### Notes on Topology

This post is the note from reading Bert Mendelson’s book Introduction to Topology, 3rd Edition. Most of the content of this note directly comes from the book. The first chapter of the book about the set theory is omitted in this note. At the moment, the note is obviously unfinished and I am still reading the book.

#### 1Metric Spaces

##### 1.1Definition of Metric Spaces

Definition (Metric Space): \langle X, d \rangle is a metric space if X is the underlying set, and d : X \times X \to \mathbb{R} is a distance function, which satisfies
• \forall x, y \in X, d(x, y) \geq 0

• \forall x, y \in X, d(x, y) = 0 \text{ only if } x = y

• \forall x, y \in X, d(x, y) = d(y, x)

• \forall x, y, z \in X, d(x, z) \leq d(x, y) + d(y, z)

The last one assert the transitivity of closeness: if x is close to y and y is close to z, then x is close to z.

Example: \langle \mathbb{R}, d \rangle is a metric space on real numbers, where d(x, y) = |x - y|.

Theorem: Given a set of metric space \langle X_1, d_1 \rangle, \langle X_2, d_2 \rangle, \cdots, \langle X_i, d_i \rangle, we can obtain a metric space on X = \Pi_{i = 1}^n X_i with d: X \times X \to \mathbb{R}. Let d(x, y) = \max_{i \leq i \leq n} {d_i(x_i, y_i)}

Proof: By verifying the four requirements in the definition. □

Remark: By instantiating the above theorem, \langle \mathbb{R}^n, d: \mathbb{R}^n \times \mathbb{R}^n \to \mathbb{R} \rangle is a metric space, where d((x_1, x_2, \cdots, x_n), (y_1, y_2, \cdots, y_n)) = \max_{i \leq i \leq n}{|x_i - y_i|}

Remark: \langle \mathbb{R}^n, d{}’: \mathbb{R}^n \times \mathbb{R}^n \to \mathbb{R} \rangle is a metric space, where d{}’((x_1, x_2, \cdots, x_n), (y_1, y_2, \cdots, y_n)) = \sqrt{\Sigma_{i=1}^n{ (x_i - y_i) }^2} Note: d{}’ is the Euclidean distance function.

##### 1.2Continuity

Definition (Continuity): A function f: \mathbb{R} \to \mathbb{R} is continuous at point a \in \mathbb{R}, if given \epsilon > 0, there exists a \delta > 0, such that if |x - a| < \delta, then |f(x) - f(a)| < \epsilon for all x \in \mathbb{R}.

The function is continuous if it is continuous at each point of \mathbb{R}.

Definition (Continuity): Given two metric spaces \langle X, d \rangle and \langle Y, d{}’ \rangle, a function f: X \to Y is continuous at point a \in X, if given \epsilon > 0, there exists a \delta > 0, such that if d(x, a) < \delta, then d{}’(f(x), f(a)) < \epsilon for all x \in X.

The function is continuous if it is continuous at each point of X.

Theorem: A constant function f : X \to Y is continuous.

Proof: It is always the case that d{}’(f(x), f(a)) = 0. □

Theorem: The identity function f : X \to X is continuous.

Proof: Choose \delta = \epsilon. □

Theorem (Composition of Continuous Functions): Let \langle X, d_1 \rangle, \langle Y, d_2 \rangle, and \langle Z, d_3 \rangle be metric spaces. Let f : X \to Y be continuous at a \in X, and g: Y \to Z be continuous at f(a) \in Y, then g \circ f : X \to Z is continuous at a \in X.

Corollary: Let \langle X, d_1 \rangle, \langle Y, d_2 \rangle, and \langle Z, d_3 \rangle be metric spaces. Let f : X \to Y and g: Y \to Z be continuous, then g \circ f : X \to Z is continuous.

##### 1.3Open Balls

Definition (Open Ball): Let \langle X, d \rangle be a metric space, and a \in X, and \delta > 0. The open ball B(a;\delta) about a of radius \delta is \{ x \in X | d(a, x) < \delta \} \subseteq X.

In other words, x \in B(a;\delta) iff x \in X and d(x, a) < \delta. Similarly, if \langle Y, d{}’ \rangle is a metric space, and f: X \to Y, then we have y \in B(f(a); \epsilon) iff y \in Y and d{}’(y, f(a)) < \epsilon.

Theorem (4.2): A function f : \langle X, d \rangle \to \langle Y, d{}’ \rangle is continuous at a point a \in X iff for all \epsilon > 0, there exists a \delta > 0 such that f(B(a; \delta)) \subset B(f(a); \epsilon)

Remark: For a function f : X \to Y, we have F(U) \subset V iff U \subset f^{-1}(V).

Theorem (4.3): A function f : \langle X, d \rangle \to \langle Y, d{}’ \rangle is continuous at a point a \in X iff for all \epsilon > 0, there exists a \delta > 0 such that B(a; \delta) \subset f^{-1}(B(f(a); \epsilon))

##### 1.4Neighborhoods

Definition (Neighborhoods): Let \langle X, d\rangle be a metric space and a \in X. A set N \subset X is called a neighborhood of a if there exists a \delta > 0 such that B(a; \delta) \subset N

The collection \mathcal{N}_a of all neighborhoods of a \in X is called a complete system of neighborhoods of the point a.

Lemma (4.5): Let \langle X, d\rangle be a metric space and a \in X. For every \delta > 0, the open ball B(a; \delta) is a neighborhood of point b \in B(a; \delta).

Proof: We need to show that for all b \in B(a; \delta), there exists \eta > 0 such that B(b; \eta) \subset B(a; \delta). Choose \eta < \delta - d(a, b). Then \forall x \in B(b; \eta), d(a, x) \leq d(a, b) + d(b, x) < d(a, b) + \eta < d(a, b) + \delta - d(a, b) = \delta Therefore, \forall x \in B(b; \eta), we have x \in B(a; \delta). □

Lemma (4.6): Let f:\langle X, d\rangle\to\langle Y, d{}’\rangle. f is continuous at point a \in X iff for for all neighborhood M of f(a), there exists a corresponding neighborhood N of a, such that f(N) \subset M \text{, or equivalently } N \subset f^{-1}(M)

Proof:
• First assume that f is continuous at a. Let M be a neighborhood of f(a). We must to show that there exists a neighborhood N of a and f(N) \subset M. 1) By the definition of neighborhoods, there exists \epsilon such that B(f(a), \epsilon) \subset M. 2) Since f is continuous, there exists a \delta > 0 such that f(B(a; \delta)) \subset B(f(a); \epsilon) (theorem 4.2). 3) Now, let N be B(a; \delta). We can conclude that f(N) = f(B(a; \delta) \subset B(f(a); \epsilon) \subset M

• To prove the other direction, assume that \forall neighborhood M of f(a), there exists a neighborhood N of a, such that f(N) \subset M. Let \epsilon > 0 be the threshold of such neighborhood M = B(f(a); \epsilon). To prove f is continuous, we must show that there exists \delta > 0 such that f(B(a; \delta)) \subset B(f(a); \epsilon) (theorem 4.2). By the hypothesis, there is neighborhood N of a such that f(N) \subset M. By the definition of neighborhood, there exists \delta such that B(a; \delta) \subset N. Therefore, we have f(B(a; \delta)) \subset f(N) \subset M = B(f(a); \epsilon)

Lemma (4.7): Let f:\langle X, d\rangle\to\langle Y, d{}’\rangle. f is continuous at point a \in X iff for each neighborhood M of f(a), f^{-1}(M) is a neighborhood of a.

Theorem (4.8): Let \langle X, d\rangle be a metric space.
• \forall a \in X, there exists at least one neighborhood of a.

• \forall a \in X and its neighborhood N, a \in N.

• \forall a \in X, if N is its neighborhood and N{}’ \supset N, then N{}’ is also a neighborhood of a.

• \forall a \in X and its neighborhoods N and M, N \cap M is also a neighborhood of a.

• \forall a \in X and each neighborhood N of a, there exists a neighborhood O of a, such that O \subset N and O is a neighborhood of points in O.

Definition (Basis of Neighborhoods): Let a be a point in metric space X. A collection \mathcal{B}_a of neighborhoods of a is called a basic for the neighborhood system at a, if every neighborhood N of a contains some element B of \mathcal{B}_a.

##### 1.5Limits

First, recall the definition of limit on real line.

Definition (Limit): a \in \mathbb{R} is the limit of the sequence a_1, a_2, \cdots if given \epsilon > 0, there exists a positive integer N such that, for all n > N, we have |a - a_n| < \epsilon. We can also say the sequence converges to a, and write \lim_n a_n = a.

Definition (Limit, generalized): Let \langle X, d\rangle be a metric space. Let a_1, a_2, \cdots be a sequence of points in X. A point a \in X is the limit of the sequence if \lim_n d(a, a_n) = 0

Corollary (5.3): Let \langle X, d\rangle be a metric space and a_1, a_2, \cdots be a sequence of points in X. \exists a \in X, \lim_n a_n = a \Leftrightarrow for each neighborhood V of a, \exists N \in \mathbb{N}, \forall n > N, a_n \in V.

Proof:
• Let V be a neighborhood of a. By definition of neighborhoods, there is \epsilon such that a \in B(a; \epsilon) \subset V. Then if \lim_{n} a_n = a, there exists an ineger N, such that d(a, a_n) < \epsilon whenever n > N. Therefore a_n \in V.

• Given \epsilon > 0 and B(a; \epsilon) is a neighborhood of a. By premises, we have N such that n > N, a_n \in B(a; \epsilon). Then d(a, a_n) < \epsilon, and therefore \lim_{n} a_n = a. □

If S is a set of infinite points, and given a proposition P, there is a finite subset X \subset S, such that \forall x \in X, P(x), then we can say the P is true for almost all the elements in S.

Therefore, \lim_{n} a_n = a if for each neighborhood V of a, almost all points a_n are in V. In other wors, for each V, there are at most finite elements in sequence a_n are not in V.

Theorem (5.4): Let \langle X, d\rangle and \langle Y, d{}’\rangle be two metric space. A function f : X \to Y is continuous at a point a \in X if and only if, if \lim_n a_n = a for a sequence of points in X, \lim_n f(a_n) = f(a).

Proof: TODO.

If \lim_n a_n = a, then by a substitution, \lim_n f(a_n) = f(a) is equivalent to \lim_n f(a_n) = f(\lim_n a_n). In other words, the continuous function commutes with the limit operation.

Definition (bounds): Let X be a set of real numbers. A number b is an upper bound of A if \forall x \in A, x \leq b. A number c is a lower bound of A if \forall x \in A, c \leq x. If A has both an upper bound and a lower bound, then A is said to be bounded. An upper bound b^* is a least upper bound of A, if for all upper bound b of A, b^* \leq b. A lower bound c^* is a greatest lower bound of A, if for all lower bound c of A, c \leq c^*.

Definition (Completeness postulate): Completeness postulateis a properties of the real number systems, stating that if a non-empty set A of real numbers has an upper bound, then it also has a lub; and if a non-empty set B of real numbers has a lower bound, then it also has a \text{glb} .

Lemma (5.6): Let b be a \text{glb} of a non-empty set A. For each \epsilon > 0, exists an element x \in A such that x - b < \epsilon.

Proof: By contradiction. □

Corollary (5.7): Let b be a \text{glb} of a non-empty set A of real numbers. There is a sequence a_n of real numbers such that a_n \in A and \lim_n a_n = b.

Definition (Distance between an element and a set): Let \langle X, d\rangle be a metric space. Let a \in X and A be a non-empty subset of X. The \text{glb} of \{ d(a, x) | x \in A \} is called the distance between a and A and is denoted by d(a, A).

##### 1.6Open Sets and Closed Sets

Definition (Open Set): A subset O of a metric space is open if O is a neighborhood of each of its points.

We shall also use "open balls" to defined "open sets".

Theorem (6.2): A subset O of a metric space \langle X, d\rangle is an open set iff it is a union of open balls.

Proof:
• For the first direction, assume that O is open. By the definition of openness, for each point a \in O, there exists a ball B(a; \delta_a) \subset O. Therefore O = \cup_{a \in O} B(a; \delta_a).

• For the other direction, assume that O is a union of open balls, whose centers form an indexed set I. Then O = \cup_{a \in I} B(a; \delta_a). If x \in O, then there exists a \in I s.t. x \in B(a; \delta_a). B(a; \delta_a) is a neighborhood of x, and sicne B(a; \delta_a) \subset O, O is also a neighborhood of every x. □

Theorem (6.3): Let f: \langle X, d \rangle \to \langle Y, d{}’ \rangle. f is continuous iff for each open set O of Y, the subset f^{-1}(O) is an open subset of X.

Proof:
• For the first direction, we need to show that f^{-1}(O) is a neighborhood of every point in it. To show this, let a \in f^{-1}(O), then f(a) \in O and O is a neighborhood of f(a). Since f is continous, by Theorem 4.7, f^{-1}(O) is a neighborhood of a.

• For the other direction, assume that \forall O \subset Y, f^{-1}(O) is open. We need to show f is continous. Let a \in X and M be a neighborhood of f(a). There exists B(f(a), \epsilon) \subset M, since B(f(a), \epsilon) is open, f^{-1}(B(f(a), \epsilon)) is open and is a neighborhood of a. Therefore, f is continous. □

Theorem (6.4): Let \langle X, d\rangle be a metric space.
• The empty set is open

• X is open

• If O_1, O_2, \cdots, O_n are open, O_1 \cap O_2 \cdots \cap O_n is also open.

• If O_1, O_2, \cdots, O_n are open, O_1 \cup O_2 \cdots \cup O_n is also open.

Definition (Closedness): A subset F of a metric space is close if its complement C(F) is open.

A set can be both open and closed. For example, given a metric space \langle X, d\rangle, \emptyset and X are open, therefore their complement X and \emptyset are closed. So, \emptyset and X are both open and closed.

Definition (Limit Point): Let A be a subset of metric space X. A point b \in X is a limit point of A if every neighborhood of b contains a point of A different from b.

Theorem (6.7): Given a metric space \langle X, d\rangle, a set F \subset X is closed iff F contains all its limit points.

Proof:
• For the first direction, assume that F is cosed and therefore C(F) is open. Let F{}’ denote the set of limit points of F. We need to proof that F{}’ \subset F by showing \forall x \notin F, x \notin F{}’. Choose a point b \in F, since C(F) is open, there exists a \delta s.t. B(b; \delta) \subset C(F). Hence, b \notin F{}’ and F{}’ \subset F.

• For the other direction, assume that F{}’ \subset F and therefore C(F) \subset C(F{}’). We need to proof F is closed by showing C(F) is open. Since F{}’ \subset F, if b \in C(F), then b \notin F{}’. C(F) is a neighborhood of each points, since B(b; \delta) \subset C(F) for some \delta. □

#### 4Compactness

History

03/28/2020 - updated until sec. 1.3

03/16/2020 - added the post