- Open and closed sets in Euclidean space
Rn.
- Details: A subset U of Rn is
open if for every point x in U, there is an
open ball around x which is completely contained in
U. More precisely, for every x in U, there
is an epsilon > 0 so that the set
Bepsilon (x) = {y : ||y-x|| < epsilon}
is contained in U. (Here, the double vertical bars mean the
length of the vector y - x.)
A subset A of Rn is
closed if its complement is not open.
After giving these definitions, I gave some examples. For instance,
the ``closed unit interval'' [0,1] is closed as a subset of
R; the ``open unit interval'' (0,1) is open as a
subset of R, but not when viewed as a subset
of R2.
- Countable and uncountable sets
- Details: A set X is countable if there is a
one-to-one onto function (i.e., a bijection)
f:Z --> X.
For example, the integers Z, the positive integers
Z+, and the rational numbers
Q are each countable.
A set X is uncountable if it is infinite and not
countable. For example, the real numbers R, the
closed interval [0,1], and the irrationals R -
Q are each uncountable.
I discussed several facts, such as: between every two distinct
rational numbers, there is an irrational number; and between every two
distinct irrationals, there is a rational. Indeed, between every two
distinct real numbers, there are countable many rationals and
uncountably many irrationals.
Lastly, I gave some examples of irrational numbers, like e,
pi, and the square root of 2. I proved that the square root
of 2 is irrational.
- Definition: a naive topological space is a subset of
Rn.
- Properties of open sets in
Rn, properties of closed sets.
- Details: Properties of open subsets of
Rn:
- The empty set ø is open.
- Rn is open.
- If U and V are open, so is their intersection.
- If you have a bunch of open sets Ux, then their union
is open.
Similarly: Properties of closed subsets of Rn:
- The empty set ø is closed.
- Rn is closed.
- If A and B are closed, so is their union.
- If you have a bunch of closed sets Ax, then their
intersection is closed.
We proved the properties of open sets using the definition of open
set. We proved the properties of closed sets by using the properties
of open sets, together with these formulas:
- (A Union B)c = (Ac) Intersection
(Bc).
- (A Intersection B)c = (Ac) Union
(Bc).
(In these formulas, the superscript ``c'' denotes the complement.
There are similar formulas for unions and intersections of any number
of elements, finite or infinite.)
- Open, closed subsets in an arbitrary naive
topological space.
- Details: Let X be a naive topological space; in particular,
suppose that X is a subset of Rn. Then a
subset U of X is open if U can be written as an intersection
of an open set in Rn with X:
- U = V Intersect X, where V is open in
Rn.
Alternatively, U is open in X if for every point in U, there is an
``open ball'' in X around that point which is contained in U:
- For every x in U, there is an epsilon so that (Bepsilon
(x) Intersect X) is in U.
As examples, we've looked at the topological spaces [0,1) (as a subset
of R2), Z (as a subset of
R), Q (as a subset of
R), and ([0,1] Union {3}) (as a subset of
R).
- Connectedness.
- Details: A topological space X is connected if the only
subsets of X which are both open and closed are X and the empty set.
(In other words, no nonempty proper subset of X is both open and
closed.) For example, R is connected, as is
Rn for any positive n (this is homework).
Z is far from connected; Q is not
connected, either, nor is ([0,1] Union {3}).
From the homework: if f: X -> Y is a continuous function and if X is
connected, then f(X) is connected. As a result, you get the
``intermediate value theorem'' from calculus.
- Neighborhoods.
- Details: Given a topological space X and a point x in X, a
neighborhood of x is an open subset (of X) containing x.
[Warning: some people use ``neighborhood'' to mean any set N
containing an open set which contains x.]
- Limit points.
- Details: Given a topological space X and a subset A of X, a point
y of X is a limit point of A if every neighborhood of y
contains a point z (different from y) with z in A. For example, when
X=R, then 0 is a limit point of the open interval
(0,1). Indeed, [0,1] is the set of limit points of (0,1). [0,1] is
also the set of limit points for ([0,1] Union {3}). The set of
integers (as a subset of R) has no limit points.
I then showed that a subset A of X is closed if and only if A contains
all of its limit points.
- Continuous functions.
- Details: Given two topological spaces X and Y, a function f: X
-> Y is continuous at a point b in X if for every epsilon
> 0, there is a delta > 0 so that whenever ||c-b|| < delta,
then ||f(c)-f(b)|| < epsilon. For example, any constant function
is continuous, as are composites of continuous functions, identity
functions, inclusion functions, and all of the familiar functions from
calculus (wherever they are defined).
- Homeomorphism.
- Details: Two topological spaces X and Y are homeomorphic
if there are continuous functions f: X -> Y and g: Y -> X which
are inverses to each other: f(g(y))=y for every y in Y, and g(f(x))=x
for every x in X. For example, every open interval in
R is homeomorphic to (0,1), and also homeomorphic to
R itself. Note that if X and Y are homeomorphic,
then the functions f and g establish a bijection between X and Y, so
that in particular X and Y have the same cardinality. For instance,
if X is countable and Y isn't, then X and Y can't be homeomorphic.
The basic idea in topology is to study properties of topological
spaces which are ``invariant under homeomorphism''; said differently,
spaces which are homeomorphic are indistinguishable to a topologist.
From the example of open intervals in R, it should be
clear that certain notions of size are not important. Indeed, since
(0,1) is homeomorphic to (0,2), length is not of interest to
topologists. Since (0,1) is homeomorphic to R, then
boundedness is not of interest, either. On the other hand, the
rationals Q cannot be homeomorphic to the irrational
R - Q, since the former is countable
and the latter is uncountable. This sort of ``size'' is important.
I discussed an example in some detail: let S1 denote the
unit circle in the complex plane, and consider the function f: [0,1)
-> S1, defined by
- f(t) = exp((2 pi t) i) = cos(2 pi t) + i sin(2 pi t).
(Here, `exp' denotes the exponential function and `pi' stands for the
eponymous Greek letter.) The function f is continuous, and it is also
a bijection; one can describe it as ``wrapping the half-open unit
interval around the circle.'' Since it is a bijection, it has a
unique inverse; this inverse takes a point cos (2 pi t) + i sin (2 pi
t) to the number t. This is not continuous at the point 1 in
the complex plane (the point corresponding to the angle 0), and
therefore f is not a homeomorphism.
- Images, preimages, continuity.
- Details: Given a function f: X-> Y and a subset A of X, the
image of A is the following subset of Y:
- f(A) = {y in Y : y=f(x) for some x in A}.
In other words, the image of A is the set of all points in Y that are
hit by points of A. Given a subset B of Y, the preimage of
B is the following subset of X:
- f-1(B) = {x in X : f(x) is in B}.
In other words, this is the set of all points that land in B after
applying the function f. Note that the preimage is always defined:
the function f does not need to have an inverse function.
For example, consider the function f: R ->
R defined by f(x)=x2. The image of the
set [-1,1] is the set of all numbers you get when you square the
numbers in [-1,1]. In other words, f([-1,1]) = [0,1]. Similarly, the
image of (-1,1) is f((-1,1)) = [0,1), and f([-2,-1]) = [1,4]. The
preimage of [0,9] is the set of all numbers whose squares lie between
0 and 9; hence, f-1([0,9]) = [-3,3]. Since no (real)
numbers have negative squares, f-1([-8,-5]) is the empty
set. Also, f-1({4}) = {-2,2}, since each of the numbers -2
and 2 have square 4.
In this example, the function f(x) is neither one-to-one nor onto, so
it does not have an inverse function. I can still talk about the
preimages of subsets of R, though. Since f(x) is not
onto, there are subsets whose preimage is empty. Since f(x) is not
one-to-one, there are single points (like 4) whose preimage consists
of more than one point. If the function f(x) did have an inverse,
i.e., if f(x) were a bijection, then every nonempty subset of the
range would have a nonempty preimage, and the preimage of each point
would consist of a single point.
Important fact: A function f:X -> Y is continuous if and only if
the preimage of every open set in Y is open in X (i.e., for every open
set U in Y, f-1(U) is open in X). This is true if and only
if the preimage of every closed set is closed. This is very
important; we could have defined continuity this way, purely in terms
of open sets (or in terms of closed sets)--we didn't have to use
epsilons and deltas.
- Open covers, compact sets.
- Details: given a subset A of a topological space X, an open
cover of A is a collection of open sets whose union contains A.
A finite subcover is a finite subcollection of these open
sets whose union still contains A.
For example, if I look at the set (0,1) in R, one
open cover would be the collection of all balls of radius 1/5 around
the rational numbers between 0 and 1 (i.e., all balls of the form
B1/5(r) where r is a rational in (0,1)). There are
certainly infinitely many such balls, but this also has a finite
subcover: the particular balls centered at 1/5, 2/5, 3/5, and 4/5
cover the whole open interval (0,1). Another open cover of (0,1)
consists of balls centered at 0 with radii 1/2, 3/4, 4/5, 5/6, ....
Every point between 0 and 1 is contained in at least one of these
balls, and so (0,1) is contained in the union. On the other hand,
this does not have a finite subcover.
The subset A of X is compact if every open cover has a finite
subcover. For example, (0,1) is not compact, because I just told you
an open cover that didn't have a finite subcover.
Theorem (Heine-Borel). Every closed interval [a,b] is compact.
I presented a proof of this. More generally:
Theorem. Every closed ``rectangle'' in Rn
is compact.
Here are some properties of compact sets:
- If X is a compact space and A is a closed subset of X, then A is
compact.
- Therefore if A is a closed and bounded subset of
Rn, then A is compact.
- If X is a naive topological space and A is a compact subset of X,
then A is closed.
- Therefore if A is a compact subset of
Rn, then A is closed and bounded.
- If f:X -> Y is a continuous function and if X is compact, then
f(X) is compact.
- As a result, you get the ``extreme value theorem'' from calculus.
- A space-filling curve.
- Details: I described a construction of a continuous
surjective function f: [0,1] -> T, where T is an
equilateral triangle. f was defined to be the limit of a sequence of
functions fn, where fn is constructed
recursively from fn-1.
- Surfaces.
- Details: I defined the notion of surface, I gave some
examples (the sphere, the torus, the torus with n hole, the projective
plane, the Klein bottle), I discussed orientability, and I
gave a classification of compact surfaces, in the orientable case and
in general.