Download closed sets, and an introduction to continuous functions

Lecture 5: closed sets, and an introduction to
continuous functions
Saul Glasman
September 16, 2016
Clarification on URL.
To warm up today, let’s talk about one more example of a topology.
Definition 1. Let X be a set. The cofinite 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 2. 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 3. Since X \ ∅ = X and X \ X = ∅, ∅ and X are always closed.
Example 4. 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 5. 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 6. 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 cofinite
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 cofinite 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.
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Although the indiscrete topology is pretty useless, the cofinite 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.
Now let’s learn how to relate topological spaces to one another: the study
of continuous functions between topological spaces. Show of hands on who’s
encountered the − δ definition of continuity before. Judging by prerequisites,
it should be everyone.
To give a brief reminder: a function f : R → R is continuous if, informally,
when x changes a small amount, f (x) only changes a small amount. This rules
out functions with “jumps”, or crazy functions like
(
1 x∈Q
χQ (x) =
0 x∈
/ Q.
The actual definition is as follows:
Definition 7. f is continuous at x ∈ R if for every > 0, there is some δ > 0
such that
f ((x − δ, x + δ)) ⊆ (f (x) − , f (x) + ).
In other words, if |x − x0 | < δ, then |f (x) − f (x0 )| < .
f is continuous if it’s continuous at x for every x ∈ R.
Our goal for the moment is to rephrase this in a way that makes sense for
any function between two topological spaces. Let’s start by noting that we can
be more flexible with the nature of our intervals:
Lemma 8. f is continuous at x if and only if for every open interval set I
containing f (x), there’s an open interval J containing x such that f (J) ⊆ I.
Proof. Suppose f is continuous. Let I be an open interval with f (x) ∈ I. Then
there’s some > 0 such that
(f (x) − , f (x) + ) ⊆ I.
Then for the corresponding δ whose existence is guaranteed by continuity, we
can take J = (x − δ, x + δ). Then f (J) ⊆ I.
Conversely, suppose f satisfies our condition. Take > 0, and let I =
(f (x)−, f (x)+). Then there’s some open interval J with x ∈ J and f (J) ⊆ I.
There is some δ such that (x − δ, x + δ) ⊆ J. So f is continuous.
With this lemma, we’ve not only given ourselves more wiggle room and made
the definition (I think) much prettier, we’ve almost banished the real numbers
from the definition statement - it’s now a statement solely about open sets and
open intervals. Let’s make this definition even snappier with the aid of a little
piece of language.
Definition 9. Let X and Y be sets and f : X → Y be a function. If U is a
subset of Y , then the preimage of U under f is
f −1 (U ) = {x ∈ X |f (x) ∈ U }.
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Don’t confuse this notation with the function f −1 , the inverse of f when
f is bijective! Preimages make perfect sense even if f is not bijective. If f is
bijective, then the preimage of U is also the image of U under f −1 , so there’s
no notational ambiguity.
Here are some facts about preimages that I’ll leave as an exercise:
Lemma 10. Let f : X → Y be a function. Then
• If (Ui )i∈I is a family of subsets of Y , then
!
[
[
f −1
Ui =
f −1 (Ui ).
i∈I
i∈I
• Similarly,
!
f
−1
\
Ui
i∈I
=
\
f −1 (Ui ).
i∈I
• If U, V ⊆ Y , then
f −1 (U \ V ) = f −1 (U ) \ f −1 (V ).
With this, we can reformulate the definition of continuity in a way that will
generalize to topological spaces:
Lemma 11. f : R → R is continuous if and only if the preimage under f of
any open set is open.
Proof. To rephrase our previous lemma: f is continuous if and only if for every
open set I and every x with f (x) ∈ I, there’s an open interval J with x ∈ J
and f (J) ⊆ I.
Rephrasing some more, this is the same as saying: for every x ∈ f −1 (I),
there’s an open interval J with x ∈ J and J ⊆ f −1 (I). But this is the same as
saying that f −1 (I) is open.
Definition 12. Let X and Y be topological spaces. A function f : X → Y is
called continuous if the preimage under f of any open subset of Y is an open
subset of X.
A continuous function is often called a continuous map, or just a map.
Remark 13. Since
f −1 (Y \ U ) = X \ f −1 (U ),
f is continuous if and only if the preimages under f of closed subsets are closed.
It’s time for some trivial examples.
Example 14. The identity function is always continuous, of course, since
id−1 (U ) = U .
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Example 15. Suppose that f : X → Y and g : Y → Z are continuous functions.
Then g ◦ f is continuous: indeed, for U ⊆ Z,
(g ◦ f )−1 (U ) = f −1 (g −1 (U )).
But g −1 (U ) is open since g is continuous, and so (g ◦ f )−1 (U ) is open, since f
is continuous.
Example 16. Let X and X 0 be two topological spaces with the same underlying
set. Then id : X → X 0 is continuous if and only if the topology on X is at least
as fine as the topology on X 0 .
Example 17. Suppose X has the discrete topology. Then for any space Y , any
function f : X → Y is continuous.
This makes intuitive sense if you regard the discrete topology as making X
into a bunch of loose points.
Example 18. Suppose Y has the indiscrete topology. Then for any space X,
any function f : X → Y is continuous.
Example 19. This one isn’t quite as immediate as the others, but it’s still very
easy, and I’ll leave the proof as an exercise: if Y has the cofinite topology, then
f : X → Y is continuous if and only if f −1 {y} is closed for each y ∈ Y .
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