Download Lecture 3 - Stony Brook Mathematics

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Lecture 3: September 3
Let me begin today’s class by talking about closed sets again. Recall from last
time that a subset A ✓ X of a topological space is called closed if its complement
X \ A is open. We also defined the closure of an arbitrary subset Y ✓ X to be the
smallest closed subset containing Y ; more precisely,
\
Y =
A
A◆Y
closed
is the intersection of all closed sets containing Y . Intuitively, taking the closure
means adding all those points of X that lie “at the edge of Y ”; the goal is to
understand this operation better.
Example 3.1. In a Hausdor↵ space X, every one-point set {x} is closed. We have
to convince ourselves that X \ {x} is an open set. Let y 2 X \ {x} be an arbitrary
point. Since X is Hausdor↵, we can find two disjoint open sets U and V with x 2 U
and y 2 V . Clearly, V ✓ X \ {x}; this shows that X \ {x} is a union of open sets,
and therefore open.
Example 3.2. Let Y be a subset of a topological space X. Then a set A ✓ Y is
closed in the subspace topology on Y if and only if A = Y \ B for some closed set
B ✓ X. Note that this is not a definition, but a (very easy) theorem. Here is the
proof: A is closed relative to Y if and only if Y \ A is open relative to Y if and only
if Y \ A = Y \ U for some open set U ✓ X if and only if A = Y \ (X \ U ). (The
last step is easiest to understand by drawing a picture.) But if U is open, X \ U is
closed, and so we get our result.
As I mentioned last time, the word “closed” comes from analysis, where it means
something like “closed under taking limits”. To help make the definition more
concrete, let us now discuss the relationship between closed sets and limit points.
If x 2 X is a point in a topological space, an open set containing x is also called a
neighborhood of x. We usually think of a neighborhood as being a “small” open set
containing x – but unless X is a metric space, this does not actually make sense,
because we do not have a way to measure distances.
Definition 3.3. Let Y be a subset of a topological space X. A point x 2 X is
called a limit point of Y if every neighborhood of x contains at least one point of
Y other than x itself.
Note that a point x 2 Y may be a limit point of Y , provided that there are
enough other points of Y nearby. An isolated point, however, is not considered to
be a limit point. The following theorem describes the closure operation in terms of
limit points.
Theorem 3.4. Let Y be a subset of a topological space X. The closure of Y is the
union of Y and all its limit points.
Proof. Let us temporarily denote by Y 0 the union of Y and all its limit points. To
prove that Y 0 = Y , we have to show two things: Y 0 ✓ Y , and Y 0 is closed. Since
Y 0 contains Y , this will be enough to give us Y 0 = Y .
Let us first prove that Y 0 ✓ Y . Of course, we only have to argue that every limit
point x of Y belongs to Y . The proof is by contradiction: if x 62 Y , then the open
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set X \ Y is a neighborhood of x, and therefore has to contain some point of Y ;
but this is not possible because Y \ (X \ Y ) = ;.
Next, let us show that Y 0 is closed, or in other words, that X \ Y 0 is open. Since
a union of open sets is open, it will be enough to prove that for every x 2 X \ Y 0 ,
some neighborhood of x is contained in X \ Y 0 . Now x is clearly not a limit point of
Y , and because of how we defined limit points, this means that some neighborhood
U of x does not contain any point of Y other than possibly x itself. Since we also
know that x 62 Y , we deduce that U \ Y = ;. But then no point of U can be a
limit point of Y (because U is open), and so U ✓ X \ Y 0 .
⇤
Sequences and limits. In metric spaces, the property of being closed can also be
expressed in terms of convergent sequences. Let X be a metric space, and suppose
that x1 , x2 , . . . is a sequence of points of X. We say that the sequence converges
to a point x 2 X, or that x is the limit of the sequence, if for every " > 0, one can
find an integer N such that
d(xn , x) < "
for every n
N.
Note that a sequence can have at most one limit: if x0 is another potential limit of
the sequence, the triangle inequality implies that
d(x, x0 )  d(x, xn ) + d(xn , x0 );
as the right-hand side can be made arbitrarily small, d(x, x0 ) = 0, which means
that x0 = x.
In view of how the metric topology is defined, we can rephrase the condition
for convergence topologically: the sequence x1 , x2 , . . . converges to x if and only if
every open set containing x contains all but finitely many of the xn . This concept
now makes sense in an arbitrary topological space.
Definition 3.5. Let x1 , x2 , . . . be a sequence of points in a topological space. We
say that the sequence converges to a point x 2 X if, for every open set U containing
x, there exists N 2 N such that xn 2 U for every n N . In that case, x is called
a limit of the sequence.
You should convince yourself that if x is a limit of a sequence x1 , x2 , . . . , then
it is also a limit point of the subset {x1 , x2 , . . . }. (Question: What about the
converse?) Unlike in metric spaces, limits are no longer necessarily unique.
Example 3.6. In the Sierpiński space, both 0 and 1 are limits of the constant
sequence 1, 1, . . . , because {0, 1} is the only open set containing the point 0.
In a Hausdor↵ space, on the other hand, limits are unique; the proof is left as
an exercise.
Lemma 3.7. In a Hausdor↵ space X, every sequence of points has at most one
limit.
The following result shows that in a metric space, “closed” really means “closed
under taking limits”.
Proposition 3.8. Let X be a metric space. The following two conditions on a
subset Y ✓ X are equivalent:
(a) Y is closed in the metric topology.
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(b) Y is sequentially closed: whenever a sequence x1 , x2 , . . . of points in Y
converges to a point x 2 X, one has x 2 Y .
Proof. Suppose first that Y is closed. Let x1 , x2 , . . . be a sequence of points in
Y that converges to a point x 2 X; we have to prove that x 2 Y . This is pretty
obvious: because Y is closed, the complement X \Y is open, and if we had x 2 X \Y ,
then all but finitely many of the xn would have to lie in X \ Y , which they don’t.
Now suppose that Y is sequentially closed. To prove that Y is closed, we have to
argue that X \Y is open. Suppose this was not the case. Because of how we defined
the metric topology, this means that there is a point x 2 X \ Y such that no open
ball Br (x) is entirely contained in X \ Y . So in each open ball B1/n (x), we can
find at least one point xn 2 Y . Now I claim that the sequence x1 , x2 , . . . converges
to x: indeed, we have d(xn , x) < 1/n by construction. Because x 2 X \ Y , this
contradicts the fact that Y is sequentially closed.
⇤
This also gives us the following description of the closure: if Y ✓ X is a subset
of a metric space, then the closure Y is the set of all limit points of convergent
sequences in Y .
Unfortunately, Proposition 3.8 does not generalize to arbitrary topological spaces;
you can find an example in this week’s homework. Closed sets are always sequentially closed – the first half of the proof works in general – but the converse is
not true. What made the second half of the proof work is that every point in a
metric space has a countable neighborhood basis: for every point x 2 X, there are
countably many open sets U1 (x), U2 (x), . . . , such that every open set containing
x contains at least one of the Un (x). In a metric space, we can take for example
Un (x) = B1/n (x). Topological spaces with this property are said to satisfy the
first countability axiom. So Proposition 3.8 is true (with the same proof) in every
topological space where the first countability axiom holds. If this axiom does not
hold in X, then there are simply “too many” open sets containing a point x 2 X
to be able to describe closed sets in terms of sequences (which are by definition
countable).
Note. If X is first countable, then the collection
B=
Un (x)
x 2 X and n
1
is a basis for the topology on X. A stronger version of this condition is that X
should have a basis consisting of countably many open sets; such spaces are said to
satisfy the second countability axiom.
Continuous functions and homeomorphisms. As suggested in the first lecture, we define continuous functions by the condition that the preimage of every
open set should be open.
Definition 3.9. Let (X, TX ) and (Y, TY ) be two topological spaces. A function
f : X ! Y is called continuous if
f
for every U 2 TY .
1
(U ) =
x2X
f (x) 2 U
2 TX
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If the topology on Y is given in terms of a basis B, then it suffices to check the
condition for U 2 B; the reason is that
!
[
[
1
f
U =
f 1 (U ).
U 2C
U 2C
We could have just as well defined continuity used closed sets; the reason is that
f
1
(Y \ A) =
x2X
f (x) 62 A
=X \f
1
(A).
We have already seen that the topological definition is equivalent to the "- one
in the case of metric spaces; so we already know many examples of continuous
functions from analysis. To convince ourselves that the topological definition is
useful, let us prove some familiar facts about continuous functions in this setting.
The first one is that the composition of continuous functions is again continuous.
Lemma 3.10. If f : X ! Y and g : Y ! Z are continous, then so is their composition g f .
Proof. Let U ✓ Z be an arbitrary open set. Since g is continuous, g
in Y ; since f is continous,
(g f )
1
(U ) = f
1
g
1
1
(U ) is open
(U )
⇤
is open in X. This proves that g f is continuous.
In analysis, we often encounter functions that are defined di↵erently on di↵erent
intervals. Here is a general criterion for checking that such functions are continuous.
Proposition 3.11 (Pasting lemma). Let X = A [ B, where both A and B are
closed sets of X. Let f : A ! Y and g : B ! Y be two continuous functions. If
f (x) = g(x) for every x 2 A \ B, then the function
®
f (x) if x 2 A,
h : X ! Y, h(x) =
g(x) if x 2 B
is well-defined and continuous on X.
Proof. We can prove the continuity of h by showing that the preimage of every
closed set in Y is closed. So let C ✓ Y be closed. We have
h
1
(C) =
=
x2X
x2A
h(x) 2 C
f (x) 2 C
[ x2B
g(x) 2 C
=f
1
(C) [ g
1
(C).
Now f is continous, and so f 1 (C) is closed in the subspace topology on A; but
because A is itself closed in X, this means that f 1 (C) is also closed in X. The
same goes for g 1 (C), and so h 1 (C) is a closed set.
⇤
Example 3.12. Let us consider the example of a product X ⇥ Y of two topological
spaces (with the product topology). Denote by
p1 : X ⇥ Y ! X
and
p2 : X ⇥ Y ! Y
the projections to the two coordinates, defined by p1 (x, y) = x and p2 (x, y) = y.
Then both p1 and p2 are continuous. This is easy to see: for instance, if U ✓ X is
an open subset, then p 1 (U ) = U ⇥ Y is open by definition of the product topology.
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Proposition 3.13. A function f : Z ! X ⇥ Y is continuous if and only if the two
coordinate functions
are continuous.
f 1 = p1 f : Z ! X
and
f 2 = p2 f : Z ! Y
Proof. One direction is easy: if f is continous, then f1 and f2 are compositions of
continous functions, hence continuous. For the other direction, we use the definition. A basis for the product topology is given by sets of the form U ⇥ V , with
U ✓ X and V ✓ Y both open. Then
f
1
(U ⇥ V ) =
z2Z
f1 (z) 2 U and f2 (z) 2 V
is the intersection of two open sets, hence open.
= f1 1 (U ) \ f2 1 (V )
⇤