$ contains $C_2$. A topological space that is both normal and $T_1$ is called T₄.

completely normal

Let $C_1$ and $C_2$ be separated sets, meaning that $\mathrm{C}\mathrm{l}(A)\cap B=A\cap \mathrm{C}\mathrm{l}(B)=\mathrm{\varnothing }$. Then there exist disjoint open sets $O_1$ and $O_2$, such that $O_1$ contains $C_1$ and $O_2$ contains $C_2$. A topological space that is both completely normal and $T_1$ is called T₅.

perfectly normal

If $C_1$ and $C_2$ are disjoint closed sets, there exists a continuous function $f:X\to [0,1]$ such that $f^{-1}(\{0\})=C_1$ and $f^{-1}(\{1\})=C_2$. A topological space that is both perfectly normal and $T_1$ is called T₆.

NOTE: Many authors treat regular, completely regular, normal, completely normal, and perfectly normal spaces as synonyms for the corresponding T_i property.

Relations among the Separation Properties

The T_i separation properties (axioms) form a hierarchy, such that if i>j, then property T_i implies property T_j. When property T_i+1 implies T_x, which in turn implies T_i, and T_x was proposed after T_i and T_i+1, T_x is designated T_i½. Other implications of these properties include:

• Complete regularity always implies regularity, even without assuming T₁;
• T₀ alone suffices to make a regular space T₃. The full T₁ property is unnecessary;
• Perfect normality implies complete normality, which in turn implies normality;
• A topological space is completely normal if and only if every subspace is normal.

Some Important Theorems

Theorem 3.1.1

A topological space $X$ is $T_1$ if and only if $\{x\}$ is closed for each $x\in X$.

Proof: Assume $X$ is $T_1$ and let $x\in X$. Now (by definition of $T_1$) for every $y_i\in X$ that is distinct from $x$ there is an open set $Y_i$ containing $y_i$ but not $x$. Clearly, $\bigcup Y_i=X\setminus \{x\}$, and (being equal to a union of opens) $X\setminus \{x\}$ is open. Therefore, $\{x\}$ is closed.

Now assume that $\{x\}$ is closed for each $x\in X$. Then if $x,y\in X$ are distinct, then $x\in X\setminus \{y\}=O_1$ and $y\in X\setminus \{x\}=O_2$ where $O_1$ and $O_2$ are open sets not containing $y$ and $x$ respectively. Thus by definition, $X$ is $T_1$. $◻$

Theorem 3.1.2

Let $X$ be a $T_2$ space and let $s_n$ be a sequence in $X$. Then $s_n$ either does not converge in $X$ or it converges to a unique limit.

Proof Assume that $s_n$ converges to two distinct values $x$ and $y$.

Since $X$ is $T_2$, there are disjoint open sets $U$ and $V$ such that $x\in U$ and $y\in V$.

Now by definition of convergence, there is an integer $M$ such that $n\ge M$ implies $s_n\in U$. Similarly there is an integer $N$ such that $n\ge N$ implies $s_n\in V$.

Take an integer $K$ that is greater than both $M$ and $N$, so that $s_K$ is in both $U$ and $V$, contradicting the fact that the two sets are disjoint. Therefore $s_n$ cannot converge to both $x$ and $y$. $◻$

Theorem 3.1.3

If X is a metric space, then X is normal.

Proof Let A and B be disjoint closed sets. Define the function f:A∪B→R⁺ where if x∈A, then f(x)=inf{p|p=p(x,y), y∈B} and where if x∈B, then f(x)=inf{p|p=p(x,y), x∈A}. This is different from 0 for any element because each one has an open ball disjoint from the other set. Let N_A be the union of all open balls of the form B_f(x)/2(x) where x is an element of A, and let N_B be the union of all open balls of the form B_f(x)/2(x) where x is an element of B. Obviously N_A contains A, and N_B contains B. They are disjoint because if x is contained in both of them, then it is contained in an open ball of the form B_f(x1)/2(x₁) where x₁ is an element of A and an open ball B_f(x2)/2(₂) where x₂ is an element of B. Then d(x₁,x₂)≤d(x₁,x)+d(x,x₂)<$\frac{f(x_1)+f(x_2)}{2}$≤max{f(x₁),f(x₂)}, a contradiction.

Theorem 3.1.4

If X is a metric space, then X is Hausdorff. Proof Let x and y be two distinct points, and let d=$\frac{d(x,y)}{3}$. Then B_d(x) and B_d(y) are open sets which are disjoint since if there is a point z within both open balls, then d(x,y)≤d(x,z)+d(x,z)<$\frac{2}{3}$d(x,y), a contradiction.

Urysohn's Lemma

A topological space $X$ is normal if and only if for any disjoint closed sets $C_1$ and $C_2$, there exists a continuous function $f:X\to [0,1]$ such that $f(C_1)=\{0\}$ and $f(C_2)=\{1\}$.

Exercises

• Proove that every Metric space is $T_2$.

It is instructive to build up a series of spaces, such that each member belongs to one class, but not the next.

• The indiscrete topology is not $T_0$.
• If $X$ is the unit interval $[0;1]$, and $𝒯=\{[0,x)|x\in [0,1]\}$, then this space is $T_0$ but not $T_1$.
• Consider an arbitrary infinite set $X$. Let $X$ and every finite subset be closed sets, and call the open sets $\mathrm{\Omega }$. Determine whether $(X,\mathrm{\Omega })$, is a topological space which is $T_1$ but not $T_2$. Hint: Consider the intersection of any two open, non-empty sets.

It is also beneficial to provide verification that

• $T_1$ implies $T_0$
• $T_2$ implies $T_1$
• $T_3$ implies $T_2$ (hint: use theorem 3.1.1)