Download Lecture 4: examples of topological spaces, coarser and finer

Lecture 4: examples of topological spaces, coarser
and finer topologies, bases and closed sets
Saul Glasman
14 September 2016
Let’s give the definition of an open subset of R.
Definition 1. Let U ⊆ R. We say U is open if for every r ∈ U , r is contained
in an open interval which is contained in U ; that is, for every r in U , there are
positive numbers
0 , 1 > 0
such that
(r − 0 , r + 1 ) ⊆ U.
Proposition 2. With this definition, the open subsets of R form a topology on
R.
We’ll deduce this result as a special case of a more general theorem. In
order to state this theorem, it’ll be helpful to introduce the idea of a basis for a
topology. As we’ll see, when defining a topology, it’s often annoying to explicitly
describe all the open sets of the topology; instead, it’s convenient to describe
“building blocks” for the topology. This is what bases are for.
Definition 3. Let X be a set. A basis for a topology on X is a collection of
subsets of X
B ⊆ P(X)1
such that
1.
[
U = X;
U ∈B
that is, every x ∈ X belongs to at least one element of the basis.
2. If U1 , U2 ∈ B and x ∈ U1 ∩ U2 , then there’s some V ∈ B with
x ∈ V, V ⊆ U1 ∩ U2 .
Let’s give a couple of examples here.
1 Recall
that P(X) is notation for the power set of X, the set of all subsets of X.
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Example 4. The set of open intervals in R (including the empty interval) is a
basis for a topology on R.
Why is this a basis, according to the definition above? Well, let’s note that
if B satisfies the first condition and is closed under finite intersections - that is,
a finite intersection of elements of B is in B - then it automatically satisfies the
second condition.
Lemma 5. If B is a collection of subsets of X such that
[
U =X
U ∈B
and such that B is closed under finite intersections, then B is a basis for a
topology on X.
Proof. We need to show that whenever U1 , U2 ∈ B and x ∈ U1 ∩ U2 , we can
find V with x ∈ V and V ⊆ U1 ∩ U2 . But U1 ∩ U2 ∈ B, so we can just take
V = U1 ∩ U2 .
The set BR of open intervals in R satisfies this condition: the intersection of
open intervals is always an open interval (possibly empty). So it’s a basis.
Example 6. An open disc in R2 is the collection of points less than distance r
away from a fixed point (x0 , y0 ):
D = {(x, y) ∈ R2 | (x − x0 )2 + (y − y0 )2 < r2 .}
We claim that set of open discs forms a basis for a topology on R2 . I won’t give
a rigorous proof of this, but I’ll give an illustrative diagram. Note that unlike
open intervals in R, the intersection of two open discs is not an open disc.
Example 7. An open rectangle in R2 is what you think it is: a set of the form
R = {{(x, y) ∈ R2 | x0 < x < x1 , y0 < y < y1 }
for some x0 , x1 , y0 , y1 in R. This is also a basis for a topology on R2 , since the
intersection of open rectangles is an open rectangle.
Example 8. For any set X, the set
Bd isc = {{x} | x ∈ X}
is a basis for a topology on X.
This is clear: the union of the points of X is X, and the intersection of
one-point sets is either a one-point set or empty.
We’ll soon see that both of the bases for topologies on R2 I’ve mentioned
generate the same topology - the so-called product topology, which we’ll get into
in general soon - and Bdisc generates the discrete topology on X. But I haven’t
yet told you how to build a topology from a basis, so let’s do that now.
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Lemma 9. Let X be a set and let B be a basis for a topology on X. Let U ⊆ X
be a subset. Then the following two conditions on U are equivalent:
1. U is a union of elements of B.
2. For each x ∈ U , there is an element V ∈ B such that x ∈ V and V ⊆ U .
Proof. First we show that 1 implies 2. Suppose
[
U=
Vi
i∈I
where Vi ∈ B for all i. Then for each x ∈ U , x ∈ Vi for some i. In checking 2,
we can take V = Vi . Now we show that 2 implies 1. For each x ∈ U , let Vx be
an element of B with x ∈ Vx , Vx ⊆ U . Then
[
U=
Vx .
x∈U
If U satisfies either of these equivalent conditions, we call it B-open. Now
by the definition I gave earlier, an open subset of R is one which is BR -open,
where BR is the set of open intervals.
Warning 10. If anyone’s taken linear algebra, this is a good time to remark that
the usage of the word “basis” here is quite different from the linear algebra usage.
In linear algebra, any vector can be written uniquely as a linear combination
of basis vectors, but in topology, it’s usually possible to write an open set as a
union of basis sets in many different ways.
Theorem 11. If B is a basis for a topology, the collection TB of B-open sets is
a topology. We call it the topology generated by B.
Proof. We have to check the axioms for a topology.
1. ∅ is B-open, since it’s the union of no elements of B.
2. X is the union of all elements of B, so it’s B-open.
3. Let (Ui )i∈I be a collection of B-open sets. We need to prove that the
union
[
U=
Ui
i∈I
is B-open. For each x ∈ U , x ∈ Ui for some i; since Ui is B-open, there is
V ∈ B such that x ∈ V , V ⊆ Ui . But V ⊆ U too, so U is B-open.
4. For each finite collection
U1 , U2 , · · · , Un
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of B-open sets, we have to show that
U=
n
\
Un
i=1
is B-open.
This is a good place for a little digression about a useful way of checking
the finite intersection condition. If you want to check that some class T is
closed under finite intersections, it’s enough to show that the intersection
of two elements of T is in T . Indeed, if U1 , U2 , U3 are elements of T , then
U1 ∩ U2 ∩ U3 = ((U1 ∩ U2 ) ∩ U3 )
But since U1 ∩ U2 ∈ T and U3 ∈ T , so is U1 ∩ U2 ∩ U3 . I could do the
same thing if I had n-sets: just iteratively intersect them one at a time.
U1 ∩ U2 ∩ · · · ∩ U3 = (((· · · (U1 ∩ U2 ) ∩ U3 ) ∩ ...) ∩ Un ).
Incidentally, in case you don’t know what a “proof by induction” is, this
is one.
So let’s check that if U1 and U2 are B-open sets, U1 ∩ U2 is B-open.
Condition 2 of the definition of a basis was designed just for this. If
x ∈ U1 ∩ U2 , then there are sets V1 , V2 ∈ B with
x ∈ V1 ⊆ U1 , x ∈ V2 ⊆ U2 .
Therefore, there’s a set V ∈ B with
x ∈ V ⊆ V1 ∩ V2 ⊆ U1 ∩ U2 .
This shows that U1 ∩ U2 is B-open.
In particular, this shows that the topology on R, as defined at the beginning
of this lecture, is actually a topology. Not bad.
For the rest of this lecture, I’d like to change tack a bit and talk about some
smaller topics concerning topologies. First, let’s establish some notation for
relating different topologies on the same set.
Definition 12. Let T , T 0 be topologies on a set X. We say that T is at least
as fine as T ’ if
T0⊆T;
that is, whenever a set is open in the topology T 0 , it’s open in the topology T .
Obviously, if T is at least as fine as T 0 and T and T 0 are not equal, we’ll say
that T is finer than T 0 . We’ll also say that T 0 is coarser than T .
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I like the metaphor that Munkres uses here, with truckloads of gravel (p.
77), although it makes a topological space seem more ”disconnected” than it
really is. I prefer to say that a finer topology is “higher resolution” than a
coarser topology: if we think of open sets as being areas of similar color on a
monitor or something, then a finer topology really does give us higher resolution
pictures.
Having a finer topology isn’t always desirable: the very finest topology on
any set X is the discrete topology on X, which is the topology in which all sets
are open (think of a monitor which is always displaying random static). The
coarsest possible topology on X is the indiscrete topology on X, which has as
few open sets as possible: only ∅ and X are open (think of a monitor which can
only display a solid field of black or white). Interesting topologies are balanced
between these two extremes.
To wrap up today, let’s talk about one more example of a topology.
Definition 13. Let X be a set. The finite complement topology Tf c on X is
the following class of subsets:
U ∈ Tf c if and only if U = ∅ or X \ U is finite.
In order to prove that Tf c is a topology, it’ll be convenient to introduce
closed sets in a topological space. The definition is simple:
Definition 14. Let X be a topological space. A set Z ⊆ X is called closed if
its complement Z \ X is open.
We’ll talk a lot more about closed sets later, but for now you should think of
closed sets as sets which are in “sharp focus”: they have no fuzzy edges. Let’s
have some examples of closed sets.
Example 15. Since X \ ∅ = X and X \ X = ∅, ∅ and X are always closed.
Example 16. In R, the “closed interval”
[a, b] = {r ∈ R | a ≤ r ≤ b}
is a closed set. To show this, we need to show that the complement
R \ [a, b] = (−∞, a) ∪ (b, ∞) = {r ∈ R | r < a or r > b}
is open. But I can write this as a union of open intervals:
[
(−∞, a) =
(a − n, a)
n∈N
and similarly for (b, ∞).
In particular, we can take a = b. Then the closed interval [a, a] is the single
point {a}, and so {a} is a closed subset of R. This is very typical behavior: for
many nice topological spaces, it’s the case that single points are closed subsets.
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Let’s now recall an extra bit of set theory.
Lemma 17. Taking complements turns intersections into unions and unions
into intersections: Let X be a set and let (Ui )i∈I be a collection of subsets of
X. Then
[
\
X \ ( Ui ) =
(X \ Ui )
i∈I
i∈I
and
X \(
\
Ui ) =
i∈I
[
(X \ Ui ).
i∈I
These two statements are known as deMorgan’s laws. If they’re not immediately obvious to you, I recommend you try to prove them at home as an
exercise.
The upshot is that we can define a topology just as well with closed sets as
with open sets:
Lemma 18. Let X be a set and let T be a collection of subsets of X. Let
T c = {X \ U | U ∈ T }
be the set of complements of elements of T . Then T is a topology if and only
if T c satisfies the following conditions:
1. ∅ ∈ T c .
2. X ∈ T c .
3. An arbitrary intersection of elements of T c is in T c .
4. A finite union of elements of T c is in T c .
Of course, if you know what the closed sets are, you also know what the
open sets are:
X \ (X \ U ) = U,
so
(T c )c = T .
The point I want to make right now is that in some cases, such as the finite
complement topology, it’s easier to check the topology axioms with closed sets
than with open sets. Indeed, a subset Z ⊆ X is closed in the finite complement
topology if and only if it’s finite or equal to X. But
1. the empty set is finite;
2. X is equal to X;
3. an arbitrary intersection of finite sets is finite;
4. a finite union of finite sets is finite.
Although the indiscrete topology is pretty useless, the finite complement topology actually shows up in the wild (for example, in algebraic geometry). It’s the
coarsest topology which satisfies the reasonable condition that one-point sets
are closed.
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