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 any finite subcollection of $\mathcal{T}$ is in $\mathcal{T}$.
A set $X$ for which a topology $\mathcal{T}$ has been specified is called a topological space. We say that a subset $U$ of $X$ is an open set of $x$ if $U$ belongs to $\mathcal{T}$.
Suppose that $\mathcal{T},\mathcal{T'}$ are two topologies on a given set $X$. If $\mathcal{T'} \supset \mathcal{T}$, then we say that $\mathcal{T'}$ is finer than $\mathcal{T}$; if $\mathcal{T'}$ properly contains $\mathcal{T}$, we say that $\mathcal{T'}$ is strictly finer than $\mathcal{T}$. We also say that $\mathcal{T}$ is coarser than $\mathcal{T'}$, or strictly coarser, in these two respective situations. We say $\mathcal{T}$ is comparable with $\mathcal{T'}$ if either $\mathcal{T'} \supset \mathcal{T}$ or $\mathcal{T} \supset \mathcal{T'}$.
If $X$ is a set, a basis for a topology on $X$ is a collection $\mathcal{B}$ of subsets of $X$ (called basis elements) 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 $B_1,B_2$, then there is a basis element $B_3$ containing $x$ such that $B_3 \subset B_1 \cap B_2$
If $\mathcal{B}$ satisfies these two conditions, then we define the topology $\mathcal{T}$ generated by $\mathcal{B}$ as follows: A subset $U$ of $X$ is said to be open in $X$ (that is, to be an element of $\mathcal{T}$) if for each $x \in U$, there is a basis element $B \in \mathcal{B}$ such that $x \in B$ and $B \subset U$. Note that each basis element is itself an element of $\mathcal{T}$.
If $\mathcal{B}$ is the collection of all open intervals in the real line $(a,b)$ for finite $a,b: a < b$, the topology generated by $\mathcal{B}$ is called the standard topology on the real line. If $\mathcal{B}'$ is the collection of all half-open intervals $[a,b)$ for finite $a,b : a < b$, the topology generated by $\mathcal{B}'$ is called the lower limit topology on $\mathbb{R}$, denoted $\mathbb{R}_l$. Finally, let $K$ defnote the set of all numbers of the form $1/n$ for $n \in \mathbb{N}$, and let $\mathcal{B}''$ be the collection of all open intervals $(a,b)$, along with all sets of the form $(a,b) - K$. The topology generated by $\mathcal{B}''$ will be called the K-topology on $\mathbb{R}$, denotes $\mathbb{R}_K$.
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 $S$.

Propositions

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 $\mathcal{B}$.
Let $X$ be a topological space. Suppose that $\mathcal{C}$ is a collection of open sets of $X$ such that for each open set $U$ of $X$ and each $x$ in $U$, there is an element $C$ of $\mathcal{C}$ such that $x \in C \subset U$. Then $\mathcal{C}$ is a basis for the topology of $X$.
Let $\mathcal{B},\mathcal{B}'$ be bases for the topologies $\mathcal{T},\mathcal{T}'$, 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 $x \in B' \subset B$.

Let $X$ be a set with a simple order relation; assume $X$ has more than 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.
Let $X,Y$ 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 $Y$.
Let $\pi_1 : X \times Y \to X$ be defined by the equation: $$\pi_1 (x,y) = x$$ and $\pi_2 : X \times Y \to Y$ be: $$\pi_2 (x,y) = y$$ the maps $\pi_1,\pi_2$ are called the projections of $X \times Y$ onto its first and second factors, respectively.
Let $X$ be a topological space with topology $\mathcal{T}$. If $Y$ is a subset of $X$, the collection $$\mathcal{T}_Y = \{Y \cap U | U \in \mathcal{T}\}$$ is a topology on $Y$, called the subspace topology. With this topology, $Y$ is called a subspace of $X$; its open sets consist of all intersections of open sets of $X$ with $Y$.

Propositions

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 $$\mathcal{D} = \{B \times C | B \in \mathcal{B} \text{ and } C \in \mathcal{C}\}$$ is a basis for the topology of $X \times Y$.
The collection $$S = \{\pi_1^{-1}(U) | U \text{ open in } X\} \cup \{p_2^{-1} (V) | V \text{ open in } Y\}$$ is a subbasis for the product topology on $X \times Y$.
If $\mathcal{B}$ is a basis for the topology on $X$ then the collection $$\mathcal{B}_Y = \{B \cap Y | B \in \mathcal{B}\}$$ is a basis for the subspace topology on $Y$.
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 as a subspace of $X \times Y$.
Let $X$ be an ordered set in the order topology; let $Y$ be a subset of $X$ that is convex in $X$. (To be convex just means that for each pair of $a,b \in Y$, $(a,b) \in Y$.) Then the order topology on $Y$ is the same as the topology $Y$ inherits as a subspace of $X$.

A subset $A$ of a topological space $X$ is said to be closed if the set $X-A$ is open.
Given a subset $A$ of a topological space $X$, the interior of $A$ is defined as the union of all open sets contained in $A$, and the closure of $A$ is defined as the intersection of all closed sets containing $A$.
If $A$ is a subset of the topological space $X$ and if $x \in X$, then $x$ is a limit point of $A$ if every neighborhood of $x$ intersects $A$ in some point other than $x$ itself.
In an arbitrary topological space, one says that a sequence $x_1,x_2,...$ of points of the space $X$ converges to the point $x \in X$ provided that, corresponding to each neighborhood $U$ of $x$, there exists $N$ such that $n \geq N \implies x_n \in U$.
A topological space $X$ is called a Hausdorff space if for each pair $x_1,x_2$ of distinct points of $X$, there exist neighborhoods $U_1,U_2$ of $x_1,x_2$ respectively that are disjoint.

Propositions

Let $Y$ be a subspace of $X$. Then a set $A$ is closed in $Y$ if and only if it equals the intersection of a closed set of $X$ with $Y$.
Let $Y$ be a subspace of $X$; let $A$ be a subset of $Y$; let $\overline{A}$ denote the closure of $A$ in $X$. Then the closure of $A$ in $Y$ equals $\overline{A} \cap Y$.
Let $A$ be a subset of the topological space $X$.
a) Then $x \in \overline{A}$ if and only if every open set $U$ containing $x$ intersects $A$
b) Supposing the topology of $X$ is given by a basis, then $x \in \overline{A}$ if and only if every basis element $B$ containing $x$ intersects $A$.
Let $A$ be a subset of the topological space $X$; let $A'$ be the set of all limit points of $A$. Then: $$\overline{A} = A \cup A'$$ Corollory: A subset of a topological space is closed if and only if it contains all its limit points.
Every finite point set in a Hausdorff space $X$ is closed.
$X$ is a space such that every finite point set is closed. Suppose $A$ is a subset of $X$. Then the point $x$ is a limit point of $A$ if and only if every neighborhood of $x$ contains infinitely many points of $A$.
If $X$ is a Hausdorff space, then a sequence of points of $X$ converges to at most one point of $X$.
Every simply ordered set is a Hausdorff space in the order topology. The product of two Hausdorff spaces is a Hausdorff space. A subspace of a Hausdorff space is a Hausdorff space.

Let $X,Y$ be topological spaces. A function $f : X \to Y$ is said to be continuous if for each oepn subset $V$ of $Y$, the set $f^{-1}(V)$ is an open subset of $X$.
Let $X,Y$ be topological spaces; let $f : X \to Y$ be a bijection. If both the function $f$ and the inverse function $f^{-1} : Y \to X$ are continuous, then $f$ is called a homeomorphism. In other words, $f(U)$ is open if and only if $U$ is open.
Suppose that $f : X \to Y$ is an injective continuous map, where $X,Y$ are topological spaces. Let $Z$ be the image set $f(X)$, considered as a subspace of $Y$; then the function $f' : X \to Z$ obtained by restricting the range of $f$ is bijective. If $f'$ happens to be a homeomorphism of $X$ with $Z$, we say that the map $f : X \to Y$ is a topological imbedding, or simply an imbedding, of $X$ in $Y$.

Propositions

Let $X,Y$ be topological spaces, let $f : X \to Y$. Then the following are equivalent:
1) $f$ is continuous
2) For every subset $A$ of $X$, one has $f(\overline{A}) \subset \overline{f(A)}$
3) For every closed set $B$ of $Y$, the set $f^{-1}(B)$ is closed in $X$
4) For each $x \in X$ and each neighborhood $V$ of $f(x)$, there is a neighborhood $U$ of $x$ such that $f(U) \subset V$.
Let $X,Y,Z$ be topological spaces.
a) (Constant function) If $f : X \to Y$ maps all of $X$ into the single point $y_0 \in Y$, then $f$ is continuous.
b) (Inclusion) If $A$ is a subspace of $X$, the inclusion function $j : A \to X$ is continuous.
c) (Composites) If $f : X \to Y$ and $g : Y \to Z$ are continuous, then the map $g \circ f : X \to Z$ is continuous.
d) (Restricting the domain) If $f : X \to Y$ is continuous, and if $A$ is a subspace of $X$, then the restricted function $f|A : A \to Y$ is continuous.
e) (Restricting or expanding the range) Let $f : X \to Y$ be continuous. If $Z$ is a subspace of $Y$ containing the image set $f(X)$, then the function $g : X \to Z$ obtained by restricting the range of $f$ is continuous. If $Z$ is a space having $Y$ as a subspace, then the function $h : X \to Z$ obtained by expanding the range of $f$ is continuous.
f) (Local formulation of continuity) The map $f : X \to Y$ is continuous if $X$ can be written as the union of open sets $U_\alpha$ such that $f|U_\alpha$ is continuous for each $\alpha$.
Let $X = A \cup B$, where $A$ and $B$ are closed in $X$. Let $f : A \to Y$ and $g : B \to Y$ be continuous. If $f(x) = g(x)$ for every $x \in A \cap B$, then $f$ and $g$ combine to give a continuous function $h : X \to Y$, defined by setting $h(x) = f(x)$ if $x \in A$, and $h(x) = g(x)$ if $x \in B$.
Let $f : A \to X \times Y$ be given by the equation $$f(a) = (f_1(a),f_2(a))$$ Then $f$ is continuous if and only if the functions $$f_1 : A \to X \text{ and } f_2: A \to Y$$ are continuous.

Let $\{A_\alpha\}_{\alpha \in J}$ be an indexed family of sets; let $X = \bigcup_{\alpha \in J} A_\alpha$. The cartesian product of this indexed family, denoted by: $$\prod_{\alpha \in J} A_\alpha$$ is defined to be the set of all $J$-tuples $(x_\alpha)_{\alpha \in J}$ of elements of $X$ such that $x_\alpha \in A_\alpha$ for each $\alpha \in J$.
Let $\{X_\alpha\}_{\alpha \in J}$ be an indexed family of topological spaces. Let us take as a basis for topology on the product space $$\prod_{\alpha \in J} X_\alpha$$ the collection of all sets of the form $$\prod_{\alpha \in J} U_\alpha$$ where $U_\alpha$ is open in $X_\alpha$, for each $\alpha \in J$. The topology generated by this basis is called the box topology.
Let $S_\beta$ denote the collection $$S_\beta = \{\pi_\beta^{-1}(U_\beta) | U_\beta \text{ open in } X_\beta\}$$ and let $S$ denote the union of these collections: $$S = \cup_{\beta \in J} S_\beta$$ The topology generated by the subbasis $S$ is called the product topology. In this topology $\prod_{\alpha \in J} X_\alpha$ is called a product space.

Propositions

The box topology on $\prod X_\alpha$ has as basis all sets of the form $\prod U_\alpha$, where $U_\alpha$ is open in $X_\alpha$ for each $\alpha$. The product topology on $\prod X_\alpha$ has as basis all sets of the form $\prod U_\alpha$, where $U_\alpha$ is open in $X_\alpha$ for each $\alpha$ and $U_\alpha$ equals $X_\alpha$ except for finitely many values of $\alpha$.
Suppose that the topology on each space $X_\alpha$ is given by a basis $\mathcal{B}_\alpha$. The collection of all sets of the form $$\prod_{\alpha \in J} B_\alpha$$ where $B_\alpha \in \mathcal{B}_\alpha$ for each $\alpha$, will serve as a basis for the box topology on $\prod_{\alpha \in J} X_\alpha$
Let $A_\alpha$ be a subspace of $X_\alpha$, for each $\alpha \in J$. Then $\prod A_\alpha$ is a subspace of $\prod X_\alpha$ if both products are given the box topology, or if both products are given the product topology.
If each space $X_\alpha$ is Hausdorff, then $\prod X_\alpha$ is Hausdorff on both the box and product topologies.
Let $\{X_\alpha\}$ be an indexed family of spaces; let $A_\alpha \subset X_\alpha$ for each $\alpha$. If $\prod X_\alpha$ is given either the product or the box topology, then $$\prod \underline{A}_\alpha = \underline{\prod A_\alpha}$$
Let $f : A \to \prod_{\alpha \in J} X_\alpha$ be given by the equation $$f(a) = (f_\alpha (a))_{\alpha \in J}$$ where $f_\alpha : A \to X_\alpha$ for each $\alpha$. Let $\prod X_\alpha$ have the product topology. Then the function $f$ is continuous if and only if each function $f_\alpha$ is continuous.

A metric on a set $X$ is a function $$d : X \times X \to R$$ having the following properties:
1) $d(x,y) \geq 0$ for all $x,y \in X$; equality holds if and only if $x = y$.
2) $d(x,y) = d(y,x)$ for all $x,y \in X$.
3) (Triangle inequality) $d(x,y) + d(y,z) \geq d(x,z)$
If $d$ is a metric on the set $X$, then the collection of all $\epsilon$-balls $B_d(x,\epsilon)$, for $x \in X$ and $\epsilon > 0$, is a basis for a topology on $X$, called the metric topology induced by $d$.
If $X$ is a topological space, $X$ is said to be metrizable if there exists a metric $d$ on the set $X$ that induces the topology of $X$. A metric space is a metrizable space $X$ together with a specific metric $d$ that gives the topology of $X$.
Let $X$ be a metric space with metric $d$. A subset $A$ of $X$ is said to be bounded if there is some number $M$ such that $$d(a_1,a_2) \leq M$$ for every pair $a_1,a_2$ of points of $A$. If $A$ is bounded and nonempty, the diameter of $A$ is defined to be the number $$\text{diam } A = \text{sup}\{d(a_1,a_2) | a_1,a_2 \in A\}$$
Given $x = (x_1,...,x_n) \in \mathbb{R}^n$, we define the norm of $x$ by the equation $$||x|| = (x_1^2 + ... + x_n^2)^{1/2}$$ and we define the euclidean metric $d$ on $\mathbb{R}^n$ by the equation $$d(x,y) = ||x - y|| = [(x_1 - y_1)^2 + ... + (x_n - y_n)^2]^{1/2}$$ We define the square metric $\rho$ by the equation $$\rho(x,y) = \text{max}\{|x_1 - y_1|, ... , |x_n - y_n|\}$$