§3.2.1 Topology, Partitions of Unity, and the Existence of Bump Functions

3.2.1 Topology, Partitions of Unity, and the Existence of Bump Functions

Definition 3.2.1.1 (Topological Space).

A topological space is a pair $(𝑋, 𝜏)$ , where $𝑋$ is a set and $𝜏$ is a collection of subsets of $𝑋$ satisfying the following properties:

$⌀ \in 𝜏$ and $𝑋 \in 𝜏$ .
The union of any (possibly infinite) collection of sets in $𝜏$ is also in $𝜏$ .
The intersection of any finite collection of sets in $𝜏$ is also in $𝜏$ .

The collection $𝜏$ is called a topology on $𝑋$ , and its elements are referred to as open sets under the topology $𝜏$ .

Obviously the statement “let $𝑋$ 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 $𝐴$ of $(𝑋, 𝜏)$ is given by the intersection of each set in $𝜏$ with $𝐴$ .

Definition 3.2.1.2.

A subset

𝐴

of a topological space

𝑋

is closed iff

𝑋 ∖ 𝐴

is open.

It is immediate from definition that the trivial sets $𝑋$ 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 $\exists 𝑈 \in 𝜏$ such that $𝑥 \in 𝑈$ , then $𝑈$ is an (open) neighborhood of $𝑥$ . If $\forall 𝑥, 𝑦 \in 𝑋$ (such that $𝑥 \neq 𝑦$ ) have disjoint neighborhoods, then $𝑋$ 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 3.2.1.3.

A topological space

𝑋

is compact iff every open cover has a finite subcover. For a topological space

𝑋

, a set

𝐴 \subseteq 𝑋

is compact iff every open cover has a finite subcover.

Proposition 3.2.1.4.

Suppose

𝑋

is a Hausdorff topological space and let

𝐴 \subseteq 𝑋

be compact. Then

𝐴

is closed in

𝑋

Proof.

Let $𝑥 \in 𝑋 ∖ 𝐴$ be fixed. For each $𝑎 \in 𝐴$ , since $𝑋$ is Hausdorff, there exist disjoint neighborhoods $𝑈_{𝑎}$ and $𝑉_{𝑎}$ with $𝑥 \in 𝑈_{𝑎}$ and $𝑎 \in 𝑉_{𝑎}$ . The set

⋃_{𝑎 \in 𝐴} 𝑉_{𝑎} \supseteq 𝐴

covers $𝐴$ , which by assumption, has a finite subcover

⋃_{𝑘 = 1}^{𝑛} 𝑉_{𝑎_{𝑘}} \supseteq 𝐴, \forall 𝑘 \in ℕ_{\leq 𝑛}, 𝑎_{𝑘} \in 𝐴 .

Moreover, the intersection

𝑈_{𝑥} = ⋂_{𝑘 = 1}^{𝑛} 𝑈_{𝑎_{𝑘}}

is an open neighborhood of $𝑥$ and by construction, it is disjoint from the finite subcover. Since it is disjoint from a superset of $𝐴$ , it lies entirely in $𝑋 ∖ 𝐴$ .

For each $𝑥 \in 𝑋 ∖ 𝐴$ , construct open $𝑈_{𝑥}$ accordingly. Then we obtain

𝑋 ∖ 𝐴 \subseteq ⋃_{𝑥 \in 𝑋 ∖ 𝐴} 𝑈_{𝑥} \subseteq 𝑋 ∖ 𝐴,

where the sandwiched union is open. Therefore, $𝐴$ is closed. ⁠ $□$

Proposition 3.2.1.5.

𝑋

is a compact space and

𝐴 \subseteq 𝑋

is closed, then

𝐴

is compact.

Proof.

Let $𝒰︀$ be an open cover of $𝐴$ in $𝑋$ . Since $𝑋 ∖ 𝐴$ is open, the set

{𝑈 \cup (𝑋 ∖ 𝐴) : 𝑈 \in 𝒰︀}

openly covers $𝑋$ . Then a finite subcover

{𝑈_{𝑘} \cup (𝑋 ∖ 𝐴)}_{𝑘 \in ℕ_{\leq 𝑛}}

exists and covers

𝑋

. The refinement

{𝑈_{𝑘}}_{𝑘 \in ℕ_{\leq 𝑛}}

then covers

𝐴

. ⁠

□

Definition 3.2.1.6.

A point

𝑎

is an accumulation point of a set

𝐴

in a topological space

𝑋

iff any open

𝑈

with

𝑎 \in 𝑈

implies that

𝑈 \cap 𝐴

contains a point other than

𝑎

Proposition 3.2.1.7.

A set

𝐴

in a topological space

𝑋

is closed iff it contains all its accumulation points.

Proof.

We first prove the forward implications under the assumption that $𝐴$ is closed. Since $𝑋 ∖ 𝐴$ is open, and suppose for contradiction, that $𝑎 \in 𝑋 ∖ 𝐴$ . Then for $𝑎 \in 𝑈 = 𝑋 ∖ 𝐴$ open, $𝑈 \cap 𝐴 = ⌀$ (and hence $𝑎$ cannot be an accumulation point by contradiction of definition).

Assume the converse assumption that $𝐴$ contains all its accumulation points. Let $𝑥 \in 𝑋 ∖ 𝐴$ be arbitrary. By assumption, $𝑥$ is not an accumulation point of $𝐴$ . Hence, for some open set $𝑈 \supseteq {𝑥}$ , $𝑈 \cap 𝐴$ does not contain a point other than $𝑥$ (which it also cannot contain), implying that $𝑈 \cap 𝐴 = ⌀$ , and hence $𝑈 \subseteq 𝑋 ∖ 𝐴$ .

For each $𝑥 \in 𝑋 ∖ 𝐴$ , we hence construct some open neighborhood fully contained in $𝑋 ∖ 𝐴$ . Together, they must union (by the definition of a topology) to an open set, being $𝑋 ∖ 𝐴$ . Therefore, $𝐴$ 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 3.2.1.8.

A function $𝑓 : 𝑋 \to 𝑌$ between two topological spaces is said to be continuous if the pre-image of every open set in $𝑌$ ,

{𝑥 \in 𝑋 : 𝑓 (𝑥) \in 𝑌},

is an open set in

𝑋

For the case of metric spaces, this generalizes the $𝜀$ – $𝛿$ notion of continuity.

Example 3.2.1.9.

Consider the function $𝑓 : ℝ \to ℝ$ defined by

𝑓 (𝑥) = {\begin{cases} 1 & if 𝑥 \geq 0, \\ 0 & if 𝑥 < 0 . \end{cases}

We equip both the domain and codomain with the standard topology on $ℝ$ . Let $𝑉 = (. 5, 1.5) \subseteq ℝ$ . Then the pre-image of $𝑉$ is

𝑓^{- 1} (𝑉) = {𝑥 \in ℝ : 𝑓 (𝑥) \in 𝑉} = ℝ_{\geq 0},

which is not an open set in the standard topology on $ℝ$ . Thus, $𝑓$ is not continuous.

For two topological spaces $𝑋$ and $𝑌$ , a function $𝑓 : 𝑋 \to 𝑌$ is a homeomorphism (also known as a bicontinuous function) if it is a bijection such that both $𝑓$ and $𝑓^{- 1}$ are continuous. If such a function exists, then $𝑋$ and $𝑌$ are homeomorphic.

The function $𝑓 : [0, 2 π) \to 𝑆^{1}$ with $𝑓 (𝑡) = (cos (𝑡), sin (𝑡))$ is indeed continuous, but the inverse $𝑓^{- 1} (𝑥_{1}, 𝑥_{2})$ is discontinuous at $(𝑥_{1}, 𝑥_{2}) = (1, 0)$ .

Proposition 3.2.1.10.

Let $(𝑋, 𝜏_{1}), (𝑌, 𝜏_{2})$ be two topological spaces. Then for $𝑓 : 𝑋 \to 𝑌$ , the following conditions are equivalent:

$𝑓$ is continuous.
If $𝐴 \subseteq 𝑌$ is closed, then the pre-image $𝑓^{- 1} (𝐴)$ is closed.
If $𝑎 \in 𝑋$ and $𝐴 \in 𝜏_{2}$ is an open neighborhood of $𝑓 (𝑎)$ in $𝑌$ , then there is some $𝑈 \in 𝜏_{1}$ that is a neighborhood of $𝑎$ such that $𝑓 (𝑈) \subseteq 𝐴$ .

Proof.

We first show that Part 1 $Part 1 Part 1$ implies Part 2 $Part 2 Part 2$ . By continuity, for $𝐴 \subseteq 𝑌$ closed, $𝑌 ∖ 𝐴 \in 𝜏_{2}$ , then

𝑓^{- 1} (𝑌 ∖ 𝐴) = {𝑥 \in 𝑋 : 𝑓 (𝑥) \in 𝑌 ∖ 𝐴} = 𝑋 ∖ 𝑓^{- 1} (𝐴) .

Assume the conditions of Part 2 $Part 2 Part 2$ for the converse. Let $𝑈 \in 𝜏_{2}$ be open, $𝑌 ∖ 𝑈$ closed, then $𝑓^{- 1} (𝑌 ∖ 𝑈)$ is closed. Similar logic shows

𝑓^{- 1} (𝑌 ∖ 𝑈) = 𝑋 ∖ 𝑓^{- 1} (𝑈),

which implies $𝑓^{- 1} (𝑈)$ is open.

Next we aim to show that continuity implies Part 3 $Part 3 Part 3$ . By assumption, the pre-image of any open $𝐴 \subseteq 𝑌$ is $𝑓^{- 1} (𝐴)$ and open and $𝑎 \in 𝑓^{- 1} (𝐴)$ . The property is complete under $𝑈 = 𝑓^{- 1} (𝐴)$ .

Assume the conditions of Part 3 $Part 3 Part 3$ for the converse. Let $𝐴 \in 𝜏_{2}$ be arbitrary. We aim to show that $𝑓^{- 1} (𝐴) \in 𝜏_{1}$ . If $𝑓^{- 1} (𝐴) = ⌀$ , then the conclusion is satisfied trivially. Hence, assume that $\exists 𝑎 \in 𝑓^{- 1} (𝐴)$ . For any such $𝑎$ , there exists a neighborhood $𝑈_{𝑎} \in 𝜏_{1}$ such that $𝑓 (𝑈_{𝑎}) \subseteq 𝐴$ . Hence $𝑈_{𝑎} \subseteq 𝑓^{- 1} (𝐴)$ for any $𝑎 \in 𝑓^{- 1} (𝐴)$ . Therefore, we obtain

\begin{matrix} ⋃_{𝑎 \in 𝑓^{- 1} (𝐴)} 𝑈_{𝑎} \subseteq 𝑓^{- 1} (𝐴), \\ ⋃_{𝑎 \in 𝑓^{- 1} (𝐴)} 𝑈_{𝑎} \supseteq ⋃_{𝑎 \in 𝑓^{- 1} (𝐴)} {𝑎} = 𝑓^{- 1} (𝐴) ⟹ ⋃_{𝑎 \in 𝑓^{- 1} (𝐴)} 𝑈_{𝑎} = 𝑓^{- 1} (𝐴) . \end{matrix}

By the definition of topologies,

⋃_{𝑎 \in 𝑓^{- 1} (𝐴)} 𝑈_{𝑎} \in 𝜏_{1} . □

Definition 3.2.1.11 (Basis for a Topology).

Let $𝑋$ be a set. A basis for a topology on $𝑋$ is a collection $𝔅$ of subsets of $𝑋$ satisfying

$⋃_{𝐵 \in 𝔅} 𝐵 = 𝑋$ .
For any $𝐵_{1}, 𝐵_{2} \in 𝔅$ and any point $𝑥 \in 𝐵_{1} \cap 𝐵_{2}$ , there exists a set $𝐵_{3} \in 𝔅$ such that
$𝑥 \in 𝐵_{3} \subseteq 𝐵_{1} \cap 𝐵_{2} .$

The topology generated by $𝔅$ is the collection of all unions of elements of $𝔅$ .

Definition 3.2.1.12 (Metric Space).

A metric space is a pair $(𝑋, 𝑑)$ , where $𝑋$ is a set and $𝑑$ is a function from $𝑋 \times 𝑋$ to $ℝ_{\geq 0}$ , called a metric, such that for all $𝑥, 𝑦, 𝑧 \in 𝑋$ the following properties hold:

$𝑑 (𝑥, 𝑦) \geq 0$ and $𝑑 (𝑥, 𝑦) = 0$ iff $𝑥 = 𝑦$ (positivity).
$𝑑 (𝑥, 𝑦) = 𝑑 (𝑦, 𝑥)$ (symmetry).
$𝑑 (𝑥, 𝑧) \leq 𝑑 (𝑥, 𝑦) + 𝑑 (𝑦, 𝑧)$ (triangle inequality).

This in turn implies the reverse triangle inequality:

𝑑 (𝑥, 𝑧) \leq 𝑑 (𝑥, 𝑦) + 𝑑 (𝑦, 𝑧) ⟹ 𝑑 (𝑥, 𝑦) \geq 𝑑 (𝑥, 𝑧) - 𝑑 (𝑦, 𝑧),

and similarly,

𝑑 (𝑦, 𝑧) \leq 𝑑 (𝑥, 𝑦) + 𝑑 (𝑥, 𝑧) ⟹ 𝑑 (𝑥, 𝑦) \geq 𝑑 (𝑦, 𝑧) - 𝑑 (𝑥, 𝑧) .

Definition 3.2.1.13.

Let $(𝑋, 𝑑)$ be a metric space. The metric topology induced by $𝑑$ is the topology $𝜏_{𝑑}$ generated by the basis

{𝐵 (𝑥, 𝑟) : 𝑥 \in 𝑋, 𝑟 > 0}

comprising the balls

𝐵 (𝑥, 𝑟) = {𝑦 \in 𝑋 : 𝑑 (𝑥, 𝑦) < 𝑟} .

The pair $(𝑋, 𝜏_{𝑑})$ is the topological space induced by the metric $𝑑$ .

Note that in most metric spaces, closedness and boundedness of sets do not generally give compactness. We now justify a claim whose triviality we have taken for granted.

Proposition 3.2.1.14.

Let

(𝑋, 𝑑)

be a metric space under the induced metric topology. Then for any open set

𝑈 \subseteq 𝑋

, any point

𝑥 \in 𝑈

, there exists a ball

𝐵 (𝑥, 𝛿)

(

𝛿 > 0

) in

𝑈

Proof.

By definition, $𝑈$ lies in the topology for $𝑋$ and is the union of (possibly infinitely many) balls. There then exists some ball $𝐵 (𝑥_{0}, 𝛿^{'})$ in $𝑈$ that contains $𝑥$ . Let

𝛿 = 𝛿^{'} - 𝑑 (𝑥_{0}, 𝑥) > 0 since 𝑑 (𝑥_{0}, 𝑥) < 𝛿^{'} .

For any $𝑦 \in 𝐵 (𝑥, 𝛿)$ , we have

𝑑 (𝑥_{0}, 𝑦) \leq 𝑑 (𝑥_{0}, 𝑥) + 𝑑 (𝑥, 𝑦) \leq 𝛿^{'} .

Hence, the open ball

𝐵 (𝑥, 𝛿)

centered at

𝑥

lies within

𝐵 (𝑥_{0}, 𝛿^{'})

. ⁠

□

Theorem 3.2.1.15.

Let

(𝑋, 𝑑_{𝑥}), (𝑌, 𝑑_{𝑦})

be two metric spaces under the metric topology. Then a function

𝑓 : 𝑋 \to 𝑌

is topologically continuous iff it is epsilon–delta continuous.

Proof.

We first imply that topological continuity implies epsilon–delta continuity. For any $𝑥 \in 𝑋$ , $\forall 𝜀 > 0$ , the ball $𝐵 (𝑓 (𝑥), 𝜀)$ is an open set (it is in the basis) in $𝑌$ . By Part 3 $Part 3 Part 3$ of Proposition 3.2.1.10 $Proposition 3.2.1.10 Proposition 3.2.1.10$ , there is some open neighborhood $𝑈$ of $𝑥$ in $𝑋$ such that $𝑓 (𝑈) \subset 𝐵 (𝑓 (𝑥), 𝜀)$ . By the previous proposition, there is a ball $𝐵 (𝑥, 𝛿) \subseteq 𝑈$ . This is equivalent to

𝜀 > 0, 𝑥 \in 𝑋 ⟹ \exists 𝛿 = 𝛿_{𝑥} > 0 : 𝑦 \in 𝐵 (𝑥, 𝛿) ⟹ 𝑓 (𝑦) \in 𝐵 (𝑓 (𝑥), 𝜀) .

Conversely, assume $𝑓$ is $𝜀$ – $𝛿$ continuous. Let $𝑉 \subseteq 𝑌$ be open and $𝑥 \in 𝑓^{- 1} (𝑉)$ . Since $𝑉$ is open in the metric topology, there exists $𝜀 > 0$ such that

𝐵 (𝑓 (𝑥), 𝜀) \subseteq 𝑉 .

By epsilon–delta continuity, there exists $𝛿 > 0$ such that

𝑑_{𝑥} (𝑥, 𝑦) < 𝛿 ⟹ 𝑑_{𝑦} (𝑓 (𝑥), 𝑓 (𝑦)) < 𝜀,

or that

𝑦 \in 𝐵 (𝑥, 𝛿) ⟹ 𝑓 (𝑦) \in 𝐵 (𝑓 (𝑥), 𝜀) \subseteq 𝑉 .

Thus, the ball $𝐵 (𝑥, 𝛿_{𝑥})$ is an open neighborhood of $𝑥$ in $𝑋$ such that

𝐵 (𝑥, 𝛿_{𝑥}) \subseteq 𝑓^{- 1} (𝑉) ⟹ 𝑓^{- 1} (𝑉) \subseteq ⋃_{𝑥 \in 𝑓^{- 1} (𝑉)} 𝐵 (𝑥, 𝛿_{𝑥}) \subseteq 𝑓^{- 1} (𝑉) .

Since the union of open sets is open, the pre-image of any open set is open, and hence $𝑓$ is topologically continuous. ⁠ $□$

Theorem 3.2.1.16 (Compact–Hausdorff).

Let

𝑋

be a compact topological space and let

𝑌

be a Hausdorff space. If

𝑓 : 𝑋 \to 𝑌

is a continuous bijection, then

𝑓

is a homeomorphism.

Proof.

If $𝐴 \subseteq 𝑋$ is compact, then the pre-images of any open cover $𝒰︀$ of $𝑓 (𝐴)$ cover $𝐴$ . Hence, there is a finite subcover

{𝑓^{- 1} (𝑈_{𝑘}) : 𝑈_{𝑘} \in 𝒰︀, 𝑘 \in ℕ_{\leq 𝑛}}

covering $𝐴$ . Then

{𝑈_{𝑘} : 𝑈_{𝑘} \in 𝒰︀, 𝑘 \in ℕ_{\leq 𝑛}}

covers $𝑓 (𝐴)$ , and hence $𝑓 (𝐴)$ is compact.

For any closed $𝐶 \subseteq 𝑋$ , Proposition 3.2.1.5 $Proposition 3.2.1.5 Proposition 3.2.1.5$ implies $𝐶$ is compact. Hence, $𝑓 (𝐶)$ is compact. By Proposition 3.2.1.4 $Proposition 3.2.1.4 Proposition 3.2.1.4$ , $𝑓 (𝐶)$ is closed. Hence, $𝑓$ maps closed sets to closed sets, and the pre-image of any closed set is closed under $𝑓^{- 1}$ . Hence, Proposition 3.2.1.10 $Proposition 3.2.1.10 Proposition 3.2.1.10$ implies $𝑓^{- 1}$ is continuous, thus $𝑓$ is a homeomorphism. ⁠ $□$

It is worth noting some motivating examples for which the conclusion fails when certain hypotheses are not satisfied.

Example 3.2.1.17.

Let $𝐼 = [0, 2 π)$ be a topological space with a metric $| \cdot |$ 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

𝑆^{1} = {𝑧 \in ℂ : | 𝑧 | = 1}

with a topology generated by the metric defined by arc-length

(𝑑_{𝑆^{1}})

. Then the continuous bijection

𝑓 : 𝐼 \to 𝑆^{1}

defined by

𝑓 (𝑡) = e^{i 𝑡}

is not a homeomorphism.

Solution to Example 3.2.1.17 $Example 3.2.1.17 Example 3.2.1.17$ .

The non-continuity of $𝑓^{- 1} : 𝑆^{1} \to 𝐼$ is easy to visually see, both topologically and by epsilon–delta. Topologically, select the relatively open interval $[0, π)$ in $𝐼$ . The pre-image of this set under $𝑓^{- 1}$ is $e^{i [0, π)}$ , which is clearly not an open set. This proves that $𝑓^{- 1}$ is not continuous (by definition).

For continuity to hold by epsilon–delta, any $𝜀$ would yield the existence of some $𝛿$ such that $\forall 𝑎, 𝑏 \in 𝑆^{1}$ with $𝑑_{𝑆^{1}} (𝑎, 𝑏) < 𝛿$ , $| 𝑓^{- 1} (𝑎) - 𝑓^{- 1} (𝑏) | < 𝜀$ .

Let $𝜀 = \frac{π}{2}$ . For any $0 < 𝛿 < 2 π$ , the points

𝑎 = e^{- i \frac{𝛿}{4}}, 𝑏 = e^{i \frac{𝛿}{4}}

satisfy

𝑑_{𝑆^{1}} (𝑎, 𝑏) = \frac{𝛿}{2} < 𝛿 .

However,

𝑓^{- 1} (𝑎) = 2 π - \frac{𝛿}{4}, 𝑓^{- 1} (𝑏) = \frac{𝛿}{4},

and

| 𝑓^{- 1} (𝑎) - 𝑓^{- 1} (𝑏) | = 2 π - \frac{𝛿}{2} > π > 𝜀 .

This contradicts the previous statement. ⁠ $□$

We now provide a formal definition of the connectivity of sets:

Definition 3.2.1.18.

A topological space

𝑋

is disconnected if it can be written as the union of two nonempty disjoint open sets. Otherwise, it is connected.

In a topological space $𝑋$ , 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 $𝑋$ are $⌀$ and $𝑋$ iff $𝑋$ is connected. A technique pertinent to many future proofs relies on the following fact:

Theorem 3.2.1.19 (Connectivity Argument).

A topological space

𝑋

is connected if and only if

𝑋

and

⌀

are the only clopen subsets of

𝑋

Example 3.2.1.20.

The topological space

ℝ

under the standard topology has only two clopen sets:

ℝ

and

⌀

Example 3.2.1.21.

Consider

𝑋 = ⋃_{𝑛 \in 2 ℤ} (𝑛, 𝑛 + 1),

equipped with the topology $𝜏$ generated by the basis

{(𝑛, 𝑛 + 1) : 𝑛 \in 2 ℤ} .

This space is disconnected. For instance, $(0, 1) \subset 𝑋$ is open (as it is in $𝜏$ ) and closed (since its complement in $𝑋$ is

⋃_{𝑛 \in (2 ℤ ∖ {0})} (𝑛, 𝑛 + 1) \in 𝜏) .

In fact, every set in

𝜏

is clopen.

Proposition 3.2.1.22.

The interval

[0, 1]

(under the subspace topology induced by

ℝ

) is connected.

Proof.

Assume, for contradiction, that there exist two disjoint nonempty open sets $𝑈, 𝑉 \subset [0, 1]$ such that

𝑈 \cup 𝑉 = [0, 1] .

Without loss of generality, assume $0 \in 𝑈$ (otherwise switch $𝑈$ and $𝑉$ ). Let

𝑎 = inf 𝑉 .

Since $𝑈, 𝑉$ are also closed in $[0, 1]$ , either $𝑎 \in 𝑉$ or $𝑎$ is an accumulation point. Either way, $𝑎$ is contained in $𝑉$ by Proposition 3.2.1.7 $Proposition 3.2.1.7 Proposition 3.2.1.7$ . Assume that $𝑎 \neq 0$ . Then by openness, there exists some $0 < 𝛿 < 𝑎$ such that

(𝑎 - 𝛿, 𝑎) \subseteq (𝑎 - 𝛿, 𝑎 + 𝛿) \cap [0, 1] \subseteq 𝑉 .

In particular,

𝑎 - \frac{𝛿}{2} \in 𝑉,

which contradicts $𝑎$ being a lower bound of $𝑉$ .

Therefore, $𝑎 = 0$ . However, since $[0, 𝛿)$ lies in $𝑈$ for some $𝛿 > 0$ , $𝑎 \geq 𝛿 > 0$ . Thus, we arrive at a contradiction, and thus $𝑉$ is the empty set. This then shows that $[0, 1]$ 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 3.2.1.23 (Path Connectivity).

A topological space

𝑋

is said to be path-connected iff for any two points

𝑎, 𝑏 \in 𝑋

, there is a continuous function

𝑓 : [0, 1] \to 𝑋

, where

[0, 1]

is equipped with the metric topology and

𝑓 (0) = 𝑎, 𝑓 (1) = 𝑏

Theorem 3.2.1.24.

A path-connected topological space

𝑋

is connected.

Proof.

Assume path-connectivity and suppose

𝑋

is disconnected. Then two open nonempty disjoint components

𝑈, 𝑉 \subset 𝑋

can be found. Let

𝑢 \in 𝑈, 𝑣 \in 𝑉

be two arbitrary points. Then there exists

𝑓 : [0, 1] \to 𝑋

such that

𝑓 (0) = 𝑢

𝑓 (1) = 𝑣

. By continuity, the pre-images of

𝑈

and

𝑉

, namely

𝑓^{- 1} (𝑈)

and

𝑓^{- 1} (𝑉)

respectively, are disjoint open subsets of

[0, 1]

. Moreover, the pre-image are nonempty as they contain

0

and

1

respectively. This contradicts the connectivity of

[0, 1]

in Proposition 3.2.1.22

Proposition 3.2.1.22 Proposition 3.2.1.22

. ⁠

□

Note that the converse here is not necessarily true.

Example 3.2.1.25.

The topologist’s sine curve under the subspace topology from $ℝ^{2}$ (with the Euclidean metric), defined by

𝑇 = {(𝑥, sin \frac{1}{𝑥}) : 𝑥 \in (0, 1]} \cup {(0, 0)},

(which can be thought to be the points on the graph of

𝑦 = sin \frac{1}{𝑥}

) is connected but not path-connected.

Solution to Example 3.2.1.25 $Example 3.2.1.25 Example 3.2.1.25$ .

Let
$𝑇^{'} = 𝑇 ∖ {0} = {(𝑥, sin \frac{1}{𝑥}) : 𝑥 \in (0, 1]}$
under the subspace topology, which is trivially path-connected. By Theorem 3.2.1.24 $Theorem 3.2.1.24 Theorem 3.2.1.24$ , this subspace is connected. If $𝑇$ is disconnected, then there exists a non-trivial clopen set $𝑈 \subset 𝑇$ including $0$ Then $𝑈 \neq {0}$ . Then $𝑈 \cap 𝑇^{'} = 𝑈 ∖ {0}$ would be a non-trivial clopen set in $𝑇^{'}$ , which is an impossibility. Thus, $𝑇$ is a connected space.
Assume for contradiction that there exists a continuous path $𝛾 (𝑡) = (𝑥 (𝑡), 𝑦 (𝑡)) : [0, 1] \to 𝑇$ with $𝛾 (0) = (0, 0)$ and $𝛾 (1) = (1, sin (1))$ .
Then $𝑥 ([0, 1]) = [0, 1]$ by the Intermediate Value Theorem. For each $𝑡$ where $𝑥 (𝑡) \neq 0$ , $𝑦 (𝑡) = sin (\frac{1}{𝑥 (𝑡)})$ . For each natural number $𝑛$ , let
$𝑥_{𝑛} = \frac{1}{\frac{π}{2} + 2 π 𝑛} and {\tilde{𝑥}}_{𝑛} = \frac{1}{- \frac{π}{2} + 2 π 𝑛},$
where $𝑥_{𝑛} \to 0$ and ${\tilde{𝑥}}_{𝑛} \to 0$ . Let $𝑡^{'} = inf 𝛾^{- 1} (𝑇^{'})$ . Then by continuity, $𝑥 (𝑡^{'}) = 0$ , and for all $𝑡 \leq 𝑡^{'}$ , $𝑥 (𝑡) = 0$ .
By continuity, $\forall 𝜀 > 0$ , $\exists 𝛿 : 0 < 𝛿 < 𝜀$ such that $\forall 𝑡 \in [𝑡^{'}, 𝑡^{'} + 𝛿)$ , $| 𝑥 (𝑡) | < 𝜀$ . By the definition of infimum, $\exists 𝑡 = 𝑡_{𝜀} \in (𝑡^{'}, 𝑡^{'} + 𝛿)$ such that $𝑥 (𝑡) \neq 0$ . As $𝜀 \to 0$ , $𝑡_{𝜀} \to 𝑡^{'}$ since $𝑡^{'} < 𝑡_{𝜀} < 𝑡^{'} + 𝛿 \leq 𝑡^{'} + 𝜀$ .
For each $𝑛 \in ℕ$ , let $𝜀 = \frac{1}{𝑛}$ , and there exist $𝑚_{𝑛} \in ℕ$ such that $𝑥_{𝑚_{𝑛}} < 𝑥 (𝑡_{𝜀})$ and ${\tilde{𝑥}}_{𝑚_{𝑛}} < 𝑥 (𝑡_{𝜀})$ . By the Intermediate Value Theorem, there exists some $𝑡_{𝑛}, {\tilde{𝑡}}_{𝑛}$ where $𝑡^{'} < 𝑡_{𝑛}, {\tilde{𝑡}}_{𝑛} < 𝑡_{𝜀} < 𝑡^{'} + \frac{1}{𝑛}$ such that $𝑥 (𝑡_{𝑛}) = 𝑥_{𝑚_{𝑛}}$ and $𝑥 ({\tilde{𝑡}}_{𝑛}) = {\tilde{𝑥}}_{𝑚_{𝑛}}$ .
For each $𝑡_{𝑛}$ , $𝑦 (𝑡_{𝑛}) = sin \frac{1}{𝑥 (𝑡_{𝑛})} = sin \frac{1}{𝑥_{𝑚_{𝑛}}} = 1$ , and for each ${\tilde{𝑡}}_{𝑛}$ , $𝑦 ({\tilde{𝑡}}_{𝑛}) = sin \frac{1}{𝑥 ({\tilde{𝑡}}_{𝑛})} = sin \frac{1}{{\tilde{𝑥}}_{𝑚_{𝑛}}} = - 1$ . Thus, there exist two sequences of $𝑡$ that approach $𝑡^{' +}$ for which $𝑦 \to 1$ and $𝑦 \to - 1$ . Thus $𝑦$ and $𝛾$ are not continuous (for any parameterization). $□$

Definition 3.2.1.26 (Exhaustion by Compact Sets).

For a topological space $𝑋$ , an exhaustion by compact sets is a nested sequence of compact sets ${𝐾_{𝑛}}_{𝑛 \in ℕ} \subseteq 𝑋$ such that $𝐾_{𝑛} \subset \overset{⚬}{𝐾_{𝑛 + 1}}$ for all $𝑛 \in ℕ$ and

𝑋 = ⋃_{𝑛 \in ℕ} 𝐾_{𝑛} .

Lemma 3.2.1.27.

Let $Ω \subseteq ℂ$ be an open set and let $𝔅$ be a basis for the topology on $Ω$ . Then there exists a collection of sets ${𝑈_{𝑛}}_{𝑛 \in ℕ} \subseteq 𝔅$ such that

$⋃_{𝑛 \in ℕ} 𝑈_{𝑛} = Ω$ .
For every compact $𝐾 ⋐ Ω$ , $𝐾$ intersects only finitely many sets in ${𝑈_{𝑛}}_{𝑛 \in ℕ}$ .

Proof.

Let ${𝐾_{𝑛}}_{𝑛 \in ℕ} \subset Ω$ be an exhaustion by compact sets with $𝐾_{0} = ⌀$ and $𝐾_{𝑛} \subseteq \overset{⚬}{𝐾_{𝑛 + 1}}$ for all $𝑛 \in ℕ$ . For each $𝑛 \in ℕ$ , define

𝑊_{𝑛} = \overset{⚬}{𝐾_{𝑛 + 1}} ∖ 𝐾_{𝑛 - 2}, 𝑉_{𝑛} = 𝐾_{𝑛} ∖ \overset{⚬}{𝐾_{𝑛 - 1}},

where $𝐾_{- 1} = ⌀$ . Each $𝑊_{𝑛}$ is open and each $𝑉_{𝑛}$ is compact, with $𝑉_{𝑛} \subseteq 𝑊_{𝑛}$ and

⋃_{𝑛 \in ℕ} 𝑉_{𝑛} = Ω .

For each $𝑛 \in ℕ$ and each $𝑧 \in 𝑉_{𝑛}$ , since $𝑊_{𝑛}$ is open and contains $𝑧$ , there exists $𝑈_{𝑧, 𝑛} \in 𝔅$ such that

𝑧 \in 𝑈_{𝑧, 𝑛} \subseteq 𝑊_{𝑛} .

The collection

{𝑈_{𝑧, 𝑛} : 𝑧 \in 𝑉_{𝑛}}

is an open cover of the compact set $𝑉_{𝑛}$ , so by Heine–Borel (Theorem 1.1.9 $Theorem 1.1.9 Theorem 1.1.9$ ) it admits a finite subcover, there exist finitely many points $𝑧_{𝑛, 1}, \dots, 𝑧_{𝑛, 𝑘_{𝑛}} \in 𝑉_{𝑛}$ such that

𝑉_{𝑛} \subset ⋃_{𝑖 = 1}^{𝑘_{𝑛}} 𝑈_{𝑧_{𝑛, 𝑖}, 𝑛} \subseteq 𝑊_{𝑛} .

Enumerate all such $𝑈_{𝑧_{𝑛, 𝑖}, 𝑛}$ over $𝑛 \in ℕ$ and $𝑖 = 1, \dots, 𝑘_{𝑛}$ to obtain a countable collection ${𝑈_{𝑗}}_{𝑗 \in ℕ} \subseteq 𝔅$ . Then

⋃_{𝑗 \in ℕ} 𝑈_{𝑗} = Ω,

proving Part 1 $Part 1 Part 1$ .

For 2, let $𝐾 ⋐ Ω$ be compact. There exists $𝑁 \in ℕ$ such that

𝐾 ⋐ \overset{⚬}{𝐾_{𝑁}},

so $𝐾$ is disjoint from $𝑉_{𝑛}$ for all $𝑛 > 𝑁 + 1$ . Since each $𝑉_{𝑛}$ intersects only finitely many $𝑈_{𝑗}$ , $𝐾$ intersects only finitely many $𝑈_{𝑗}$ . Thus the collection is locally finite. ⁠ $□$

Figure 3.4: Geometry of the finite subcover of

𝑉_{𝑛} \subset 𝑊_{𝑛}

for some

𝑛 \in ℕ

Remark.

The property of local finiteness of an open collection $𝑆$ in $Ω$ is commonly stated as: for every $𝑧 \in Ω$ , there exists an open neighborhood of $𝑧$ that intersects only finitely many sets in $𝑆$ .

This is equivalent to Part 2 $Part 2 Part 2$ in Lemma 3.2.1.27 $Lemma 3.2.1.27 Lemma 3.2.1.27$ . Indeed, if every point has such a neighborhood, then any compact $𝐾 \subset Ω$ admits a finite subcover of these neighborhoods by Heine–Borel (Theorem 1.1.9 $Theorem 1.1.9 Theorem 1.1.9$ ), so $𝐾$ intersects finitely many sets in $𝑆$ . Conversely, for any $𝑧 \in Ω$ , take an open neighborhood $𝑉$ with $𝑧 \in 𝑉$ and with relatively compact closure in $Ω$ ; then $\overline{𝑉}$ intersects finitely many sets in $𝑆$ , and so does $𝑉$ .

Theorem 3.2.1.28 (Partition of Unity).

Let $Ω \subseteq ℂ$ be a nonempty open set and let ${Ω_{𝑘}}_{𝑘 \in ℕ}$ be an open cover of $Ω$ . Then there exists a collection of bump functions ${𝛼_{𝑗}}_{𝑗 \in ℕ} \subseteq 𝐶^{\infty} (ℂ)$ , each with compact support in $Ω$ , satisfying:

For each $𝑗 \in ℕ$ , there exists $𝑘 \in ℕ$ such that $supp (𝛼_{𝑗}) \subseteq Ω_{𝑘}$ .
The collection ${supp (𝛼_{𝑗})}_{𝑗 \in ℕ}$ is locally finite.
For each $𝑗 \in ℕ$ , $0 \leq 𝛼_{𝑗} \leq 1$ .
$\sum_{𝑗 = 1}^{\infty} 𝛼_{𝑗} \equiv 1$ on $Ω$ .

Then ${𝛼_{𝑗}}_{𝑗 \in ℕ}$ is called a $𝐶^{\infty}$ partition of unity subordinate to ${Ω_{𝑘}}_{𝑘 \in ℕ}$ .

Proof.

For each $𝑧 \in Ω$ there exists $𝑟_{𝑧} > 0$ and $𝑘_{𝑧} \in ℕ$ such that

\overline{𝐷 (𝑧, 𝑟_{𝑧})} \subset Ω_{𝑘_{𝑧}} .

The collection

{𝐷 (𝑧, 𝑟) : 𝑧 \in Ω \land 0 < 𝑟 < 𝑟_{𝑧}}

is an open basis for $Ω$ . By Lemma 3.2.1.27 $Lemma 3.2.1.27 Lemma 3.2.1.27$ there exists a locally finite open cover

{𝐷 (𝑧_{𝑗}, 𝑟_{𝑧_{𝑗}})}_{𝑗 \in ℕ} \subseteq 𝔅

of $Ω$ with

𝐷 (𝑧_{𝑗}, 𝑟_{𝑧_{𝑗}}) \subset \overline{𝐷 (𝑧_{𝑗}, 𝑟_{𝑧_{𝑗}})} ⋐ Ω_{𝑘_{𝑧_{𝑗}}}, \forall 𝑗 \in ℕ .

Define the standard bump function

𝜃 (𝑧) = {\begin{cases} e^{\frac{1}{{| 𝑧 |}^{2} - 1}} & if | 𝑧 | < 1 \\ 0 & if | 𝑧 | \geq 1 . \end{cases}

For $𝜀 > 0$ let

𝜃_{𝜀 (𝑧)} = 𝜃 (\frac{𝑧}{𝜀}),

which has support $\overline{𝐷 (0, 𝜀)}$ . Define

𝛽_{𝑗} (𝑧) = 𝜃_{𝑟_{𝑧_{𝑗}}} (𝑧 - 𝑧_{𝑗}),

supp (𝛽_{𝑗}) = \overline{𝐷 (𝑧_{𝑗}, 𝑟_{𝑧_{𝑗}})} \subset Ω_{𝑘_{𝑧_{𝑗}}} .

By local finiteness of ${𝐷 (𝑧_{𝑗}, 𝑟_{𝑧_{𝑗}})}_{𝑗 \in ℕ}$ , for each $𝑧 \in Ω$ there exists an open neighborhood $𝑉$ with $𝑧 \in 𝑉$ intersecting only finitely many $\overline{𝐷 (𝑧_{𝑗}, 𝑟_{𝑧_{𝑗}})}$ . Thus ${supp (𝛽_{𝑗})}_{𝑗 \in ℕ}$ is locally finite on $Ω$ . Then the sum

𝑆 (𝑧) = \sum_{𝑗 = 1}^{\infty} 𝛽_{𝑗} (𝑧)

defined for $𝑧 \in Ω$ involves only finitely many nonzero terms (by local finiteness) on a neighborhood of every point $𝑧$ . Hence $𝑆 \in 𝐶^{\infty} (Ω)$ and $𝑆 (𝑧) > 0$ (since ${𝐷 (𝑧_{𝑗}, 𝑟_{𝑧_{𝑗}})}_{𝑗 \in ℕ}$ covers $Ω$ ). Define

𝛼_{𝑗} (𝑧) = \frac{𝛽_{𝑗} (𝑧)}{𝑆 (𝑧)}, \forall 𝑗 \in ℕ .

Each $𝛼_{𝑗} \in 𝐶^{\infty} (ℂ)$ has compact support in $Ω$ , $0 \leq 𝛼_{𝑗} \leq 1$ , the supports are locally finite, and

\sum_{𝑗 = 1}^{\infty} 𝛼_{𝑗} (𝑧) = 1

for all $𝑧 \in Ω$ . Moreover

supp (𝛼_{𝑗}) \subseteq Ω_{𝑘_{𝑧_{𝑗}}},

proving subordination. ⁠ $□$

Theorem 3.2.1.29 (Existence of Bump Functions).

Let $𝐾 \subset ℂ$ be compact and $𝑉 \subset ℂ$ an open neighborhood of $𝐾$ . Then there exists a compactly supported $𝜑 \in 𝐶^{\infty} (ℂ)$ such that

0 \leq 𝜑 (𝑧) \leq 1 \forall 𝑧 \in ℂ,

supp (𝜑) \subset 𝑉

, and

𝜑 \equiv 1

on some open neighborhood of

𝐾

Proof.

Let

𝑉 (𝐾, 𝜀) = {𝑧 \in ℂ : inf_{𝜁 \in 𝐾} | 𝑧 - 𝜁 | < 𝜀}

denote the open $𝜀$ -neighborhood of $𝐾$ . Since $𝑉$ is an open neighborhood of $𝐾$ , $\exists 𝜀 > 0$ such that

𝐾 \subset 𝑉 (𝐾, 𝜀) \subset 𝑉 (𝐾, 2 𝜀) \subset 𝑉,

where $𝐴 \subset 𝐵$ means that the closure of $𝐴$ is compact and contained in $𝐵$ .

Define the open sets

Ω_{1} = 𝑉 (𝐾, 2 𝜀), Ω_{2} = ℂ ∖ \overline{𝑉 (𝐾, 𝜀)} .

Then ${Ω_{1}, Ω_{2}}$ is an open cover of $ℂ$ .

By the Partition of Unity Theorem (Theorem 3.2.1.28 $Theorem 3.2.1.28 Theorem 3.2.1.28$ ), there exist compactly supported functions ${𝛼_{𝑗}}_{𝑗 \in ℕ} \subseteq 𝐶^{\infty} (ℂ)$ forming a partition of unity subordinate to this cover. That is,

0 \leq 𝛼_{𝑗} \leq 1, supp (𝛼_{𝑗}) \subseteq Ω_{𝑖_{𝑗}} for some 𝑖_{𝑗} \in {1, 2}, \sum_{𝑗 = 1}^{\infty} 𝛼_{𝑗} \equiv 1 on ℂ .

Let

𝐽 = {𝑗 \in ℕ : supp (𝛼_{𝑗}) \subseteq Ω_{1}} .

Define

𝜑 (𝑧) = \sum_{𝑗 \in 𝐽} 𝛼_{𝑗} (𝑧) .

Then $𝜑 \in 𝐶^{\infty} (ℂ)$ is compactly supported within $Ω_{1}$ , and since only finitely many $𝛼_{𝑗}$ are nonzero on a neighborhood of each point, $𝜑 \in 𝐶^{\infty} (ℂ)$ . Moreover,

supp (𝜑) \subset Ω_{1} \subset 𝑉 .

For $𝑧 \in 𝑉 (𝐾, 𝜀)$ , all functions with support in $Ω_{2}$ vanish at $𝑧$ , so

𝜑 (𝑧) = \sum_{𝑗 \in 𝐽} 𝛼_{𝑗} (𝑧) = \sum_{𝑗 = 1}^{\infty} 𝛼_{𝑗} (𝑧) = 1 .

Hence, $𝜑 \equiv 1$ on the open neighborhood $𝑉 (𝐾, 𝜀)$ of $𝐾$ . Outside $𝑉 (𝐾, 2 𝜀)$ , all terms with support in $Ω_{1}$ vanish, so $𝜑 (𝑧) = 0$ . Finally, $0 \leq 𝜑 \leq 1$ everywhere by construction. Thus $𝜑$ satisfies all required properties. ⁠ $□$