Why would one want to generalize notions such as convergence and continuity to a setting even more abstract than metric spaces?
Topology deals with the relative position of objects to each other and their features. It is not about their concrete length, volume, and so on. Hence, topological features will not change if continuous transformations are applied to these objects. That is, topological features are preserved under stretching, squeezing, bending, and so on but they are not preserved under non-continuous transformations such as tearing apart, cutting and so on. Objects such as a circle, a rectangle and a triangle are from a topological point of view “equal” / homeomorphic even though the shapes are geometrically rather different.
What features are therefore of interest such that it is worth studying topology?
Assume that is a closed curve, i.e. a circle, a rectangle, a triangle or something like we can see in the graph above. As long as we transform a shape
in a continuous fashion, the relative positions of the points
and
will be similar: for instance, points that have been inside of
will still be inside after the continuous transformation. Points that have been on the boundary will still be on the boundary.
Hence, the generalization of continuity and the concept of convergence (i.e. points being ‘close to each other’) are the two most characterizing features in topological spaces.
However, convergence and continuity in a metric space were based on a notion of a distance function
for points
. Set-theoretic topology generalizes the features of topological metric space and ought to be based on an axiomatized notion of “closeness“.
This post is based on the literature [1] to [5]. For English-speaking beginners, [5] is recommended. The German lecture notes [3] is also a good introduction to topology.
Topology on a Set
The term “topology on a set” is based on an axiomatic description of so-called “open sets” with respect to some set-theoretic operators. It will turn out, that a topology is a set that has just enough structure to meaningful speak of convergence and continuous functions on it.
Open Sets
Definition 1.1 (Topological Space)
A topological space is a pair , where
is a set and
is a family of subsets that satisfies
(i) ;
(ii) if
for an arbitrary index set
;
(iii) if
for a finite index set
.
Let — short
— be a topological space in this post.
The following video provides a rather unorthodox way of thinking about a topology. However, it might help to get a heuristic understanding. The connection between metrics and topologies is also mentioned.
Some examples will further support the understanding.
Example 1.1 (Topologies)
(a) Let be a set and
then
is the so-called trivial, chaotic or indiscrete topology. The only open sets of the trivial topology are
and
.
(b) The power set of a set
is the so-called discrete topology. In this topology every subset is open.
(c) There are four topologies on the set , i.e.
,
,
,
.
It is still an open problem, which is related to combinatorics and lattice theory, to find a simple formula for the number of topologies on a finite set.
(d) Let be a topological space, and let
. The relative topology on
(or the topology inherited from
) is the collection
of subsets of . It is clearly a topology on
. The space
is then called a subspace of
.
(e) Let be a non-empty infinite set and
be the family of open sets. Then
is the so-called finite complement topology. First note, that an element
,
is of infinite cardinality. Think about what happens if we remove finitely many elements from an infinite set.
Apparently, since
can be considered as finite.
due to the definition of the set
even though
is not finite.
The union of arbitrary open as well as the intersection of a finitely many open sets are open again. The corresponding proof employs De Morgan’s Laws.
(f) Let be a non-empty countable set and
be the family of open sets. Then
is the so-called countable complement topology.
Apparently, since
is countable and
due to the definition. The corresponding proofs of the union and intersection of open sets also employs De Morgan’s Laws.
According to Proposition 1.1 – Fundamentals of Topology & Metric Spaces, the set of all open sets in a metric space
complies with the definition of a topology.
Definition 1.2 (Induced & Equivalent Topology)
A topology induced by a metric space
is defined as the set
of all open sets in
.
Two metrics and
on the same basic set
are called topologically equivalent if
.
Let us define a very basic but important term.
Definition 1.3 (Neighborhood)
A neighborhood of a point
in a topological space
is any set containing
as well as an open set
of
, i.e.
.
Some examples will improve the understanding.
Example 1.2 (Metrizable and Equivalent Topologies)
(a) If is a metric space and
is the set of all open sets, then
is a topology. The topology
does not depend on the particular metric
since the proof of the referred Proposition 1.1 can also be done using the term neighborhood instead of the distance function. Hence, any metric on
equivalent to
yields the same topology. Topological spaces of this kind are called metrizable.
(b) Let us consider equipped with the natural topology by taking the topology
induced by the Euclidean metric
with and
. Instead of using the Euclidean metric, we could also employ the following distance functions:
All three induced topologies would be equivalent, i.e. since
for all . The corresponding unit open balls centered at
are illustrated as follows.

An open ball with
as shown in Fig. 1 is actually a set of points where each point has a distance of max. 1 to the origin
. This, however, is nothing but the corresponding norm
. For instance, the point
is not an element of the unit ball induced by
since
. However, the same point
is element of the unit balls induced by
and
.
From a topological point of view the shapes in Fig. 1 are all equivalent.
Closed Sets
Directly linked via the definition to open sets and equivalent in their explanatory power are closed sets.
A set is open in an Euclidean metric space if and only if for every
an open ball
exists such that
. Let us therefore consider the situation in the real plane
employing the topology induced by the Euclidean metric. The open balls
and
are both contained in
.
However, the open ball cannot be fully contained in
no matter how small we pick
. Thus,
is not an element of the open set
.
Let us now study sets whenever
is open.
Definition 1.4 (Closed Sets)
A set is called closed in
if
is open in
.
The definition actually tells us that one just needs to consider the complement set to figure whether the set
is closed.
Example 1.3 (Closed Sets)
(a) Let b a metric space. Then
is closed in
if and only if
is closed in
. For instance, an arbitrary open ball
with
is open. Thus, the set
is closed. The situation for
and the Euclidean topology is sketched in the next figure.
(b) The sets are not only open but also closed for any topological space
since
and
.
The topological space is trivial / indiscrete if and only if these two sets are the only closed sets in . The closed sets of the indiscrete / trivial topology are the complements of the open sets. Hence, the closed sets are also
and
.
(c) The topology is discrete if and only if every subset
is closed. This can be seen by
.
(d) The subset of
is closed because its complement
is open. Similarly,
is closed, because its complement
is open.
The subsets of
are neither open nor closed.
(e) In the finite complement topology on an infinite set , the closed sets consists of
itself and all finite subsets of
. This follows directly from the definition of the set
as set out in Example 1.1 (e). According to this definition, the sets
with
need to be finite with the exception of
.
Let us characterize closed sets.
Proposition 1.1: (Characterization of Closed Sets)
(1) The set of all closed sets of
complies with the following conditions:
(i) and
.
(ii) implies
.
(iii) implies
.
(2) Let be a family of sets that complies with (i), (ii) and (iii) then there exists a topology
, such that
is the set of all closed sets in
.
Proof.
(1) This follows directly from the definitions and applying the rules ,
,
as well as
.
(2) The family of closed sets fully determines the topology
on the same basic set
since
. Its existence follows from the fact that
actually is a topology but this is clear given (1).
The family of closed sets of a topology could also be used to define a topological space, i.e. the set of all closed sets contains exactly the same information as the set of all open sets that actually define the topology.
Points, that lie at the boundary between and
, are crucial for the understanding and distinction of open and closed sets. The intuitive idea of a boundary point
is depicted in the next figure.
Interior, Closure & Boundary
The points that lie close to both the “inside” and the “outside” of the set play an important role. However, before we can define a boundary point, we need to clarify what is meant by “inside” and “outside”.
Definition 1.5 (Interior & Closure)
Let be a topology. The interior of
is defined as the union of all open sets contained in
, i.e.
The closure of is defined as the intersection of all closed sets containing
, i.e.
Apparently, the interior of is open and a subset of
while the closure of
is closed and contains
. Thus, the following set relation is valid for any set
in a topological space:
The following theorem provides some useful relationships. However, we will not prove all of the statements and refer to section 2.1 in [5], for instance.
Theorem 1.1 (Properties of Closure and Interior)
For sets in a topological space
, the following statements hold:
(i) If is an open set in
and
, then
Int
;
(ii) If is a closed set in
and
, then Cl
;
(iii) If then Int
Int
and Cl
Cl
;
(iv) is open if and only if Int
;
(v) is closed if and only if Cl
.
Proof. (i) Since Int is the union of all of the open sets that are contained in
, it follows that
is one of the sets making up this union and therefore is a subset of the union. That is,
Int
.
(ii) Since Cl is the intersection of all of the closed sets that contain
, it follows that
is one of the sets making up this intersection and therefore is Cl
is contained in
. That is, Cl
.
(iii) Since , Int
is an open set contained in
. According to (i) every open set contained in
is contained in Int
. Therefore, Int
Int
. One can use (ii) to show the second statement of (iii).
(v) If Int
, then
is an open set since by definition Int
is an open set. Now assume that
is open. We show that
Int
. First, Int
by definition of Int
. Furthermore, since
is an open set contained in
, it follows by (i) that
Int
. Thus,
Int
as we wished to show.
Example 1.4 (Closure and Interior)
(a) Consider in the standard topology on
. Then
and Int
.
(b) Consider as a subset of
with the discrete topology
. Since all subsets are open and closed we have Int
Cl
.
(c) Consider in the finite complement topology on
. Since a closd set in this topology is either
or finite, it is clear that only
is a closed set containing the infinite set
. Hence, Cl
. The open sets are precisely the empty set and the cofinite subsets, i.e., the subsets whose complements are finite subsets of
. Hence, Int
since there are no open sets in
contained in
.
The last example highlights that not only the actual sets matter but also the sourounding topology. The next theorem provides a simple means for determining when a particular point is in the interior or in the closure of a given set
.
Theorem 1.2 (Closure, Interior and Open Sets)
Let be a topological space,
be a subset of
, and
.
(i) Then Int
if and only if there exists an open set
such that
;
(ii) Then Cl
if and only if every open set containing
intersects
.
Proof. (i) First, suppose that there exists an open set such that
. Then, since
is open and contained in
, it follows that
Int
. Thus,
Int
.
Next, if Int
, and we set
Int
, it follows that
is an open set such that
.
(ii) Considering the contrapositive: Cl
if and only if there is one neighborhood that does not intersect
. If
Cl
then the set
Cl
is open and does contain
, which is as claimed in the contrapositive.
Conversely, if there is a neighborhood of
which does not intersects
, then its complement
is a closed set that contains
. By definition of the closure Cl
, the set
must contain Cl
. Since
it follows that
Cl
. Contraction.
Theorem 1.3 (Relations of Closure and Interior)
For sets in a topological space
, the following statements hold:
(i) Int =
Cl
;
(ii) Cl =
Int
;
(iii) Int Int
Int
, and in general equality does not hold;
(iv) Int Int
= Int
.
Proof. Refer to Theorem 2.6 in [5] and to Folgerung 1.2.25 in [3].
Now, we have all ingredients to define the so-called boundary.
Definition 1.6 (Boundary Set & Points)
Let ,
. The boundary of
, denoted by
. A point of
is called boundary point.
There are situations that challenge or defy our intuitive understanding of boundary sets. For example, what is the boundary of as a subset of
in the standard topology?
Example 1.4 (Boundary Sets & Points)
(a) Consider in the standard topology on
. The boundary set
equals
.
(b) Consider in the standard topology on
. Since
, and
, it follows that
. The entire real line is therefore the boundary of the rational numbers, which makes sense. Every real number is arbitrarily close to the set of rational numbers and to its complement, the set of irrational numbers.
(c) Let in
with the discrete topology
. That is, every subset is open and closed at the same time. Hence,
and
.
In a metric space, a point is a boundary point of
if
(1)
for all .
Let us now prove the generalized statement about boundary sets.
Proposition 1.2 (Characterization of Boundary Points)
Let be a subset of a topological space
and let
. Then
if and only if every neighborhood of
intersects both
and
.
Proof. Suppose , then
and
due to the definition. Since
, it follows that every neighborhood of
intersects
. Note that a neighborhood of
contains
. Furthermore, since
, it follows that every neighborhood of
intersects
. Thus, every neighborhood of
intersects
and
.
Now suppose that every neighborhood of intersects
and
. It follows that
and
Cl
. By
Theorem 1.4 (Open Sets and Neighborhoods)
Let be a topological space and let
be a subset of
. Then
is open in
if and only if for each
, there is a neighborhood
of
such that
.
Proof. First, suppose that is open in
and
. If we let
then
is a neighborhood of
for which
.
Now suppose that for every there exists a neighborhood
of
such that
. Since we know that the union
of open sets is open, the assertion follows.
The set can be contained in
or in
:
- If all boundary points
are outside of
, i.e. if
, then it is an open set.
- If all boundary points are contained within the set, then it is a closed set.
Proposition 1.2 (Closure and Open Sets)
Let be a topological space,
be a subset of
, and
be an element of
. Then
if and only if every open set containing
intersects
.
Proof. The closure is the smallest closed set containing
as a subset. Let
and let
be an open neighborhood of
. If
then
, and the latter set is closed (by definition), which leads to a contradiction. So
.
The property of the closure hitting an open set is so important that usually a new term is defined. Note, however, that we will not further use it in this post.
Definition 1.7 (Adherent Point)
Let ,
and
. The point
is called adherent point if every open
of
has a non-empty intersection with
, i.e.
for all
with
.
Let us come back to Example 1.2 (a) – considering this example we could ask whether every topological space is metrizable?
The answer is no, and the root-cause is that topological spaces have different types of separation properties.
Separation Properties
A metric enables us to separate points in a metric space since any two distinct points have a strictly positive distance. In general topological spaces, separating points from each other is more subtle.
Hausdorff Space
Hausdorff spaces and the Hausdorff condition are named after Felix Hausdorff, one of the founders of topology. Let us first check out the formal definition.
Definition 2.1: (Hausdorff Space, Spaces)
A topological space is a Hausdorff or
-space if, for any pair of distinct points
,
there are disjoint open sets
with
,
and
.
Every Euclidean space is Hausdorff since we can use the Euclidean metric to separate two distinct points. The following video outlines the Hausdorff condition and it provides a simple example of a Hausdorff space.
Example 2.1 (Metric Space is Hausdorff)
(a) Let be a metric space, and let
be such that
. It follows that
. Let
and
. Then,
is Hausdorff since
.
(b) Consider the indiscrete / trivial topology with
. By definition, the only neighborhood of any two points
is the entire set
. Thus, every neighborhood of
will contain
and vice verca. It follows that the indiscrete / trivial topology is neither Hausdorff nor metrizable.
In a Hausdorff space, distinct points can be separated by open sets.
The situation in along with the topology implied by the Euclidean metric is illustrated in the graph above. For two distinct points
, we take half (or less) the distance to define
to come up with two distinct open balls, that can also be seen as disjoint neighborhoods.
Proposition 2.1: (Subset of -spaces)
Let be a topological Hausdorff space. Then, each subset
of a Hausdorff space is Hausdorff.
Proof. Let be in
. The space being Hausdorff, let
and
be the two open separating sets as required in Definition 1.1. Then
as well as
are open since the difference of two open sets is open. In addition,
and
.
Spaces with Weaker Separation Property
The following separation properties are weaker than the Hausdorff (-) condition. This is also indicated by the index of the corresponding names of the separation axioms (from
to
).
Definition 2.2. ( Space)
A topological space is called a
– or Kolmogorov space if, for any
with
, there is an open set
with
and
or
and
.
The most striking difference between a Hausdorff / – and
-space is that only one open set
, that contains only one of two distinct points, is required to fulfill the definition of a
-space. Apparently, every
-space is also a
-space.
Example 2.2:
(a) Let be any set with at least two elements equipped with the so-called chaotic topology
. Then, there is no open
that separates the two distinct elements. Hence, this topology is not a
-space and definitely also not a Hausdorff space. Hence, it is also not metrizable.
(b) Let be any set with the discrete topology
. Then,
separates the two elements
and
.
In a -space two open sets are required to separate two distinct points, however, the two sets don’t need to be disjoint.
Definition 2.3. ( Space)
A topological space is called a
-space if, for any
with
, there are open sets
with
and
and
and
.
The main difference between a – and a
-space is that the two required open sets do not have to be disjoint. However, one open set only contains one of the two distinct points.
Hence, every -space is also a
-space. Just take one of the two open sets of the
-space and it fulfills all requirements of a
-space.
Proposition 2.2: (Characterization of -spaces)
(a) Let be a topological space. Then,
is a
-space if and only if
is a closed set for each
;
(b) Each Hausdorff space is a -space.
Proof. (a) Suppose is a
-space, and let
. For any
with
, there is an open subset
of
with
, but
. It follows that
.
Conversely, suppose that all singleton subsets of are closed, and let
be such that
. Then,
and
fulfill the requirements of a
-space.
(b) Let be a given point. By assumption, each
belongs to an open set
such that
. Consequently,
. Thus,
is open, and
is closed.
Convergent Sequences
One of the key features of topological spaces is the generalization of the convergence concept.
A sequence in a (metric) space is a function
that we also denote by
, in particular, if we want to refer to the elements of the sequence. Given a sequence
in a metric space, a sub-sequence is the restriction of
to an infinite subset
. If we exhibit
as
, then we write the subsequence
.
We say that a sequence converges to
if given
, there exist
such that for all
, we have
.
In other words, for all , we have
as illustrated in Figure above. The finitely many elements
of the sequence
are, however, not contained in
. We take this property to define what a convergent sequence in a topological space is.
Definition 3.1. (Convergent Sequence)
Let be a topological space. A sequence
converges to
if
, the set
is finite for any open set
.
The point is then called the limit of the sequence
and we denote it by
or by
.
Note that the set of is infinite for all open sets
while (at the same time) the set
is finite. Both sets/conditions matter in this situation as we will see further below!
Lemma 3.1. (Limit of a sequence is unique)
The limit of a convergent sequence
in a Hausdorff space
is unique.
Proof. Assume that this is not the case and as well as
with
holds true. A metric space is Hausdorff, that is, we find two disjoint open balls
and
. Given that
and
are the limit points almost all elements must lie in the disjoint balls, which contradicts the initial assumption of
.
Lemma 1.1 is false in arbitrary topological spaces.
Every of a topological space
is the limit of a certain sequence
. Apparently, we could simply use the constant sequence
or we could define
for all
,
. This fact should be also considered in the following examples.
Example 3.1:
(a) Let be the discrete topology with
. Further, let
. Recall that in this topology every set is open by definition. Hence, also
is an open set that must be contained in all other (open) supersets. Hence, the set
has to be finite and
has to be infinite.
(b) Let be the indiscrete topology with
. Further, let
. Since
is the only set that contains
, the set
has to be finite for every sequence
. Hence, every sequence converges to every point in
.
Closely related to converging sequences and their limits are accumulation points.
Definition 3.2. (Accumulation Point)
An element of a sequence
is called accumulation point (sometimes also cluster or limit point) if
is infinite for every open set
of
.
The subtle but important difference between an accumulation point and a limit is that the complement set of can also be infinite. Let us consider a simple example.
Example 3.2:
Let us consider the sequence in the topology induced by the Euclidean space
on the real line. There are two accumulation points
but no limit of the sequence. Note that the sequence is alternating between
and
, such that
and
are both infinite but disjoint to each other. In addition, the set
for an open set
of
are both infinite.
A sub-sequence of a convergent sequence
converges to the same limit
. This is evident since if the condition of a convergent sequence is fulfilled for all elements
,
of the series
. Hence, the condition is also fulfilled for a subset
that represents the sub-sequence.
Due to the fact that the finiteness of implies the infiniteness of
in
every limit is an accumulation point. The converse is not true as we can see in Example 3.2.
Theorem 3.1 (Convergence in Topological Spaces)
Let be a topological space.
(i) Every limit of a convergent sequence is also the limit of any sub-sequence.
(ii) Every accumulation point of any sub-sequence is also an accumulation point of
.
(iii) Every accumulation point of a sequence in
is an adherence point of the set
.
Proof. (i) If then
is finite for all open
of
. In particular, this holds true for any sub-sequence
and thus
.
(ii) Let be an accumulation point of the sub-sequence, i.e. the set
is infinite for every open set
of
. Since the sub-sequence is only a subset of the element of the sequence, the assertion follows directly.
(iii) Let be an accumulation point of
and
be an open set of
. Apparently,
intersects
since the set
is infinite. Thus,
will also intersect
, which implies that
Cl
.
Let us now assume that and
is a converging sequence with
for all
. Then, the limit point
is also contained in
. A closed set contains all its limit points.
Compactness
The concept of compactness is not as intuitive as others topics such as continuity. In , the compact sets are the closed and bounded sets, but in a general topology compact sets are not as simple to describe.
Compact sets are so important since they possess important properties, that are known from finite sets:
- Set is bounded;
- Set contains a maximal and minimal element;
- An infinite sequence contains a constant subsequence.
The famous Heine-Borel Theorem shows that compact sets in metric spaces do indeed have these properties. This analogy is also outlined in this really nice video (in German only) by Prof. Dr. Edmund Weitz.
Let be a topological space.
Definition 4.1 (Cover)
The collection is said to cover a set
or to be a cover of
if the union of the elements of
contains
.
If covers
, and each set in
is open, then we call
an open cover of
.
A sub-collection of a cover is called a subcover of
.
A cover of is a collection of possibly overlapping sets in
which, after considering their union globally, contains the set
inside.
Example 4.1 (Real Line)
Let us consider the set of the topological space
, where
are the open sets of
. Note that any nonempty open subset of
can be written as a finite or countable union of open mutually disjoint intervals.
Then, is an open cover of
because
. This cover apparently contains infinitely many open sets
.
is an open subcover of
since
and
. This subcover, however, still contains infinitely many open sets.
Now, we have the ingredients for the central definition of this section.
Definition 4.2 (Compact Set)
A subset of a topological space
is said to be compact if every open cover
of
has a finite (open) subcover.
A topological space is called compact if
is compact.
It is clear that every finite set is compact and the following example is going to illustrate that.
Example 4.2 (Finite Set & Compactness)
Let be a finite set in the standard Euclidean topological space
. By setting
with , an open set
is assigned bijectively to each point
. Hence, the set
is an open and finite cover for any set
. Each point of the finite set
is contained in one of the elements of
. Given any open cover
, we can always use
with
small enough simply because there are only finitely many elements in
.
Let us consider another simple example.
Example 4.3 (Real Line)
The real line in the standard Euclidean topology is not compact since the set
of Example 5.1 is an open cover, but no finite sub-collection of
covers
. If we picked finitely many open sets of
the points before the minimum / after the maximum of this sub-collection would not be covered.
The last example directly used the definition of a compact set to show that it is not compact since every open cover needs to have a finite sub-cover such that the set can be compact.
Even though is a finite open cover of
, this does not mean that
is compact: if it were compact, all open covers (including the one defined in Example 5.2/5.3) would have to have a finite sub-cover.
Example 4.4 (Converging Sequence & Compact Set)
(a) The subset
is compact in the standard topology on
.
Given any open cover of
, there is an element set
containing
. The set
contains either all points or all but finitely many of
. If
contains all of the points in
, then
, by itself, is a finite subcover of
. Otherwise, let
be the smallest of the points in
that are not in
. Then we can pick open sets
for
,
for
,
, and
for
such that
is a finite sub-cover of
. Hence,
is compact.
(b) The compact sets of the discrete topology are finite. To see this realize that all subsets of
are open and closed. Thus, all possible families of subsets can be open covers. For instance, the family
is an open cover and if
is of infinite cardinality, so is
. If we leave out one of the elements of
it would not be a cover. Hence, there can be no open subcover, which implies that the compact sets are finite. Refer also to this formal proof.
Let us now extend the definition of compactness to subsets of topological spaces.
Definition 4.1 (Subspace Topology)
Let be a topological space. If
, the collection
is a topology on , called the subspace topology. With this topology,
is called topological subspace of
.
Let us check that is indeed a topology. It contains
and
because
and
, where
and
on the right-hand side of the
-symbol are elements of
. The fact that it is closed under finite intersections and arbitrary unions follows from the equations
Lemma 4.1 (Compactness & Subspaces)
Let be a subspace of
. Then
is compact in
if and only if every open cover of
by open sets in
contains a finite subcover of
.
Proof. Suppose that is compact and
is a cover of
by sets open in
. Then the collection
is a covering of
by sets open in
. Due to the assumption that
is compact in
, there exists a finite subcover
in . For each
chose a set
. Then
covers .
Suppose that every open cover of by sets open in
contains a finite open subcover of
. We would like to show that
is compact in
. Let
be an open cover of
by sets open in
(and thus in
). By hypothesis, some finite subcollection
exists that covers
. Then, by definition of the subspace topology, for each
,
we can chose an open set
via
. It follows that
is a finite subcover of
in
. Hence,
is compact in
.
A closed subspace is a subspace , that when treated as a subset of the original space
is a closed set in the original topology
.
Theorem 4.1 (Compactness & Closed Spaces)
Every closed subspace of a compact space
is compact.
Proof. Let be a closed subspace of the compact space
. Given a covering
of
by sets open in
, let us form an open cover
by adjoining to
the single open set
. Note that
is closed such that its complement needs to be open.
Thus, is an open cover of
. Due to the fact that
is compact, some finite subcollection of
covers
. If this subcollection contains the set
, discard
. Otherwise, leave the subcollection alone. The resulting collection is a finite cover of
that covers
. Hence,
is compact.
Theorem 4.2 (Compactness in Hausdorff Spaces)
Every compact subspace of a Hausdorff space
is closed.
Proof. Let be a compact subspace of a Hausdorff space
. We are going to prove that
is open, so that
is closed.
Let . We show there is a neighborhood of
that is disjoint from
. For each point
, let us choose disjoint neighborhoods
and
of the points
and
, respectively (using the Hausdorff condition). The collection
is a covering of
by sets open in
. Therefore, finitely many of them
cover
(since
is assumed to be compact). The open set
contains
. The open set
contains all points that are disjoint from any of the open sets
and thus
. Hence,
is a neighborhood of some arbitrary
and thus
is closed.
Example 4.5 (Theorems 4.1 and 4.2)
(a) Once we prove that the interval in
is compact, it follows from Theorem 4.1 that any closed subspace of
is compact. On the other hand, it follows from Theorem 4.2 that the intervals
and
in
cannot be compact because they are not closed in the Hausdorff space
.
(b) The Hausdorff condition in Theorem 4.2 is necessary. Consider the finite complement topology on the real line. The only proper subsets of that are closed in this topology are the finite sets. But every subset of
is compact in this topology since you can always find an open finite subcover given any open cover.
(c) The interval is not compact in the Euclidean standard topology of
. The open cover
contains no finite sub-collection covering
.
Continuous Functions
Topological spaces have been introduced because they are the natural habitat for continuous functions. These spaces have been built such that the topological structure is respected. Continuous functions therefore take the same role on topological spaces as linear maps within vector spaces.
The notion of continuity is particularly easy to formulate in terms of open (and closed) sets and the following version is called the open set definition of continuity.
Definition 5.1: (Continuous Function on Topological Space)
Let and
be two topological spaces. A function
is continuous if
is open in
for every open set
in
.
Before we will illustrate this definition let us recall the definitions of an image and a preimage
of a function
with
,
:
The condition that be continuous says that for each open set
of
, the inverse image of
under the map
is open in
Example 5.1: (Simple Continuous Function)
Let and
be two topological spaces defined by
Let be three functions defined by
The functions is continuous since the pre-image of each open set in
is an element of
. Similarly,
is continuous but note that
is not an open set. The function
, however, is not continuous since
is open in
, but
is not open in
.
Let us now study how the functions map closed sets. By definitions these are the complements of and
, i.e.
Recognize that the image of a closed set in under
and
is contained in a closed set in
. For instance,
and
and
.
The last example made the definition of a continuous function between simple topological spaces rather clear.
Sometimes it is also helpful to study the properties that will not be preserved: a continuous function does not necessarily map open sets to open sets.
For example, the function , given by
, is continuous, but the image of the open set
is
, which is neither open nor closed. Let us also double-check that in Example 4.1. The function
maps the open set
to
, which is not open.
Now, let us have a look at more general examples.
Example 5.2: (Identity and Constant Function)
(a) The identity function id:, given by id
, is continuous in all topological spaces. If a function
is continuous at
, i.e. if
is open so is its preimage
. This argument can be generalized onto subsets
.
(b) The constant function defined by
for every
. Suppose
is open in
, then
if
, and
if
. In either case, the preimage is open in
, and therefore
is continuous.
Continuous functions preserve proximity as we can see in the next theorem. Also refer to Example 4.1.
Theorem 5.1 (Continuous Functions & Closeness)
Let be continuous and assume that
. If
, then
.
Proof. Suppose that is continuous,
, and
. We prove that if
, then
.
Hence, suppose that . By Proposition 1.2 there exists an open set
containing
, but not intersecting
. It follows that
is an open set containing
that does not intersect
. Thus,
, and the result follows.
The next theorem translates the well-known –
definition of continuity with Definition 4.1.
Theorem 5.2 (Continuity & –
Condition)
A function is continuous in the open set definition of continuity if and only if, for every
and every open set
containing
, there exists a neighborhood
of
such that
.
Proof. First, suppose that the open set definition holds for functions . Let
and an open set
containing
be given. Set
. It follows that
and that
is open in
since
is continuous by the open set definition 4.1. Clearly,
, and therefore we have shown the desired result.
Now assume that for every and every open set
containing
, there exist a neighborhood
of
such that
. We show that
is open in
for every open set
in
. Hence, let
be an arbitrary open set in
. To show that
is open in
, choose an arbitrary
. It follows that
, and therefore exists a neighborhood
of
in
such that
, or, equivalently, such that
. Thus, for an arbitrary
there exists an open set
such that
. Then, the assertion follows applying Theorem 1.1.
Theorem 4.2 generalizes this idea of continuity in metrizable topological spaces to general topological spaces. In a metric space, we can consider an open ball as an open set and therefore as a neighborhood. That is, for each -ball
there need to be a suitable
-ball
, such that
. For further details please refer to this deep-mind.org post and this Wikipedia article.
The second important property that is preserved by continuous functions is the concept of convergence.
Theorem 5.3 (Continuity & Convergent Sequences)
Assume that is continuous. If a sequence
in
converges to a point
, then the sequence
in
converges to
.
Proof. Let be an arbitrary neighborhood of
in
. Since
is continuous,
is open in
. Furthermore,
implies that
. The sequence
converges to
; thus, there exists
such that
for all
. It follows that
for all
, and therefore the sequence
converges to
.
Another important property of continuous functions is directly linked to the actual definition.
Theorem 5.4 (Continuity & Pre-Image of Closed Sets)
Let and
be topological spaces. A function
is continuous if and only if
is closed in
for every closed set
.
Proof. Let be closed in
and let
. We wish to prove that
is closed in
. We show that Cl
. By elementary set theory, we have
. Therefore, if
then
. Hence,
and Cl
and thus Cl
.
Theorem 5.5 (Continuous Functions & Compact Sets)
Let be continuous, and let
be compact in
. Then
is compact in
.
Proof. Let be continuous, and assume that
is compact in
. To show that
is compact in
, let
be a cover of
by open sets in
. Then
is open in
for every set
. Hence,
is a cover of
by open sets in
. Since
is compact there is a finite sub-collection of
that covers
. Thus,
has a finite sub-cover, implying that
is compact in
.
Appendix
A really nice introduction to the abstract concept of a topology, however, in German language only.
Literature
[1]
[2]
[3]
[4]
[5]