Topological Spaces and Continuous Functions

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Topological Spaces

Defintions

Topology: A topology on a set $X$ is a collection $\mathcal{T}$ of subsets of $X$ having the following properties.

1. $\emptyset$ and $X$ are in $\mathcal{T}$.
2. The union of the elements of any subcollection of $\mathcal{T}$ is in $\mathcal{T}$.
3. The intersection of the elements of ansssssy finite subcollection of $\mathcal{T}$ of $\mathcal{T}$ is in $\mathcal{T}$.

Topological Space: A set $X$ for which a topology $\mathcal{T}$ has been specified is called a topological space and is denoted as the ordered pair $(X, \mathcal{T})$.

Open Subset: Let $(X, \mathcal{T})$ be a topological space. A subset $U$ of $X$ is an open set of $X$ if $U$ belongs to the collection $\mathcal{T}$.

Oodroo Fact

Oodroo Bum personalities form a topology since any union of any personalities forms a new Oodroo Bum personality and all personalities intersect to a single cuteness Bum.

Coarser and Finer: Let $\mathcal{T}$ and $\mathcal{T}'$ be two topologies on set $X$. If $\mathcal{T}' \supseteq \mathcal{T}$, then $\mathcal{T}'$ is finer than $\mathcal{T}$ and $\mathcal{T}$ is coarser than $\mathcal{T}'$. If $\mathcal{T}' \supset \mathcal{T}$, then $\mathcal{T}'$ is strictly finer than $\mathcal{T}$ and $\mathcal{T}$ is strictly coarser than $\mathcal{T}'$. If $\mathcal{T}' \supseteq \mathcal{T}$ or $\mathcal{T}' \subseteq \mathcal{T}$, then $\mathcal{T}$ is comparable to $\mathcal{T}'$.

Common Toplogies

Let $X$ be a set. Then the following are a few common topologies on $X$.

• The discrete topology is the collection of all subsets of $X$.
• The indiscrete topology, or trivial topology, is the collection of subsets $\emptyset$ and $X$.
• The finite complement topology is the collection of all subsets $U$ such that $X - U$ is finite or is all of $X$.

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Basis for a Topology

Defintions

Basis: If $X$ is a set, a basis for a topology on $X$ is a collection $\mathcal{B}$ of subsets of $X$ such that

1. For each $x \in X$, there is at least one basis element $B$ containing $x$.
2. If $x$ belongs to the intersection of two basis elements $B_1$ and $B_2$, then there is a basis element $B_3$ containing $x$ such that $B_3 \subset B_1 \cap B_2$.

Each subset of $X$ is called a basis element. The topology $\mathcal{T}$ generated by $\mathcal{B}$ defines open subsets as follows:

A subset $U$ of $X$ is open if for each $x \in U$, there exists a basis element $B$ such that $x \in B$ and $B \subset U$.

Subbasis: A subbasis $\mathcal{S}$ for a topology on $X$ is a collection of subsets of $X$ whose union equals $X$. The topology generated by the subbasis $\mathcal{S}$ is defined to be the collection $\mathcal{T}$ of all unions of finite intersections of elements of $\mathcal{S}$.

Oodroo Fact

The Oodroo Bum topology is formed by a basis of bounciness. Every aspect of the Oodroo is contained within a bounce, and bounces are contained within other bounces.

Real Line Topologies

• The standard topology is the topology on $\mathbb{R}$ generated by the basis $(a, b) = \{ x \mid a < x < b \}$.
• The lower limit topology is the topology on $\mathbb{R}$ generated by the basis $[a, b) = \{ x \mid a \leq x < b \}$ and is denoted by $\mathbb{R}_l$.
• Let set $K = \{ \frac{1}{n} \mid n \in \mathbb{Z}_+ \}$. Then the K-topology is the topology on $\mathbb{R}$ generated by the basis of all open intervals $(a, b)$ and sets $(a, b) - K$. The topology is denoted by $\mathbb{R}_K$.

The topologies $\mathbb{R}_l$ and $\mathbb{R}_K$ are not comparable. Note that open set $[0, 1) \in \mathbb{R}_l$, but $[0, 1) \notin \mathbb{R}_K$.

Theorems

Lemma 13.1: Let $X$ be a set. Let $\mathcal{B}$ be a basis for a topology $\mathcal{T}$ on $X$. Then $\mathcal{T}$ equals the collection of all unions of elements of $\mathcal{B}$.

Lemma 13.2: Let $X$ be a topological space. Let $\mathcal{C}$ be a collection of open sets of $X$ such that for each open set $U$ of $X$ and each $x \in U$, there is and element $C$ of $\mathcal{C}$ such that $x \in C \subseteq U$. Then $\mathcal{C}$ is a basis for the topology of $X$.

Lemma 13.3: Let $\mathcal{B}$ and $\mathcal{B}'$ be bases for the topologies $\mathcal{T}$ and $\mathcal{T}'$, respectively on $X$. Then the following are equivalent:

1. $\mathcal{T}'$ is finer than $\mathcal{T}$.
2. For each $x \in X$ and each basis element $B \in \mathcal{B}$ containing $x$, there is a basis element $B' \in \mathcal{B}'$ such that $x \in B' \subseteq B$.

Lemma 13.4: The topologies of $\mathbb{R}_l$ and $\mathbb{R}_K$ are strictly finer than the standard topology on $\mathbb{R}$, but are not comparable with one another.

Exercises

Exercise 1

Let $X$ be a topological space; let $A$ be a subset of $X$. Suppose that for each $x \in A$ there is an open set $U$ containing $x$ such that $U \subseteq A$. Show that $A$ is open in $X$.

For each $x \in A$, let $U_x$ be the open set that contains $x$ such that $U_x \subseteq A$. Then $\cup_{ x \in A } U_x = A$ and $A$ is an open set since it is the union of open sets.

Exercise 2

Consider the nine topologies on the set $X = \{ a, b, c \}$ indicated in Example 1 of section 12. Compare them; that is, for each pair of topologies, determine whether they are comparable, and if so, which is the finer.

Exercise 3

Show that the collection $\mathcal{T}_c$ fiven in Example 4 of section 12 is a topology on the set X. Is the collection $\mathcal{T}_\infty = \{ U \mid X - U \text{ is infinite or empty or all of } X \}$ a topology on $X$?

Let $X$ be a set. Let $\mathcal{T}_C$ be the collection of all subsets $U$ of $X$ such $X - U$ is countable or is all of $X$. Let $U_c$ be a collection of nonempty sets from $\mathcal{T}_C$. Thus each set $X - U_c$ is countable.

1. $\emptyset$ and $X$ are in $\mathcal{T}_C$ since $X - \emptyset = X$ and $X - X = \emptyset$, which is countable.
2. By De Morgan's set difference union law, $X - \cup_c U_c = \cap (X - U_c)$. Since the intersection of a collection of countable sets is countable, $\mathcal{T}_C$ is closed under set union.
3. By De Morgan's set difference intersection law, $X - \cap_c U_c = \cup (X - U_c)$. Since the finite union of countable sets is countable, $\mathcal{T}_C$ is closed under finite set intersection.

Therefore $\mathcal{T}_C$ is a topology.

Exercise 4

(a) If $\{ \mathcal{T}_\alpha \}$ is a family of topologies on $X$, show that $\cap \mathcal{T}_\alpha$ is a topology on $X$. Is $\cup \mathcal{T}_\alpha$ a topology on $X$?
(b) Let $\{ \mathcal{T}_\alpha \}$ be a family of topologies on $X$. Show that there is a unique smallest topology on $X$ containing all the collections $\mathcal{T}_\alpha$, and unique largest topology conatined in all $\mathcal{T}_\alpha$.
(c) If $X = \{ a, b, c \}$, let
$$ \mathcal{T}_1 = \{ \emptyset, X, { a }, { a, b } \} \text{ and } \mathcal{T}_2 = \{ \emptyset, X, { a }, { b, c } \} $$ Find the smallest topology containing $\mathcal{T}_1$ and $\mathcal{T}_2$, and the largest topology contained in $\mathcal{T}_1$ and $\mathcal{T}_2$.

Exercise 5

Show that if $\mathcal{A}$ is a basis for a topology on $X$, then the topology generated by $\mathcal{A}$ equals the intersection of all topologies on $X$ that contain $\mathcal{A}$. Prove the same if $\mathcal{A}$ is a subbasis.

Let $\{ \mathcal{T}_c \}_{c \in C}$ be the collection of all topologies such that $\mathcal{A} \subseteq \mathcal{T}_c$ for each topology $\mathcal{T}_c$. Let $T_{\mathcal{A} }$ be the topology generated by $\mathcal{A}$.

$T_{\mathcal{A} }$ is topology that contains $\mathcal{A}$ and thus is in $\{\mathcal{T}_c \}_{c \in C}$. As a result $\cap_{c \in C} \mathcal{T}_c$ is a subset of $T_{\mathcal{A} }$.

Every topology from $\{ \mathcal{T}_c \}_{c \in C}$ contains $\mathcal{A}$ and topologies are closed under set union. Thus the collection of all set unions from $\mathcal{A}$ must be a subset of every topology from $\{ \mathcal{T}_c \}_{c \in C}$. By Lemma 13.1, $T_{ \mathcal{A} }$ is a subset of $\cap_{c \in C} \mathcal{T}_c$.

Therefore, $T_{ \mathcal{A} } = \cap_{c \in C} \mathcal{T}_c$.

Exercise 6

Show that the topologies of $\mathbb{R}_l$ and$\mathbb{R}_K$ are not comparable.

Exercise 7

Consider the following topologies on $\mathbb{R}$,
  $\mathcal{T}_1 = \text{ the standard topology }$,
  $\mathcal{T}_2 = \text{ the topology of } \mathbb{R}_K$,
  $\mathcal{T}_3 = \text{ the finite complement topology, }$,
  $\mathcal{T}_4 = \text{ the upper limit topology, having all sets } (a, b] \text{ as basis }$,
  $\mathcal{T}_5 = \text { the topology having all sets } (-\infty, a) = \{ x \mid x < a \} \text { as basis }$,
Determine, for each of these topologies, which of the others it contains.

Since each topology is on $\mathbb{R}$, by lemma 13.1 any topology $\mathcal{T}_x \subseteq \mathcal{T}_y$ if for any basis element $Y \in \mathcal{T}_y, Y \in \mathcal{T}_x$. Since the exercise requires several proofs, each proof will be split into its own section.

• $\mathcal{T}_5 \subseteq \mathcal{T}_1 \wedge \mathcal{T}_1 \not\subseteq \mathcal{T}_5$

  Let set $(\infty, a)$ be any basis element from $\mathcal{T}_5$. Then $\cup_{ x < a } (x, a) = (-\infty, a)$, and as follows every basis element from $\mathcal{T}_5$ is in $\mathcal{T}_1$. Thus $\mathcal{T}_1 \subseteq \mathcal{T}_5$.

  Consider the open set (0, 1) from $\mathcal{T}_1$. Let $X \subseteq \mathbb{R}$ and note that $\cap_{x \in \mathbb{R} } (-\infty, x) = (-\infty, \inf(X))$ if $\inf(X)$ exists else $\emptyset$. Similarly, $\cup_{x \in \mathbb{R} } (-\infty, x) = (-\infty, \sup(X))$ if $\sup(X)$ exists else $\mathbb{R}$. As follows, open sets from $\mathcal{T}_5$ can only be of the form $(-\infty, x)$ for $x \in \mathbb{R}$. Thus $(0, 1)$ does not belong to $\mathcal{T}_5$ and $\mathcal{T}_5 \not\subseteq \mathcal{T}_1$.

• $\mathcal{T}_1 \subseteq \mathcal{T}_2 \wedge \mathcal{T}_2 \not\subseteq \mathcal{T}_1$

  By definition, $\mathcal{T}_1 \subseteq \mathcal{T}_2$.

  Let set $K = \{\frac{1}{n} \mid n \in \mathbb{Z}_+ \}$ and as follows $\lim_{n \rightarrow \infty} \frac{1}{n} = 0$. As follows any interval $(a, b)$, where $a < 0 < b$, contains elements from $K$. Since the union and intersection of open intervals on $\mathbb{R}$ must be equal to the union of disjoint open intervals, any set from $\mathcal{T}_1$ which contains 0 must be an open interval around 0. As follows, every set from $\mathcal{T}_1$ containing 0, must also contain an element from $K$. Thus $(-1, 1) -K$ does not belong to $\mathcal{T}_1$ and $\mathcal{T}_2 \not\subseteq \mathcal{T}_1$.

• $\mathcal{T}_3 \subseteq \mathcal{T}_1 \wedge \mathcal{T}_1 \not\subseteq \mathcal{T}_3$

  By definition, $\emptyset$ and $\mathbb{R}$ are in each topology. So let set $X$ be any open set in $\mathcal{T}_3$ that is neither $\emptyset$ or $\mathbb{R}$. Then set $U = \mathbb{R} - X$ must be finite. Note that $\cap_{u \in U} [(-\infty, u) \cup (u, \infty)] = X$ which is in $\mathcal{T}_1$ since topologies are closed under finite intersection and union. Thus $\mathcal{T}_1 \subseteq \mathcal{T}_3$.

  Consider open set $(0, 1)$ from $\mathcal{T}_1$. Since $\mathbb{R} - (0, 1)$ is not finite, it does not belong to $\mathcal{T}_3$. Thus $\mathcal{T}_1 \not\subseteq \mathcal{T}_3$.

• $\mathcal{T}_2 \subseteq \mathcal{T}_4 \wedge \mathcal{T}_4 \not\subseteq \mathcal{T}_2$

  Let $a, b \in \mathbb{R}$ and note that the set $\cup_{ \{ x \mid a < x < b \} } (a, x] = (a, b)$ is in $\mathcal{T}_4$ since topologies are closed under arbitrary union. Let set $K = \{\frac{1}{n} \mid n \in \mathbb{Z}_+ \}$ and let set $\mathcal{K} = \cup_{ k \in \mathbb{Z}_+ } (\frac{1}{k + 1}, \frac{1}{k})$. Then $\mathcal{K}$ contains no points from $K$ and $(a, b) - K = (a, 0] \cup \mathcal{K} \cap (0, b)$. Thus $\mathcal{T}_2 \subseteq \mathcal{T}_4$.

  Consider open set $(2, 3]$ from $\mathcal{T}_4$. Since the arbitrary union and finite intersection of intervals from $\mathbb{R}$ is equal to the union of disjoint open intervals and $(2, 3]$ is not open in the standard topology, $(2, 3]$ cannot be in $\mathcal{T}_2$. Thus $\mathcal{T}_4 \not\subseteq \mathcal{T}_2$.

• $\mathcal{T}_3 \not\subseteq \mathcal{T}_5 \wedge \mathcal{T}_5 \not\subseteq \mathcal{T}_3$

  Consider open set $(-\infty, 0) \cup (0, \infty)$ from $\mathcal{T}_3$. From the $\mathcal{T}_1 \not\subseteq \mathcal{T}_5$ proof, $\mathcal{T}_5$ cannot contain the set $(-\infty, 0) \cup (0, \infty)$. Thus $\mathcal{T}_3 \not\subseteq \mathcal{T}_5$.

  Consider open set $(-\infty, 0)$ from $\mathcal{T}_5$. Since $\mathbb{R} - (-\infty, 0)$ is not finite, it does not belong to $\mathcal{T}_3$. Thus $\mathcal{T}_5 \not\subseteq \mathcal{T}_3$.

Therefore the finer relationships between the topologies are

$$\mathcal{T}_ 5 \subseteq \mathcal{T}_1 \subseteq \mathcal{T}_2 \subseteq \mathcal{T}_4 \\ \mathcal{T}_ 3 \subseteq \mathcal{T}_1 \subseteq \mathcal{T}_2 \subseteq \mathcal{T}_4 \\ \mathcal{T}_ 3 \not\subseteq \mathcal{T}_ 5 \wedge \mathcal{T}_ 5 \not\subseteq \mathcal{T}_ 3$$

Exercise 8

(a) Apply Lemma 13.2 to show that the countable collection $$ \mathcal{B} = \{ (a, b) \mid a < b, a \text{ and } b \text{ rational } \} $$ is a basis that generates the standard topology on $\mathbb{R}$. (b) Show the collection $$ \mathcal{C} = \{ [a, b) \mid a < b, a \text{ and } b \text{ rational } \} $$ is a basis that generates a topology different from the lower limit topology on $\mathbb{R}$.

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The Order Topology

Definitions

Order Topology: Let $X$ be a set with a simple order relation; assume $X$ has more thatn one element. Let $\mathcal{B}$ be the collection of all sets of the following types:

1. All open intervals $(a, b)$ in $X$.
2. All intervals of the form $[a_0, b)$, where $a_0$ is the smallest element (if any) of $X$.
3. All intervals of the form $(a, b_0]$, where $b_0$ is the largest element (if any) of $X$.

The collection $\mathcal{B}$ is a basis for a topology on $X$, which is called the order topology.

Ray: If $X$ is an ordered set, and $a$ is an element of $X$, there are four subsets of $X$ that are called the rays determined by $a$. They are the following:

$$(a, +\infty) = \{ x \mid x > a \} \\ (-\infty, a) = \{ x \mid x < a \} \\ [a, +\infty) = \{ x \mid x \geq a \} \\ (-\infty, a] = \{ x \mid x \leq a \}$$

Sets of the first two types are called open rays, and sets of the last two types are called closed rays.

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The Product Topology on $X \times Y$

Definitions

Product Topology: Let $X$ and $Y$ be topological spaces. The product topology on $X \times Y$ is the topology having as basis the collection $\mathcal{B}$ of all sets of the form $U \times V$, where $U$ is an open subset of $X$ and $V$ is an open subset of $Y$.

Projection: Let $\pi_1: X \times Y \rightarrow X$ be defined by the equation

$$\pi_1(x, y) = x$$

Let $\pi_2: X \times Y \rightarrow Y$ be defined by the equation

$$\pi_2(x, y) = y$$

The maps $\pi_1$ and $\pi_2$ are called projections of $X \times Y$ onto its first and second factors, respectively.

Theorems

Theorem 15.1: If $\mathcal{B}$ is a basis for the topology of $X$ and $\mathcal{C}$ is a basis for the topology of $Y$, then the collection

$$\mathcal{D} = \{ B \times C \mid B \in \mathcal{B} \text{ and } C \in \mathcal{C} \}$$

is a basis for the topology of $X \times Y$.

Theorem 15.2: The collection

$$S = \{ \pi_1^{-1}(U) \mid U \text{ open in } X \} \cup \{ \pi_2^{-1}(V) \mid V \text{ open in } Y \}$$

is a subbasis for the product topology on $X \times Y$.

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The Subspace Topology

Definitions

Subspace Topology: Let $X$ be a topological space with topology $\mathcal{T}$. If $Y$ is a subset of $X$, the collection

$$\mathcal{T}_Y = \{ Y \cap U \mid U \in \mathcal{T} \}$$

is a topology on $Y$, called the subspace topology and $Y$ is called a subspace of $X$.

Convex: Let $X$ be an ordered set. A subset $Y$ of $X$ is convex in $X$ if for each pair of points $a < b$ of $Y$, the entire interval $(a, b)$ of points of $X$ lies in $Y$.

Theorems

Lemma 16.1: If $\mathcal{B}$ is a basis for the topology of $X$ then the collection

$$\mathcal{B}_Y = \{ B \cap Y \mid B \in \mathcal{B} \}$$

is a basis for the subspace topology on Y.

Lemma 16.2: Let $Y$ be a subspace of $X$. If $U$ is open in $Y$ and $Y$ is open in $X$, then $U$ is open in $X$.

Theorem 16.3: If $A$ is a subspace of $X$ and $B$ is a subspace of $Y$, then the product topology on $A \times B$ is the same as the topology $A \times B$ inherits a a subspace of $X \times Y$.

Theorem 16.4: Let $X$ be an ordered set in the order topology. Let $Y$ be a subset of $X$ that is convex in $X$. Then the order topology on $Y$ is the same as the topology $Y$ inherits as a subspace of $X$.

Exercises

Exercise 1

Show that if $Y$ is a subspace of $X$, and $A$ is a subset of $Y$, then the topology $A$ inherits as a subspace of $Y$ is the same as the topology it inherits as a subspace of $X$.

Exercise 3

Consider the set $Y = [-1, 1]$ as a subspace of $\mathbb{R}$. Which of the following sets are open in $Y$? Which are open in $\mathbb{R}$? $$ A = \{ x \mid \frac{1}{2} < |x| < 1 \}, \\ B = \{ x \mid \frac{1}{2} < |x| \leq 1 \}, \\ C = \{ x \mid \frac{1}{2} \leq |x| < 1 \}, \\ D = \{ x \mid \frac{1}{2} \leq |x| \leq 1 \}, \\ E = \{ x \mid 0 < |x| < 1 \text{ and } \frac{1}{x} \notin \mathbb{Z}_+ \} $$

Exercise 5

Let $X$ adn $X'$ denote a single set in the topologies $\mathcal{T}$ and $\mathcal{T}'$, respectively. Let $Y$ and $Y'$ denote a single set in the topologies $\mathcal{U}$ and $\mathcal{U}'$, respectively. Assume these sets are nonempty.
(a) Show that if $\mathcal{T}' \supseteq \mathcal{T}$ and $\mathcal{U}' \supseteq \mathcal{U}$, then the product topology on $X' \times Y'$ is finer than the product topology on $X \times Y$.
(b) Does the converse of (a) hold? Justify your answer.

Exercise 7

Let $X$ be an ordered set. If $Y$ is a proper subset of $X$ that is convex in $X$, does it follow that $Y$ is an interval or a ray in $X$?

Exercise 9

Show that the dictionary order topology on the set $\mathbb{R} \times \mathbb{R}$ is the same as the product topology $\mathbb{R}_d \times \mathbb{R}$, where $\mathbb{R}_d$ deontes $\mathbb{R}$ in the discrete topology. Compare this topology with the standard topology on $\mathbb{R}^2$.

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