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 $(X,\mathcal{T})$ 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 $(X,\mathcal{T})$:

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 O₁ and O₂, such that O₁ contains x but not y and O₂ contains y but not x.

T₂ 

For every x, y, there exist disjoint open sets O₁ and O₂, such that O₁ contains x and O₂ contains y. T₂ spaces are also called Hausdorff spaces.

T_2½ 

For every x, y, there exist disjoint closed neighborhoods O₁ and O₂ 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 O₁ and O₂, such that O₁ contains C and O₂ contains z. A topological space that is both regular and T₁ is called T₃.

completely regular 

If C is a closed set, and z is a point not in C, there exists a continuous function $f:X\rightarrow[0,1]$ 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 T₁ is called T_3½.

normal 

If C₁ and C₂ are disjoint closed sets, there exist disjoint open sets O₁ and O₂, such that O₁ contains C₁ and O₂ contains C₂. A topological space that is both normal and T₁ is called T₄.

completely normal 

Let C₁ and C₂ be separated sets, meaning that $\mathrm{Cl}(A)\cap B=A\cap\mathrm{Cl}(B)=\O$. Then there exist disjoint open sets O₁ and O₂, such that O₁ contains C₁ and O₂ contains C₂. A topological space that is both completely normal and T₁ is called T₅.

perfectly normal 

If C₁ and C₂ are disjoint closed sets, there exists a continuous function $f:X\rightarrow[0,1]$ such that f ^{− 1}({0}) = C₁ and f ^{− 1}({1}) = C₂. A topological space that is both perfectly normal and T₁ 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.

Exercises

1. Suppose that a topological space X is T₁. Given $x\in X$, show that $X\setminus\{x\}$ is open to conclude that X is closed.
2. Given a topological space X, and given that for all $x\in X$, {x} is closed, show that X is T₁.

Some Important Theorems

Theorem 3.1.2

Let X be a T₂ 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 T₂, 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 $n\geq M$ implies $s_n\in U$. Similarly there is an integer N such that $n\geq N$ 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. $\square$

Theorem 3.1.3

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

Proof

Given any $A\subset X$, and any point $x\in X$, define the distance, d(x,A) from x to A by $d(x,A):=\inf\{d(x,a)|a\in A\}$, where d(x,a) is the distance function supplied by definition of a metric space. Observe that $x\mapsto d(x,A)$ is continuous.

Fix closed, disjoint subsets A,B of X, and define $f:X\rightarrow [-1,1]$ by $f(x):=\frac{d(x,B)-d(x,A)}{d(x,A)+d(x,B)}.$ (Note f well--defined, since for any $S\subset X$, we have d(x,S) = 0 iff x in the closure of S.) Observe f is 1 on A, -1 on B, and in the open interval ( − 1,1) elsewhere. Also, f continuous by the continuity of $x\mapsto d(x,A)$. Therefore, $f^{-1}([-1,-1/2))\supset B$ and $f^{-1}((1/2,1])\supset A$ are the preimages of open sets (open in [ − 1,1], that is) and therefore open, and they're disjoint as the preimages of disjoint sets.

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 $d=\tfrac{d(x,y)}{3}$. 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 $d(x,y)\leq d(x,z)+d(x,z)<\tfrac{2}{3}d(x,y)$, a contradiction.

Urysohn's Lemma

A topological space X is normal if and only if for any disjoint closed sets C₁ and C₂, there exists a continuous function $f:X\rightarrow[0,1]$ such that f(C₁) = {0} and f(C₂) = {1}.

Proof
In order to prove Urysohn's Lemma, we first prove the following result:

Let X be a topological space. X is normal if and only if for every closed set U, and open set V containing U, there is a set S containing U whose closure is within V.

Suppose that X is normal. If V is X, then X is a set containing U whose closure is within V. Otherwise, the complement of V is a nonempty closed set, which is disjoint from U. Thus, by the normality of X, there are two disjoint open sets A and B, where A contains U and B contains the complement of V. The closure of A does not meet B because all points in B have a neighborhood that is entirely within B and thus does not meet A (since they are disjoint), so all points in B are not within the closure of A. Thus, the set A is a set containing U, and whose closure does not meet B, and therefore does not meet the complement of V, and therefore is entirely contained in V. Conversely, take any two disjoint closed sets U and V. The complement of V is an open set containing the closed set U. Therefore, there is a set S₁ containing U whose closure is within the complement of V, which is the same thing as being disjoint from V. Then the complement of V is an open set containing the closed set Cl(S₁). Therefore, there is a set S₂ containing S₁ such that Cl(S₂) is within the complement of V i. e. is disjoint from V. Then S₁ and the complement of S₂ are open sets which respectively contain U and V, which are disjoint.

Now we prove Urysohn's Lemma.

Let X be a normal space, and let U and V be two closed sets. Set U₀ to be U, and set U₁ to be X.

Let $U_{\frac{1}{2}}$ be a set containing U_0 whose closure is contained in U₁. In general, inductively define for all natural numbers n and for all natural numbers a < 2^{n − 1}

$U_{\frac{2a+1}{2^n}}$ to be a set containing $U_{\frac{a}{2^{n-1}}}$ whose closure is contained within the complement of $U_{\frac{a+1}{2^{n-1}}}$. This defines U_p where p is a rational number in the interval [0,1] expressible in the form $\frac{a}{2^n}$ where a and n are whole numbers.

Now define the function $f:X \rightarrow$ [0,1] to be f(p)=inf{x|$p \in U_x$}.

Consider any element x within the normal space X, and and consider any open interval (a,b) around f(x). There exists rational numbers c and d in that open interval expressible in the form $\frac{p}{2^n}$ where p and n are whole numbers, such that c<f(x)<d. If c<0, then replace it with 0, and if d>1, then replace it with 1. Then the intersection of the complement of the set U_c and the set U_d is an open neighborhood of f(x) with an image within (a,b), proving that the function is continuous.

Conversely, suppose that for any two disjoint closed sets, there is a continuous function f from X to [0,1] such that f(x)=0 when x is an element of U, and that f(x)=1 when x is an element of V. Then since the disjoint sets [0,.5) and (.5,1] are open and under the subspace topology, the inverses f ^{− 1}([0,.5)), which contains X, and f ^{− 1}((.5,1]), which contains Y, are also open and disjoint.

Exercises

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 T₀.
• If X is the unit interval [0;1], and $\mathcal{T} = \{ [0,x) | x\in [0,1] \}$, then this space is T₀ but not T₁.
• Consider an arbitrary infinite set X. Let X and every finite subset be closed sets, and call the open sets Ω. Determine whether (X,Ω), is a topological space which is T₁ but not T₂. Hint: Consider the intersection of any two open, non-empty sets.

Verify that

• T₁ implies T₀
• T₂ implies T₁
• T₃ implies T₂ (hint: use theorem 3.1.1)