3.2.1 Topology, Partitions of Unity, and the Existence of Bump Functions
Definition 3.2.2 (Topological Space).
A topological space is a pair , where is a set and is a collection of subsets of satisfying the following properties:
- and .
- The union of any (possibly infinite) collection of sets in is also in .
- The intersection of any finite collection of sets in is also in .
The collection is called a topology on , and its elements are referred to as open sets under the topology .
Obviously the statement “let be a topological space” itself has little meaning. However, when the topology is implicitly obvious or the space is describable without it, then it may be verbally elided.
The implied topology of a subspace of is given by the intersection of each set in with .
Definition 3.2.3.
A subset of a topological space is closed iff is open.
It is immediate from definition that the trivial sets and are always closed. It is equally trivial from definition that the union of finitely many closed sets is closed, and the intersection of any collection of closed sets is closed.
If such that , then is an (open) neighborhood of . If (such that ) have disjoint neighborhoods, then is a Hausdorff space.
The following discussions involved with topological spaces here will always be of Hausdorff spaces, although making such distinction is important for future extensibility.
Definition 3.2.4.
A topological space is compact iff every open cover has a finite subcover. For a topological space , a set is compact iff every open cover has a finite subcover.
Proposition 3.2.1.
Suppose is a Hausdorff topological space and let be compact. Then is closed in .
Proof.
Let be fixed. For each , since is Hausdorff, there exist disjoint neighborhoods and with and . The set
covers , which by assumption, has a finite subcover
Moreover, the intersection
is an open neighborhood of and by construction, it is disjoint from the finite subcover. Since it is disjoint from a superset of , it lies entirely in .
For each , construct open accordingly. Then we obtain
where the sandwiched union is open. Therefore, is closed.
Proposition 3.2.2.
If is a compact space and is closed, then is compact.
Proof.
Let be an open cover of in . Since is open, the set
openly covers . Then a finite subcover
exists and covers . The refinement then covers .
Definition 3.2.5.
A point is an accumulation point of a set in a topological space iff any open with implies that contains a point other than .
Proposition 3.2.3.
A set in a topological space is closed iff it contains all its accumulation points.
Proof.
We first prove the forward implications under the assumption that is closed. Since is open, and suppose for contradiction, that . Then for open, (and hence cannot be an accumulation point by contradiction of definition).
Assume the converse assumption that contains all its accumulation points. Let be arbitrary. By assumption, is not an accumulation point of . Hence, for some open set , does not contain a point other than (which it also cannot contain), implying that , and hence .
For each , we hence construct some open neighborhood fully contained in . Together, they must union (by the definition of a topology) to an open set, being . Therefore, is closed.
A topology allows the definition and general conceptualization of continuity, convergence, and connectivity in a general setting, without necessarily relying on a notion of distance (a metric).
Definition 3.2.6.
A function between two topological spaces is said to be continuous if the pre-image of every open set in ,
is an open set in .
For the case of metric spaces, this generalizes the epsilon–delta notion of continuity.
Example 3.2.3.
Consider the function defined by
We equip both the domain and codomain with the standard topology on . Let . Then the pre-image of is
which is not an open set in the standard topology on . Thus, is not continuous.
For two topological spaces and , a function is a homeomorphism (also known as a bicontinuous function) if it is a bijection such that both and are continuous. If such a function exists, then and are homeomorphic.
The function with is indeed continuous, but the inverse is discontinuous at .
Proposition 3.2.4.
Let be two topological spaces. Then for , the following conditions are equivalent:
- is continuous.
- If is closed, then the pre-image is closed.
- If and is an open neighborhood of in , then there is some that is a neighborhood of such that .
Proof.
We first show that Part 1 implies Part 2. By continuity, for closed, , then
Assume the conditions of Part 2 for the converse. Let be open, closed, then is closed. Similar logic shows
which implies is open.
Next we aim to show that continuity implies Part 3. By assumption, the pre-image of any open is and open and . The property is complete under .
Assume the conditions of Part 3 for the converse. Let be arbitrary. We aim to show that . If , then the conclusion is satisfied trivially. Hence, assume that . For any such , there exists a neighborhood such that . Hence for any . Therefore, we obtain
By the definition of topologies,
Definition 3.2.7 (Basis for a Topology).
Let be a set. A basis for a topology on is a collection of subsets of satisfying
- .
-
For any and any point , there exists a set such that
The topology generated by is the collection of all unions of elements of .
Definition 3.2.8.
A metric space is a pair , where is a set and is a function from to , called a metric, such that for all the following properties hold:
- and iff (positivity).
- (symmetry).
- (triangle inequality).
This in turn implies the reverse triangle inequality:
and similarly,
Definition 3.2.9.
Let be a metric space. The metric topology induced by is the topology generated by the basis
comprising the balls
The pair is the topological space induced by the metric .
We now justify a claim whose triviality we have taken for granted.
Proposition 3.2.5.
Let be a metric space under the induced metric topology. Then for any open set , any point , there exists a ball () in .
Proof.
By definition, lies in the topology for and is the union of (possibly infinitely many) balls. There then exists some ball in that contains . Let
Since , for any , we have
Hence, the open ball centered at lies within .
Theorem 3.2.7.
Let be two metric spaces under the metric topology. Then a function is topologically continuous iff it is epsilon–delta continuous.
Proof.
We first imply that topological continuity implies epsilon–delta continuity. For any , , the ball is an open set (it is in the basis) in . By Part 3 of Proposition 3.2.4, there is some open neighborhood of in such that . By the previous proposition, there is a ball . This is equivalent to
Conversely, assume is – continuous. Let be open and . Since is open in the metric topology, there exists such that
By epsilon–delta continuity, there exists such that
or that
Thus, the ball is an open neighborhood of in such that
Since the union of open sets is open, the pre-image of any open set is open, and hence is topologically continuous.
Theorem 3.2.8.
Let be a compact topological space and let be a Hausdorff space. If is a continuous bijection, then is a homeomorphism.
Proof.
If is compact, then the pre-images of any open cover of cover . Hence, there is a finite subcover
covering . Then
covers , and hence is compact.
For any closed , Proposition 3.2.2 implies is compact. Hence, is compact. By Proposition 3.2.1, is closed. Hence, maps closed sets to closed sets, and the pre-image of any closed set is closed under . Hence, Proposition 3.2.4 implies is continuous, thus is a homeomorphism.
It is worth noting some motivating examples for which the conclusion fails when certain hypotheses are not satisfied.
Example 3.2.4.
Let be a topological space with a metric under the standard topology (the subspace topology induced by the basis formed with open “balls” or symmetric intervals around each point). Equip the unit circle
generated by the metric defined by arc-length . Then the continuous bijection defined by is not a homeomorphism.
Proof.
The non-continuity of is easy to visually see, both topologically and by epsilon–delta. Topologically, select the relatively open interval in . The pre-image of this set under is , which is clearly not an open set. This proves that is not continuous (by definition).
For continuity to hold by epsilon–delta, any would yield the existence of some such that with , .
Let . For any , the points
satisfy
However,
and
This contradicts the previous statement.
We now provide a formal definition of the connectivity of sets:
Definition 3.2.10.
A topological space is disconnected if it can be written as the union of two nonempty disjoint open sets. Otherwise, it is connected.
In a topological space , a subset can be open, closed (the complement of some open set), both (clopen), or neither. The only clopen sets that exist in any topological space are and iff is connected. A technique pertinent to many future proofs relies on the following fact:
Theorem 3.2.9 (Connectivity Argument).
A topological space is connected if and only if and are the only clopen subsets of .
Proof.
Suppose is connected and let be clopen. Then and are both open in , disjoint, and their union is . Thus, either or (i.e. ).
Conversely, suppose is disconnected. Then there exist nonempty open sets such that and . Thus,
and
are both clopen, contradicting the assumption that and are the only clopen subsets. Hence, must be connected.
Example 3.2.5.
The topological space under the standard topology has only two clopen sets: and .
Now consider
equipped with the topology generated by the basis
This space is disconnected. For instance, is open (as it is in ) and closed (since its complement in is
In fact, every set in is clopen.
Proposition 3.2.6.
The interval (under the subspace topology induced by ) is connected.
Proof.
Assume, for contradiction, that there exist two disjoint nonempty open sets such that
Without loss of generality, assume (otherwise switch and ). Let
Since are also closed in , either or is an accumulation point. Either way, is contained in by Proposition 3.2.3. Assume that . Then by openness, there exists some such that
In particular,
which contradicts being a lower bound of .
Therefore, . However, since lies in for some , . Thus, we arrive at a contradiction, and thus is the empty set. This then shows that is connected.
Connectivity intuitively means that a space cannot be split into two disjoint open subsets, but is not meaningful in terms of how points within the space relate to each other. In many geometric situations, the notion of path-connectivity requires that any two points be joined by a continuous path. We will see that this more concrete condition forces the space to be topologically connected.
Definition 3.2.11.
A topological space is said to be path-connected iff for any two points , there is a continuous function , where is equipped with the metric topology and .
Theorem 3.2.10.
A path-connected topological space is connected.
Proof.
Assume path-connectivity and suppose is disconnected. Then two open nonempty disjoint components can be found. Let be two arbitrary points. Then there exists such that , . By continuity, the pre-images of and , namely and respectively, are disjoint open subsets of . Moreover, the pre-image are nonempty as they contain and respectively. This contradicts the connectivity of in Proposition 3.2.6.
Definition 3.2.12 (Exhaustion by Compact Sets).
For a topological space , an exhaustion by compact sets is a nested sequence of compact sets such that for all and
Lemma 3.2.1.
Let be an open set and let be a basis for the topology on . Then there exists a collection of sets such that
- .
- For every compact , intersects only finitely many sets in .
Proof.
Let be an exhaustion by compact sets with and for all . For each , define
where . Each is open and each is compact, with and
For each and each , since is open and contains , there exists such that
The collection
is an open cover of the compact set , so by Heine–Borel (Theorem 1.1.3) it admits a finite subcover, there exist finitely many points such that
Enumerate all such over and to obtain a countable collection . Then
proving Part 1.
For 2, let be compact. There exists such that
so is disjoint from for all . Since each intersects only finitely many , intersects only finitely many . Thus the collection is locally finite.
Remark.
The property of local finiteness of an open collection in is commonly stated as: for every , there exists an open neighborhood of that intersects only finitely many sets in .
This is equivalent to Part 2 in Lemma 3.2.1. Indeed, if every point has such a neighborhood, then any compact admits a finite subcover of these neighborhoods by Heine–Borel (Theorem 1.1.3), so intersects finitely many sets in . Conversely, for any , take an open neighborhood with and with relatively compact closure in ; then intersects finitely many sets in , and so does .
Theorem 3.2.11 (Partition of Unity).
Let be a nonempty open set and let be an open cover of . Then there exists a collection of bump functions , each with compact support in , satisfying:
- For each , there exists such that .
- The collection is locally finite.
- For each , .
- on .
Then is called a partition of unity subordinate to .
Proof.
For each there exists and such that
The collection
is an open basis for . By Lemma 3.2.1 there exists a locally finite open cover
of with
Define the standard bump function
For let
which has support . Define
so
By local finiteness of , for each there exists an open neighborhood with intersecting only finitely many . Thus is locally finite on . Then the sum
defined for involves only finitely many nonzero terms (by local finiteness) on a neighborhood of every point . Hence and (since covers ). Define
Each has compact support in , , the supports are locally finite, and
for all . Moreover
proving subordination.
Theorem 3.2.12 (Existence of Bump Functions).
Let be compact and an open neighborhood of . Then there exists a compactly supported such that
, and on some open neighborhood of .
Proof.
Let
denote the open -neighborhood of . Since is an open neighborhood of , such that
where means that the closure of is compact and contained in .
Define the open sets
Then is an open cover of .
By the Partition of Unity Theorem (Theorem 3.2.11), there exist compactly supported functions forming a partition of unity subordinate to this cover. That is,
Let
Define
Then is compactly supported within , and since only finitely many are nonzero on a neighborhood of each point, . Moreover,
For , all functions with support in vanish at , so
Hence, on the open neighborhood of . Outside , all terms with support in vanish, so . Finally, everywhere by construction. Thus satisfies all required properties.