math 404 lecture notes

7/29/2019 Math 404 Lecture Notes

1/31

Math 404a/576a, Fall 2007

General TopologySynopsis: This course on general topology covers the following:

1. Topological spaces: basic definitions, operations on subsets, neighbour-hoods, bases

2. Constructions for top. spaces: Subspaces, quotient spaces, productspaces

3. Properties of top. spaces: Connectedness, compactness, countability,separation axioms

4. Function spaces

Lecture 1: Topological Spaces - Basic Definitions (Refer to Munkres12,13)

Motivation: The idea of a topological space is intended to capture intu-itions about properties of space which do not depend on distance, angles, etc.The key idea is an open set, which can be thought of as a neighborhoodsaying which points are near each other.

Definition 1 A topological space is a setX together with a topology T,a collection of subsets of X such that:

1. T includes the empty set and X itself

2. If O1, . . . , On are in T, so is the intersection

ni=1Oi

3. For any subset S T, so also the union S T

The sets included in T are called the open sets of the topology.

The concept of an open set is a generalization of the concept of an openinterval (or union of open intervals) in the real line. That is, they tell uswhich points should be thought of as near each other in some sense. Later,we will see that the open intervals actually give a topology. For now, letslook at some more elementary examples.

1


2/31


3/31

Lecture 2: Topological Spaces -Definitions and Examples (Refer to

Munkres 12,13)

Recall: Last time we defined a topological space as a set X with a topology- which is a collection of subsets of X satisfying some axioms. We also saidwhat it meant for one topology to be coarser or finer than another. All ofthese concepts are related to that of a partial order.

Definition 3 Given a set S, a partial order relation on S is a relation satisfying:

1. s s for all s S (reflexivity)

2. s t and t s only when s = t (antisymmetry)

3. If r s and s t then r t (transitivity)

If either s t or t s, then we say s and t are comparable.

Given any set X, there is a partial relation on S = P(X), the set of allsubsets of X, given by defining s t if and only if s t (the containmentrelation). This relation is inherited by any subset of P(X).

In particular, a topology on the set X is a subset ofP(X), so the contain-ment relation makes the collection of open sets of the topology into a partialorder.

There is also a partial order on topologies themselves - inherited fromthe partial order on P(P(X)), since each topology is a subset of P(X)! Therelation that says topology T1 is finer than topology T2 means that T2 iscontained in T1.

A topology is a special kind of partial order - not any subset ofP(X) willdo, since the axioms of a topology imply that it has some special properties:

1. There is a maximal and a minimal element (X and ).

2. Given any finite collection of elements (i.e. open sets), there is a uniquegreatest lower bound - a largest element which is less than (con-tained in) all of them. (Namely, the intersection).

3


4/31

3. Given any collection of elements, there is a unique least upper bound

- a smallest element which is greater than (contains) all of them. (Namely,the union).

Notice again that the definition of a topology is not symmetric! We onlyexpect to find finite intersections, but we can hope for infinite unions. In alater exercise, youll investigate properties like these for the partial order ofall topologies on X.

Why is this strangely asymmetric definition the interesting one? Wellsee the motivation once we have a better way to describe some more familiarexamples.

Motivation: A topological space usually has too many open sets to list,or even conveniently describe, since any union of open sets is open. Instead,one often gives a simpler collection of open sets, so that all others can befound from these.

Definition 4 A basis for a topology on X is a collection B P(X) ofsubsets of X, with:

1. Every x X is in some B B

2. For any x B1 B2, there is some B3 with x B3 B1 B2 (allBi B)

Given a basis B, the topology generated by B consists of all sets O Xwith the property that x O implies there is a B B with x B O.

The idea is that a basis for a topology provides enough open sets so thatevery other open set in the topology is given as a union of sets in the basis.

Example 2 See handout A. The given B is a basis for the topology T. A

basis for the discrete topology on any setX is the collection of all one-elementsubsets of X. A basis for the indiscrete topology on X contains just X itself.

Exercise 3 Check these claims.

A more familiar example is the usual topology on the line:

4


5/31

Example 3 Let X = be the real line, and B be the collection of all open

intervals, (a, b) for a, b .

This is the basis for a topology on , since:

1. Every real number x is in some interval in - for instance, the interval(x 1, x + 1).

2. If x I1 I2 where I1 = (a1, b1) and I2 = (a2, b2), then we musthave that a2 < b1 (by transitivity). So there is a well-defined intervalI = (a, b), where a = max(a1, a2) and b = min(b1, b2). It is easy tocheck that x I and that I I1 I2.

The topology this generates is the usual one on , as we will see later.We will also see that this is a special example of the order topology on atotally ordered set.

A similar example deals with the Euclidean plane:

Example 4 Let X be the plane 2, and B be the set {D((a, b), r)|x, y , r +} where D((a, b), r) = {(x, y)|(x a)2 + (y b)2 < r2}. ThenB is

the basis for a topology on X.

Exercise 4 Check this. Also: check that the corresponding definition worksfor any n.

We will see later that this last example is also special case - in this case,of the metric topology on a metric space.

Remark: In this last example, the topology generated by B consists ofall sets O 2 such that, for any x O implies there is an open disc ofsome radius, containing x, and contained in O. This agrees with the usualdefinition of an open set.

The basic idea is that the description of any open set has some leeway

in it - if a point is in that set, then sufficiently close points are also in theset. In the case where there is a metric to define close, this is exactly right.If there is not, a basis for a topology gives the same information.

These standard examples also help to answer our earlier question - whyis the definition of a topology asymmetric?

5


6/31

Claim: Consider 2 with the topology generated by open discs. There

is a countable collection of open sets in 2 whose intersection is not an openset. Proof: Take any point (x, y) 2. Then take the collection of sets

On = D((x, y),1

2n)

the disc of radius 12n

centred at (x, y).Then the intersection of all these discs is

nOn = {(x, y) 2|(x x)2 + (y y)2 x}

contains x2. These are obviously disjoint. Since the xi were arbitrary, O isHausdorff.

22


23/31

Example 18 By the above theorem, is a Hausdorff space. By a similar

argument, n with the metric topology is also Hausdorff, since we can findopen balls with radius less than 1

2(d(x1, x2)) around x1 and x2 which will be

disjoint.

Exercise 16 Refer to handout A again. Check that X is not Hausdorff.

In fact, not even every space that appears in real applications is Hausdorff:

Example 19 The Zarizki topologyon 2 is defined as follows. The closedsets are all sets of the form:

{(x, y)|f(x, y) = 0, f I}where I is some set of algebraic functions (e.g. I could be the set {f1(x, y) =x2 + y3, f2(x, y) = (x

3 y5)(x y)}).

The Zariski topology is important in algebraic geometry. In fact, it canbe defined in many more settings than just 2, or n - this definition makessense anywhere there is a concept of algebraic function.

Exercise 17 For HWK Due Oct 2: Check that the Zariski topology on 2

is a topology, and that it is not Hausdorff.

Hausdorff spaces have useful properties:

Theorem 12 If X is a Hausdorff space, every finite set S X is closed.

Proof: Since a finite union of closed sets is closed, we only need to checkthis for single points. But then, since X is Hausdorff, given a point x0, andany other point x, we can find disjoint open neighborhoods U and V of thetwo points. Since V around x does not include x0, we know that x is not alimit point of the set {x0}. Since this works for any x, {x0} must be closed.

Again, we really only used the fact that, given x0 and x, there is an openneighborhood V of x which does not include x0: the neighborhood U of x0plays no role here. The condition that such a V exists for any x and a fixedx0 is called the T1-axiom: it is one of the separation axioms, which wewill say more about later on.

23


24/31

Weve talked about the concept of a limit point of a set - this is related to

the notion of convergence, which we can also speak of in a general topologicalspace:

Definition 18 A sequence of points x1, x2, . . . is said to converge to a pointx if, for every open neighborhood U of x, elements of the sequence eventuallylie in U.

Theorem 13 If X is a Hausdorff space, then any sequence of points {xn}converges to at most one point of X.

Proof: Suppose {xn} converged to more than one point, say to x and y.Then since X is Hausdorff, there are open neighborhoods U of x and V of ywhich are disjoint. By assumption, there is some NX for which, if n > NX ,we must have xn U. Similarly, there is some NY so that if n > NY wehave xn V. But then if n > max(NX , NY), we have xn U V, which isimpossible since they are disjoint.

Notice that this may not be true in a non-Hausdorff space (see, e.g.Soin a Hausdorff space, if a sequence converges to anything, it converges toa unique point, and we can speak of the limit of the sequence, and writexn x, or x = limn xn.

Exercise 18 Try: Munkres: p100, no. 3,8For HWK due Tues, Oct 2, do Munkres: p100, no. 5,9,13

Continuous Functions

Now that we have raised the subject of convergence of a sequence, youshould be reminded of matters in the calculus of functions of a real variable.Not all ideas in the calculus make sense in a general topological space, butone which does is the idea of continuous functions:

Definition 19 A continuous function(or continuous map) f : X Ybetween topological spaces X and Y is a function of the underlying sets, suchthat, if V Y is open, so is f1(V) X.

Continuous functions are the maps which are typical of topologicalspaces. Commonly when a mathematical structure is defined, there is a

24


25/31

typical kind of map which takes one such structure to another. For example:

linear maps between vector spaces; homomorphisms of groups or of rings,and so on. Often, the maps are the ones which preserve the structure ofthe objects involved.

Notice that the definition of a continuous function actually says that itis f1, rather than f, which must preserve the structure of a topologicalspace, in the sense that the image of an open set is an open set - but theimage under f1. A map for which f(U) is open for every open set U is anopen map - also a useful idea, but not to be confused with continuity.

Theorem 14 If the toplogy on Y is given by a basis B, then a function f is

continuous if and only if f1(B) is open for every B B.

Proof: Any open set is of the form

V =

B

So thenf1(V) =

f1(B)

so if all the f1(B) are open, then so is f1(V).

Example 20 Suppose X and X

denote two topologies on the same under-lying setX (i.e. this is a shorthand way of talking about(X, T) and (X, T))and suppose the topology onX is strictly finer than that onX. Then considerthe identity functions:

I : X X and I : X X

(both are the identity function on the underlying set). Then the first is notcontinuous, and the second is not! This is because the preimage of any set Vis just itself, but thought of in the other topology. Since the topology on X

is strictly finer, it has open sets which are not open in X, whereas any open

set in X is also open in X

.

There are several different, equivalent, statements which mean the sameas continuity of a function, and make sense in any topological space, andsome other common ones which only make sense in metric spaces. Thegeneral statements:

25


26/31

Theorem 15 If X and Y are topological spaces, and f : X Y is a func-

tion, the following are equivalent (TFAE):

1. f is continuous

2. A X, f(A) f(A)

3. B Y closed f1(B) X closed

4. x X, and V neighborhood of f(x), U neighborhood of x withf(U) V

Well prove the first three equivalent right now (Munkres does all four:

for the proofs of (4) (1) and (1) (4), see Munkres p105.Proof: (1) (2): Suppose f is continuous, and A X. We need to

show that if x (A), f(x) f(A) (the image of a limit point is a limitpoint). But suppose V is a neighborhood of f(x): since f is continuous,f1(V) is an open neighborhood of x in X. Since x A, this must intersectA at some y. But then f(y) f(A) V, so f(x) must be a limit point off(A), since V was arbitrary.

(2) (3): If B is a closed set in Y, we want to show A = f1(B) isclosed in X. We know A is closed iff A = A. Now, if x A, then by (2), wehave f(x) f(A) B since f(A) B. So then x A = f1(B). But this

means A = A. That is, f1

(B) is closed.(3) (1): Suppose the preimage of any closed set in Y is closed. To

show the preimage of any open set is open, suppose A is an open set in Y,with Y A = B. Then since f1(B) = f1(Y) f1(A) = X f1(A) isclosed, it must be that f1(A) is open.

Condition (2) here shows us how to understand continuous maps as mapswhich preserve structure in the sense we discussed before. That is continu-ous maps preserve limit points ! Of course, in order to define a limit point, wehad to have the concept of an open set - so we originally defined continuityin terms of open sets. By preserving limit points, we mean the familiar

idea that, for continuous real-valued functions f, we have

limn

f(xn) = f( limn

xn)

Moreover, condition (4) gives a condition a bit like the definition forcontinuous real-valued functions, without a metric.

26


27/31

Once one knows what the good maps are between topological spaces -

namely, the continuous ones - we can do things which may be familiar fromother settings. One of these is to define an isomorphism in the world oftopological spaces.

An isomorphism of sets is a bijection - a set function which has an in-verse. In the world of vector spaces and linear maps, we can talk aboutlinear isomorphisms: a linear map with a linear inverse. An isomorphism ofgroups, h, is a group homomorphism from A to B, for which there is anotherhomomorphism h1 from B to A which is its inverse. The correspondingnotion for topological spaces is this:

Definition 20 A homeomorphism is a continuous function f : X Ywith a continuous inverse. (This implies f is a bijection of the underlyingsets).

In other words, f gives a bijection of the underlying sets X and Y so thata subset A X is open iff f(A) Y is open. Now, a bijection f : X Ygives a corresponding bijection P(Y) P(X) (preimages of subsets of Yare all distinct). So a homeomorphism gives not only a bijection of theunderlying sets, but a bijection of the topologies as well: f : TX TY.

So any property of X which can be expressed in terms of open sets (e.g.X is Hausdorff, etc.) will be true of Y and vice versa. Properties which

are preserved by homeomorphism are topological properties. (In the world oftopological spaces, two spaces which are related by a homeomorphism are,for all practical purposes, the same.)

Example 21 The function f(x) = tan(x) is a continuous function from(

2, 2

) to . It has a continuous inverse, f1(y) = arctan(y). So this openinterval is homeomorphic to . It is easy to find a linear function showingthat any open interval is homeomorphic to any other, so any open interval ishomeomorphic to .

Exercise 19 For HWK due Tues, Oct 2:1. Show that n is homeomorphic to an open ball in n.

2. Do Munkres p111, no. 1,3,

27


28/31

Lecture 7: Cts Fns concluded; More on Product Topology

(Refer to Munkres 18,19)

Weve defined and characterized continuous functions, but well need toknow some facts about them, given here for reference without proof (but seeMunkres p108):

1. Constant functions are continuous

2. An inclusion function i : A X is continuous

3. Iff : X Y and g : Y Z are continuous, so is g f : X Z

4. If A X, and f : X Y is continuous, then f|A : A Y iscontinuous

5. IfX =

U, a function f : X Y is continuous if f|U : U Y is

continuous for each .

Having defined continuous functions, we might want to know how thisidea relates to the way we know how to build new topological spaces fromold ones - the product topology.

Theorem 16 If f : A X Y is given by coordinate functions f1 andf2 as f(a) = (f1(a), f2(a)), then f is continuous if and only if f1 : A Xand f2 : A Y are both continuous.

Proof: The product X Y has the two projection maps 1 : X Y Xand 2 : X Y Y, which are both continuous, since

11 (U) = U Y

which is open if U is (and similarly for 2). Now the coordinate functionsare just

fi(a) = i(f(a))Now if f is continuous, then these are composites of continuous functions,hence continuous.

On the other hand, if each fi is continuous, then we want to show that f1

of a basis element in the product topology is open in A. Such an element isUV for open sets U X and V Y. Now, a f1(UV) ifff(a) UV,

28


29/31

which means f1(a) U and f2(a) V. So f1(U V) = f11 (U) f

1

2 (V).

This is the intersection of two open sets, hence open.

This theorem is closely related to the universal property of the product incategory theory: a product of X and Y is an object X Y together withprojection maps 1 and 2 such that any other object with maps f1 and f2into X and Y has a unique map f into X Y with fi = i f. When weare talking about topological spaces, all these maps must be continuous.

Our aim in this lecture is to state a similar theorem for much biggerproduct spaces...

We can define a product topology on the product of more than two sets

- indeed, of an infinite collection of sets!

Definition 21 If {X}I is a collection of sets, and X =

X, then wecan define the cartesian product of the X to be:

I

X = {f : I X|f() X}

and the th projection map is a function

:

I

X X

given by (f) = f().

That is, the cartesian product is the set of all ways of choosing, for eachindex , an element of X (by the function f). The projection maps pickout the value of the function at some index. If I = {1, . . . , n} this agreeswith the usual idea that a product consists of n-tuples, and the projectionmaps pick out the component in some position. Now what about a topologyon this?

Definition 22 If the X in the above product are topological spaces, thenthe box topology is the one generated by the basis consisting of all sets

I

U

where U X is open for each .The product topology is the one generated by the subbasis consisting of

all sets of the form 1 (U for any , and U an open set is X.

29


30/31

Notice that we already proved that for the product of two sets, these are

the same. However, generating a topology from a subbasis means takingall unions of all finite intersections. So while the box topology is generatedby boxes - which are products of any choice of open sets in each X, theproduct topology is generated by a basis of boxes which are products ofthe whole space X for all but finitely many , and any open set in the rest.

The product topology is the usual one on a general product of topologicalspaces - although it is coarser, in general, for infinite collections.

Now here are some important facts about the box and product topologies:

Theorem 17 If the topology on the space X is given by a basis B, then abasis for the box topology is the collection of all sets

I

B where B B.A basis for the product topology is the collection of all sets of this form

where all but finitely many of the B are just X.

Exercise 20 Convince yourself that these are bases, and that they generatethe right topologies.

Next, we point out some results that show both the box and producttopologies get along with subspaces, closure, and the property of being Haus-

dorff:

Theorem 18 If A X for each , then

I

A

I

X

is a subspace if both have the box/product topology.

(That is, the subspace topology in the box topology is the box topologyon the subspace, and similar for the product topology.)

Exercise 21 Check this.

Theorem 19 A product of Hausdorff spaces is still a Hausdorff space ineither the box or product topologies.

Exercise 22 Check this.

30


31/31

Theorem 20 If A X for each , and the product of the X is given

either the box or product topology then the closures satsify:

A =

A

(See Munkres for a proof).

Finally, we have a theorem like the one we began with - and this onedistinguishes between the box and product topologies!

Theorem 21 If f : A

I X is given by component functions as:

f(a) = (f(a))I

where f : A X, and

IX has the product topology, then f is contin-

uous if and only if each f is continuous.

(The proof is similar to that for a product of two spaces - see also Munkresp117).

However, the analogous theorem is false for the box topology!

31

math 404 lecture notes

Documents