Lectures on Order and Topology
Antonino Salibra February 25, 2015
1 Preliminaries
Let (X, ≤) be a partially ordered set (poset, for short) and Y ⊆ X. We say that:
(a) Y is an upper set if
y ∈ Y ∧ x ∈ X ∧ (x ≥ y) ⇒ x ∈ Y.
(b) Y is a down set if
y ∈ Y ∧ x ∈ X ∧ (x ≤ y) ⇒ x ∈ Y.
(c) Y is directed if, for all x, y ∈ Y , there exists z ∈ Y such that x ≤ z and y ≤ z.
A point a ∈ X is an upper bound of Y if y ≤ a for every y ∈ Y . The least upper bound (lub, for short) of Y , denoted by tY , is the minimum element of the following set {x : x is an upper bound of Y }, when it exists.
Example 1 Let N be the set of natural numbers, totally ordered by n ≤ k ⇔ (∃r ∈ N) k = n + r.
Then the set N has no upper bound.
Example 2 Let P(N) be the power set of the set of natural numbers, partially ordered by the relation “to be contained within or equal to”⊆. The set N is an upper bound of {A ⊆ N : 5 /∈ A}. The set N \ {5} is the least upper bound of {A ⊆ N : 5 /∈ A}.
Example 3 Let P(N) be the power set of the set of natural numbers, partially ordered by the relation “to be contained within or equal to”⊆. The set of all finite subsets of N is a directed set.
2 Topology: main definitions and notation
Topology explains the meaning of the term “neighbourhood of a point”. Given a set X, a topol- ogy is defined on X if, for every subset V of X, it is defined a greatest subset U of V , called the interiorof V , such that V is a neighbourhood of any point of U and V is not a neighbourhood of any point of V \ U . An open set U is a set which is a neighbourhood of each of its points. A neighbourhoodof a point x is a set V whose interior contains x.
Definition 2.1 A (topological) space is a pair (X, OX) where X is a nonempty set and OX is a family of subsets ofX, whose elements are called open sets, satisfying the following properties:
(i) ∅, X ∈ OX;
(ii) If(Ui∈ OX : i ∈ I) is a family of opens, thenS
i∈IUi∈ OX;
(iii) IfU, V ∈ OX then U ∩ V ∈ OX.
The elements of OX are called open sets. A closed set is the complement of an open set (i.e., A ⊆ X is closed iff X \ A is open). The set of all closed sets is denoted by CX.
The set CX of closed sets satisfies the following conditions:
(i) ∅, X ∈ CX;
(ii) If (Ui∈ CX : i ∈ I) is a family of closed sets, thenT
i∈IUi∈ CX;
(iii) If U, V ∈ CX then U ∪ V ∈ CX.
A clopen set is a set which is both open and closed. The clopen subsets of a space constitutes a Boolean algebra.
There is an alternative way to define a topology, by first defining a neighbourhood system, and then open sets as those sets containing a neighbourhood of each of their points.
Definition 2.2 Let X be a set. A neighbourhood system on X is the assignment of a family Nx
of subsets ofX to each x ∈ X, such that 1. x ∈ U for every U ∈ Nx;
2. U, V ∈ Nx⇒ U ∩ V ∈ Nx; 3. U ∈ Nx∧ U ⊆ V ⇒ V ∈ Nx;
4. (∀U ∈ Nx)(∃V ∈ Nx) V ⊆ U ∧ (∀y ∈ V ) V ∈ Ny.
Let X be a space. A set U is a neighbourhood of x (w.r.t. the topology OX) if there exists an open V such that x ∈ V ⊆ U . We denote by N Oxthe set of all neighbourhoods of x. An open neighborhoodof x is any neighbourhood U of x such that U is open.
N OX = (N Ox: x ∈ X) denotes the family of all neighbourhoods of a space X.
Proposition 4 1. IfX is a space, then the family N OX = (N Ox : x ∈ X) of neighbour- hoods of a point (w.r.t. the topologyOX) constitutes a neighbourhood system on X.
2. IfN = (Ny : y ∈ Y ) is a neighbourhood system on a set Y , then the family τN of setsU such that
∀y(y ∈ U ⇒ U ∈ Ny) constitutes a topology onY .
Both definitions are compatible:
(i) If(X, OX) is a space, then the topology τN O obtained from the neighbourhood system N OX = (N Ox: x ∈ X) as in item (2) is the original one OX (i.e., τN OX = OX).
(ii) starting out from a neighbourhood systemN = (Ny : y ∈ Y ), we get a space (Y, τN) such thatN OτN = N .
2.1 Examples of topological spaces
Example 5 (Euclidean topology on R) Let R be the set of real numbers. Given a < b ∈ R, we denote by(a, b) = {x ∈ R : a < x < b} the interval of all reals between a and b. The family of all setsU ⊆ R satisfying the following property
(∀x ∈ U )(∃a∃b) (a < b) ∧ x ∈ (a, b) ∧ (a, b) ⊆ U
constitutes a topology on R. It is the usual euclidean topology of real line. The following are examples of open sets: an interval(a, b); an infinite interval (a, +∞); The following are exam- ples of closed sets: a closed interval[a, b]; a closed infinite interval [a, +∞). An interval [a, b) is neither open neither closed.
Let R be the real line with the euclidean topology. Then we have: U ∈ N O0is a neighbour- hood of the real0 iff there are two reals r, s > 0 such that the open interval (−r, +s) ⊆ U . For example, the setU = {x ∈ R : x ≥ −1} is a neighbourhood of the real 0.
Example 6 (Euclidean topology on the three-dimensional space) Let R3be the three-dimensional space. Given a pointa = (a0, a1, a2) and r > 0, we denote by B(a, r) = {(x0, x1, x2) ∈ R3 : x20+ x21+ x22 < r} the open sphere of radius r. The family of all sets U ⊆ R3 satisfying the following property
(∀x ∈ U )(∃a ∈ R3∃r > 0) x ∈ B(a, r) ∧ B(a, r) ⊆ U
constitutes a topology on R3. It is the usual euclidean topology of the three-dimensional space.
The following are examples of open sets: an open sphereB(a, r); an infinite semispace {(x0, x1, x2) : x0 > 0}; The following are examples of closed sets: a closed sphere {(x0, x1, x2) ∈ R3 : x20+ x21+ x22≤ r}; a closed infinite interval {(x0, x1, x2) : x0≥ 0}.
Example 7 (Discrete topology) Let X be a set. Then (X, P(X)), where P(X) is the set of all subsets ofX, is a topological space. Every subset of X is open. This topology is called the discrete topology. Every subset of X is clopen. The discrete topology is generated by the family of singleton sets({x} : x ∈ X).
Example 8 (Indiscrete topology) Let X be a set. Then (X, {∅, X}) is a topological space. This topology is called theindiscrete topology. The only open sets are the empty set and the set X.
Example 9 (Cofinite topology) Let X be a set. A subset Y of X is cofinite if X \ Y is finite. The family of all cofinite subsets ofX constitutes the so-called cofinite topology of X.
Example 10 (Alexandrov topology) Let X = (X, ≤) be a poset. The family of all upper sets of X is a topology, called the Alexandrov topology.
Example 11 (Sierpinski topology) Let 2 = {0, 1}. We can define two interesting topologies on 2:
• The Sierpinski topology, whose open sets are: ∅, {1}, {0, 1}.
• The co-Sierpinski topology, whose open sets are : ∅, {0}, {0, 1}.
The Sierpinski topology coincides with the Alexandrov topology if we order the set2 as follows:
0 < 1. The co-Sierpinski topology coincides with the Alexandrov topology if we order the set 2 as follows:1 < 0.
Let2 be the Sierpinski space. Then we have: U ∈ N O0iffU = {0, 1}. In other words, the full space is the only neighbourhood of the point0. U ∈ N O1iff eitherU = {0, 1} or U = {1}.
Example 12 (Scott topology) Let (X, ≤) be a poset. The Scott topology is the topology, with the following closed sets:Y ⊆ X is Scott closed if Y is a down set, and, for every directed set A ⊆ Y such that tA exists, we have tA ∈ Y .
Y ⊆ X is Scott open if Y is an upper set, and, for every directed set A ⊆ X such that tA exists, we have:tA ∈ Y ⇒ Y ∩ A 6= ∅.
2.2 Metric spaces
The euclidean topology on Rnis defined by a distance or metric.
Definition 2.3 A metric on a set X is a map dX : X × X → R+ satisfying the following properties:
(i) dX(x, y) = 0 iff x = y;
(ii) dX(x, y) = dX(y, x);
(iii) dX(x, y) ≤ dX(x, z) + dX(z, y).
If x ∈ X and r > 0, then B(x, r) = {y : dX(x, y) < r} is the open ball of center x and radius r.
Example 13 (Metric topology) Let (X, dX) be a metric space. The topology induced by the metricdXonX is defined as follows:
U ⊆ X is open iff, for every x ∈ U there exists r > 0 such that B(x, r) ⊆ U .
The euclidean topology on R of Example 6 is induced by the metric dR(x, y) = |x − y|, where |x − y| is the absolute value of x − y.
The euclidean topology on R3is induced by the metric dR3(x, y) =p
(x0− y0)2+ (x1− y1)2+ (x2− y2)2, where x = (x0, x1, x2) and y = (y0, y1, y2).
Example 14 (Metric on strings) Let A be a finite alphabet, A∗ be the set of finite strings of alphabet A and the empty string. We define a metric d on A∗. Given two distinct strings α, β ∈ A∗we define:
d(α, β) = 2−k,
wherek is the number of states of the least automata distinguishing α and β.
We recall that a deterministic automata (without initial state and final states) is a triple A = (Q, A, δ), where Q is a finite set of states, A is an alphabet and δ : Q × A → Q is a map.
The mapδ can be extended by induction to a map δ∗ : Q × A∗→ Q, where α ∈ A∗anda ∈ A:
δ∗(q, ) = q; δ∗(q, αa) = δ(δ∗(q, α), a).
The stringsα, β ∈ A∗are indistinguishable by the automaA iff δ∗(q, α) = δ∗(q, β) for every q ∈ Q.
For example, letA = {0, 1} be the bits. What is the distance d(00, 11)? Let Q = {q, q0} be a set of two states. Defineδ(q, 0) = q, δ(q, 1) = q0,δ(q0, 0) = q0andδ(q0, 1) = q0. Then δ∗(q, 00) = q 6= δ∗(q, 11). It follows that d(00, 11) = 1/4.
Lemma 15 Let (X, dX) be a metric space. Then for all x 6= y ∈ X there exists r > 0 such that B(x, r) ∩ B(y, r) = ∅.
Proof. Define r = dX(x, y)/2.
The following is an example of topology which cannot be induced by a metric.
Example 16 Let N be a the set of natural numbers. We denote by [n) = {x ∈ N : x ≥ n} the set of naturals greater than or equal ton. Then (N, {∅} ∪ {[n) : n ∈ N}) is a space, whose topology cannot be induced by a metric. The intersection of two nonempty open sets is always nonempty:[n) ∩ [k) = [max(k, n)) is infinite. For example, [5) ∩ [2) = [5). Then the conclusion follows from Lemma 15.
2.3 Bases
Let X be a space. A family B of open sets is a base (subbase) if every open set OX is union of elements of B (is union of finite intersections of elements in B).
Example 17 The collection of all open intervals (a, b) (a, b ∈ R) in the real line forms a base for the euclidean topology, because (i) the intersection of any two open intervals is itself an open interval or empty; (ii) every open is union of intervals.
The collection of all semi-infinite intervals of the forms(−∞, a) and (a, +∞), where a is a real number, form a subbase. Any interval(a, b) = (a, +∞) ∩ (−∞, b).
A base is not unique. Many bases, even of different sizes, may generate the same topology.
For example, the open intervals with rational endpoints are also a base for the euclidean real topology, as are the open intervals with irrational endpoints, but these two sets are completely disjoint and both properly contained in the base of all open intervals.
Definition 2.4 A family B of subsets of a set X is called a base if it satisfies the following two conditions:
1. For everyx ∈ X, there is B ∈ B such that x ∈ B;
2. IfA, B ∈ B and x ∈ A ∩ B, then there is C ∈ B such that x ∈ C ⊆ A ∩ B.
If B is a base on X then B generates the following topology: U is open iff there exists D ⊆ B such that U =S D.
Any family B of subsets of a set X, such thatS B = X, generates a topology on X. In fact, B generates the base constituted by the sets B1∩ · · · ∩ Bn(Bi ∈ B).
Example 18 Let Z be the set of all integers. We define a topology on Z to provide a topological proof that there exist infinite prime numbers. Forb > 0 and 0 ≤ a < b, let
Nab= {x ∈ Z : x ≡ a (mod b)} = {a, a ± b, a ± 2b, . . . , a ± kb, . . . }.
In other words,Nabis the set of all integers that are congruent toa modulo b.
• The sets Nab(b > 0, 0 ≤ a < b) constitute a subbase and generate a base for a topology on Z. An element of the base is the set Nab11∩ · · · ∩ Nabnn. This set is either empty or infinite for the Chinese remainder theorem:x ≡ a1(mod b1); . . . ; x ≡ an(mod bn) admits a solution iffai ≡ ajmodgcd(b1, . . . , bn) for all i, j. If e is the least positive solution, then the set of all solution isNelcm(b1,...,bn).
• All elements of the base are clopen. The complement of Nabis equal to Z \ Nab= ∪0≤i6=a<bNib.
• Except for −1, +1 and 0, all integers have prime factors. Therefore each is contained in one or moreN0,p, where p is prime. We thus arrive at the following identity, whereP is the set of prime numbers: Z \ {−1, +1} = ∪p∈PN0,p. If the setP were finite, then the right hand side would be closed as the union of a finite number of closed sets. Then the set{−1, +1} would be open as a complement of a closed set. This would contradict that every open set is infinite.
2.4 Interior, Closure, adherent and limit point
Let X be a space.
1. The interior ˚A of a set A ⊆ X is the largest open subset of A:
A =˚ [
{U : U ∈ OX ∧ U ⊆ A}.
We often write int(A) for ˚A.
Example 19 • In any space, the interior of the empty set is the empty set.
• If X is the Euclidean space R of real numbers, then int[0, 1] = (0, 1).
• If X is the Euclidean space R, then the interior of the set Q of rational numbers is empty.
• If X is the complex plane R2, thenint({(x, y) ∈ R2 : x2+ y2 ≤ 1}) = {(x, y) ∈ R2: x2+ y2< 1}.
• In the Euclidean space, the interior of any finite set is the empty set.
• If one considers on R the discrete topology in which every set is open, then int([0, 1]) = [0, 1].
• If one considers on R the indiscrete topology in which the only open sets are the empty set and R itself, then int([0, 1]) is the empty set.
2. The closure A of a subset A of X is the intersection of all closed sets containing A:
A =\
{U : U ∈ CX ∧ A ⊆ U }.
({x} will be denoted by x). We sometimes write cl(A) for A.
Proposition 20 A is the set of all points y such that U ∩A 6= ∅ for every open set U ∈ Ny. Proof. If y ∈ A \ A, y ∈ U open and U ∩ A = ∅, then A ⊆ X \ U . The set X \ U is closed, contains A but it does not contains y. We get the contradiction y /∈ A.
Example 21 • Consider a sphere in 3-dimensions. Implicitly there are two regions of interest created by this sphere; the sphere itself and its interior (which is called an open3-ball). It is useful to be able to distinguish between the interior of 3-ball and the surface, so we distinguish between the open 3-ball, and the closed 3-ball - the closure of the 3-ball. The closure of the open 3-ball is the open 3-ball plus the surface.
• In any space, ∅ = cl(∅).
• In any space X, X = X.
• If X is the Euclidean space R of real numbers, then (0, 1) = [0, 1].
• If X is the Euclidean space R, then the closure of the set Q of rational numbers is the whole space R. We say that Q is dense in R.
• If X is the complex plane R2, thencl({z ∈ C : |z| > 1}) = {z ∈ C : |z| ≥ 1}.
• If S is a finite subset of a Euclidean space, then S = S.
• If one considers on R the discrete topology in which every set is closed (open), then (0, 1) = (0, 1).
• If one considers on R the indiscrete topology in which the only closed (open) sets are the empty set and R itself, then (0, 1) = R.
• In the Sierpinski space 2 (see Example 11) we have 0 = {0} and 1 = {0, 1}.
These examples show that the closure of a set depends upon the topology of the underlying space. The closure of a set also depends upon in which space we are taking the closure.
For example, if X is the set of rational numbers, with the usual relative topology induced by the Euclidean metric, and ifS = {q ∈ Q : q2 > 2, q > 0}, then S is closed in Q;
however, the closure ofS in the Euclidean space R is the set of all real numbers greater than or equal to√
2. Recall that√
2 is irrational.
3. A point y is a adherent to A if (U ∩ A) 6= ∅ for every U ∈ Ny.
4. A point y is a limit point of A if (U ∩ A) \ {y} 6= ∅ for every U ∈ Ny. Every point in A \ A is a limit point. The closure of A is the union of A and its limit points.
Let A = {(x, y) : x2+ y2 < 1} ∪ {(2, 0)} in the plane with the Euclidean topology.
Then A = {(x, y) : x2+ y2 ≤ 1} ∪ {(2, 0)}, while the set of limit points is equal to {(x, y) : x2+ y2≤ 1}.
5. The boundary (or frontier) of A is the closed set A ∩ X \ A.
For example, the frontier of {(x, y) : x2+ y2 < 1} in the euclidean plane is the set {(x, y) : x2+ y2= 1}.
3 Continuous Functions
Definition 3.1 Let X and Y be spaces. A function f : X → Y is continuous at point x0∈ X if, for every neighbourhoodU of f (x0), there exists a neighbourhood V of x0such thatf [V ] ⊆ U . A functionf : X → Y is continuous if it is continuous at any point of X.
Proposition 22 The following conditions are equivalent for a function f : X → Y : 1. f is continuous;
2. For every open setU ∈ OY , f−1[U ] ∈ OX.
Proof. (1) ⇒ (2): Let U ∈ OY be open and let x ∈ f−1[U ]. Then f (x) ∈ U . By (1) there is a neighbourhood Vxof x such that f [Vx] ⊆ U . By definition of neighbourhood, there exists an open set Vx0⊆ Vxsuch that x ∈ Vx0. It follows that
f−1[U ] =[
{Vx0: x ∈ Vx0∈ OX, f [Vx0] ⊆ U } is open because it is union of open sets.
(2) ⇒ (1): If U is a neighbourhood of f (x0), then there exists an open set U0 such that f (x0) ∈ U0⊆ U . By (2) we have that f−1[U0] ∈ OX and f [f−1[U0]] ⊆ U .
Proposition 23 The constant functions and the identity function are continuous. Continuous functions are closed under composition.
If B is a base of the topology on Y , then f : X → Y is continuous if and only if, for every open set U of the base B, f−1[U ] ∈ OX.
Example 24 Let N be the set of natural numbers with the Alexandrov topology (i.e., the open sets are the infinite intervals[n)). Is the function f (x) = x2continuous from N into N? YES.
For example, isf−1[ [57) ] open? YES. We have:
f−1[ [57) ] = {x ∈ N : x2≥ 57} = {x ∈ N : x ≥ 8} = [8) In a similar way, we have thatf−1[ [n) ] is open for every n. Then, f is continuous.
Example 25 The function f : R → R, defined by f (x) = x2+ x + 2, is continuous at 0.
We havef (0) = 2. Consider the open interval (2 − , 2 + ) with 0 < < 1. Then we have 2− < x2+x+2 < 2+ for every x ∈ (−/2, +/2), because x2+x+2 < (/2)2+/2+2 =
2/4 + /2 + 2 < /4 + /2 + 2 = 2 + 3/4 < 2 + . Similarly for the other direction. The functionf is continuous in every point of the real line.
3.1 Continuous functions of metric spaces
If X, Y are metric spaces, then f : X → Y is continuous at x0if for all > 0 there exists δ such that dX(x, x0) < δ implies dY(f (x), f (x0)) < .
Proposition 26 If (X, d1) and (Y, d2) are metric spaces, then every isometry f : X → Y (i.e., d2(f (x1), f (x2)) = d1(x1, x2)) is continuous.
Proof. Every isometry is injective: if f (x1) = f (x2) then d2(f (x1), f (x2)) = 0 = d(x1, x2), that is, x1= x2.
The balls B(y, ) with y ∈ Y and > 0 constitute a base of the topology of the metric space Y . If f is a isometry, then
f−1[B(y, )] = [
x∈X,f (x)=y
B(x, ).
is open.
Example 27 An isometry of the euclidean plane is either a rotation or a translation or a reflec- tion with respect to a line.
3.2 Continuous functions of ordered spaces
Proposition 28 Let (X, ≤X) and (Y, ≤Y) be posets. Then a function f : X → Y is continuous w.r.t. the Alexandrov topology if and only iff is monotone:
∀x∀y(x ≤X y ⇒ f (x) ≤Y f (y)).
4 The construction of spaces from given spaces
4.1 Subspaces
If X is a space and Y ⊆ X then the family of sets (U ∩ Y : U ∈ OX) is the topology induced by the space X on the subset Y . Example 29 Consider the euclidean plane and the square
Q = {(x, y) : x, y ∈ [0, 1]}.
The set{(x, y) : x2+ y2< 1/2, x, y ∈ [0, 1]} = Q ∩ B(0, 1/2) is open in Q with respect to the induced topology.
Example 30 The Sierpinski space 2 induces:
(i) the discrete topology on{1};
(ii) the indiscrete topology on{0}.
4.2 Product topology
Let X and Y be spaces. The product topology on X × Y is the topology generated by the following base:
{U × V : U ∈ OX ∧ V ∈ OY }.
In general, if we have a family (Xi : i ∈ I) the product topology onQ
i∈IXi is the topology generated by the following base:
{Y
i∈I
Ui: ∃J ⊆finI (i ∈ J → Ui∈ OXi) ∧ (i ∈ I \ J → Ui= Xi}.
Example 31 The Scott Topology on P(N) is the product topology of Sierpinski space. Let N be the set of natural numbers. Then the powerset P(N) of N is in bijective correspon- dence with the set2N of all sequences (an : n ∈ N). For example, the set of even num- bers is related to the sequence (1, 0, 1, 0, 1, 0, . . . ). The set of all squares to the sequence (1, 1, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1 . . . ). The product of the Sierpinski topology on 2 = {0, 1} is generated by the positive information. For example, {A ⊆ N : 5, 3 ∈ A} is base open.
5 Separation
Separation axioms in topology stipulate the degree to which distinct points may be separated by open sets or by closed neighborhoods of open sets, while connectedness axioms in topology examine the structure of a topological space in an orthogonal way with respect to separation axioms. They deny the existence of certain subsets of a topological space with properties of separation.
5.1 Specialisation preorder
Let X be a space. The specialization preorder ≤X is defined by:
x ≤Xy iff (∀U ∈ OX) x ∈ U ⇒ y ∈ U.
The open sets are upward closed w.r.t. ≤X, while the closed sets are downward closed.
Example 32 • Sierpinski topology on 2: we have 0 ≤ 1.
• Euclideal topology: we have x ≤ y iff x = y.
• Product Topology on 2Nof Sierpinski topology:A ≤ B iff A ⊆ B.
5.2 T
0-spaces
A space X is T0iff ≤X is a partial order. This means that there are no x and y with the same system of neighborhoods.
The Scott topology (see Example 12) is T0but not T1.
Proposition 33 1. Every countable basedT0-spaceX can be embedded into P(N) equipped with Scott topology.
2. Every continuous mapf : X → Y between subspaces X and Y of P(N) admits a Scott continuous extensionF : P(N) → P(N).
Proof. (1) If (Un : n ∈ N) is an enumeration of the basis of X then the map x ∈ X → f (x) = {n : x ∈ Un} is an embedding. Then, for example, f−1({0}↑) = U0.
(2) Define F x = S{T{f (z) : z ∈ X ∩ y↑} : y ⊆f x}, where y ⊆f x means that y is a finite subset of x.
5.3 T
1-spaces
A space X is T1if, for all x, y ∈ X, there exist open sets U and V such that U ∩ {x, y} = {x}
and V ∩ {x, y} = {y}.
Theorem 34 The following conditions are equivalent for a space X:
• X is T1.
• For every x ∈ X, the intersection of all neighborhood of x is the singleton set {x}.
• Every point x of X is closed.
• The specialization preorder of X agrees with the equality relation.
If X is T1and x is an accumulation point of A, then U ∩ A is infinite for every neighborhood of x. The set of accumulation points of A is closed.
Example 35 Let T be a recursively enumerable lambda theory. The Visser topology over Λ/T isT1but notT2.
Example 36 Let X be a countable set. We define A ⊆ X open if X \ A is finite. This topology, called thecofinite topology, is T1but notT2.
5.4 Hausdorff spaces (or T
2-spaces)
A space X is T2or Hausdorff if, for all x, y ∈ X, there exist open sets U and V such that x ∈ U , y ∈ V and U ∩ V = ∅.
Theorem 37 The following conditions are equivalent for a space X:
• X is T2.
• For every x ∈ X, the intersection of all closed neighborhood of x is the singleton set {x}.
• The diagonal ∆ = {(x, x) : x ∈ X} is a closed subset of X × X.
• Every point x ∈ X has a closed neighborhood, which is T2w.r.t. the induced topology.
Example 38 ([?, Example 75]) The irrational slope topology is T2but notT21/2.
5.5 Completely Hausdorff spaces (or T
21/2-spaces)
A space X is T21/2(or completely Hausdorff) if for all distinct a, b ∈ X there exist open sets U and V with a ∈ U , b ∈ V and Uc∩ Vc= ∅.
The previous axioms of separation can be relativized to pairs of elements. For example, a and b are T21/2-separable, if there exist open sets U and V with a ∈ U , b ∈ V and Uc∩ Vc = ∅.
T2-, T1-, T0-separability for pairs of elements are similarly defined.
Example 39 ([?, Example 78]) The half-disc topology is T21/2 but notT3. LetP = {(x, y) : x, y ∈ R, y > 0} be the upper half-plane and let L = {(x, 0) : x ∈ R} be the real axis. We generate a topology onP ∪ L by defining a basis of neighborhoods: a basis element for an elementp ∈ P is an open disc contained in P , while a basis element for an element a ∈ L consists of an open half-disc centered ata together with a itself. For example, Z = {(x, y) : (x − a)2+ y2 < , y > 0} ∪ {(a, 0)} is a basis element for (a, 0). The closure of Z is the set Zc = {(x, y) : (x − a)2+ y2 ≤ , y ≥ 0}. It is now easy to show that the half-disc topology isT21/2. The half-disc topology is notT3:(P ∪ L) − Z is a closed set. Every neighborhood of (P ∪ L) − Z intersects every neighborhood of (a, 0).
5.6 Regular spaces (or T
3-spaces)
A space X is regular (or T3) if it is T1and it holds one of the following equivalent conditions:
• For every closed set A and for every x /∈ A, there exists disjoint open sets U and V such that A ⊆ U and x ∈ V .
• The closed neighborhoods of a point constitute a fundamental system of neighborhoods.
Example 40 Let R be the set of real numbers. The topology generated by the open intervals (a, b) and the sets (a, b) ∩ Q, where Q is the set of rational numbers, is T2but notT3. The set Q is open, while the setF = {π/n : n ≥ 1} of irrational numbers is closed. The point 0 and the closed setF cannot be separated.
5.7 Normal spaces (or T
4-spaces)
A space X is normal (or T4) if it is T1and it holds the following condition: For all disjoint closed sets A and B, there are disjoint open sets U and V such that A ⊆ U , B ⊆ V .
Example 41 The Euclidean line is T4.
5.8 Totally separated spaces
A space is totally separated if arbitrary distinct points are separated by a clopen set.
6 Connected spaces
The notion of a connected space is orthogonal to that of totally separated space.
6.1 Connected spaces vs totally separated spaces
A space is connected if there are not distinct points separated by a clopen set (in other words, if there are no nontrivial clopen sets or X is not the union of two nonempty disjoint open sets).
Theorem 42 The following conditions are equivalent for a space X:
1. X is connected.
2. X is not the union of two nonempty disjoint closed sets.
3. IfX = A ∪ B with A, B arbitrary nonempty sets, then (A ∩ B) ∪ (A ∩ B) 6= ∅.
Proof. (3 ⇒ 1) Obvious.
(1 ⇒ 3) Assume that there exists nonempty sets A and B such that X = A ∪ B, (A ∩ B) = ∅ and (A ∩ B) = ∅. We now show that A and B are disjoint. If x is adherent to both A and B, then by X = A ∪ B we have that, for example, x ∈ A. Then x belongs to the open set X \ B, so that it is not adherent to B. In conclusion, X = A ∪ B and A ∩ B = ∅. This contradicts the hypothesis of connectedness.
A subset Y of X is connected if Y is connected as a subspace of X.
Proposition 43 The following properties hold:
1. The closure of a point is connected.
2. IfA is a non-trivial clopen of a space X, then either Y ⊆ A or Y ⊆ X \ A, for every connected subsetY of X.
3. The union of arbitrary connected subsets, having pairwise nonempty intersection, is con- nected.
4. IfY is connected, then the closure Y of Y is also connected.
Proof. (1) If there are two open sets A and B such that x = (A ∩ x) ∪ (B ∩ x) and (A ∩ x) ∩ (B ∩ x) = ∅, then, without loss of generality, we may assume that x ∈ (A ∩ x). Then, every point y ∈ (B ∩ x) cannot be in the closure of x, so that B ∩ x = ∅. Contradiction.
(3) Let Y = ∪i∈IYi. If Y = A ∪ B with A, B open and disjoint, then by (2) we get Yi ⊆ A for every i or Yi⊆ B for every i. Contradiction.
(4) Let Y = A ∪ B, where A and B are open in the subspace Y with A ∩ B = ∅. Since Y is connected, then we have that, for example, Y ⊆ A and B ⊆ Y \ Y . If B is nonempty, this contradicts B ∩ Y = ∅.
6.2 Connected components
• The connected component of x ∈ X is the greatest connected subset containing x. The connected component of x contains the closure of x. The connected component of x is closed. Connected components constitute a partition of X. If there is a unique connected component then the space is connected.
• Each clopen set contains the connected components of all of its points. The quasicom- ponentof x ∈ X is the intersection of all clopen sets containing x. Quasicomponents constitute a partition of X. If there is a unique quasicomponent then the space is con- nected.
• A path (arc) is a (one-to-one) continuous function from the unit interval into X. Two points are path (arc) connected if there is a path (arc) f such that f (0) = a and f (1) = b.
A space is path (arc) connected if two arbitrary points are path (arc) connected. A path (arc) component is a maximal subset with respect to path (arc) connectedness.
We have:
Arc component ⊂ Path component ⊂ Connected component ⊂ Quasicomponent.
Example 44 ([?, Example 12.point13]) Let X be a set and p ∈ X be a point. The particular point topology onX is constituted by all the sets containing the particular point p. This topology is path connected (ifq 6= p then f (1) = q and f ([0, 1)) = p is a path), but it is not arc connected (the inverse image of the open set{p} w.r.t. a one-to-one continuous map would be one point, which is not open in the interval[0, 1]).
