Topology/Separation Axioms

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Separation Axioms for Topological Spaces[edit]

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 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 
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\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 T_1 is called T.
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{Cl}(A)\cap B=A\cap\mathrm{Cl}(B)=\O. 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\rightarrow[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 Ti property.

Relations among the Separation Properties[edit]

The Ti separation properties (axioms) form a hierarchy, such that if i>j, then property Ti implies property Tj. When property Ti+1 implies Tx, which in turn implies Ti, and Tx was proposed after Ti and Ti+1, Tx is designated T. 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[edit]

  1. Suppose that a topological space X is T_1. 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 T_1.

Some Important Theorems[edit]

Theorem 3.1.2[edit]

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\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[edit]

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[edit]

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[edit]

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\rightarrow[0,1] such that f(C_1)=\{0\} and f(C_2)=\{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 an open 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 S_1 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(S_1). Therefore, there is a set S_2 containing S_1 such that Cl(S_2) is within the complement of V i. e. is disjoint from V. Then S_1 and the complement of S_2 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 U_0 to be U, and set U_1 to be X.

Let U_{\frac{1}{2}} be a set containing U_0 whose closure is contained in U_1. 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[edit]

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 \mathcal{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 \Omega. Determine whether (X, \Omega), is a topological space which is T_1 but not T_2. Hint: Consider the intersection of any two open, non-empty sets.

Verify that

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


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