A topology $(X, \tau)$ is defined as a collection $\tau$ of subsets of $X$ with the following properties:

- $X$ and $\phi$ are in $\tau$
- Arbitrary unions and finite intersections (AUFI) of $\tau$ are in $\tau$

The sets in $\tau$ is termed as `open sets`; then the collection of closed sets is the complement of the set of open sets.

## Basis of a topology

The basis of a topology $(X, \tau)$ is defined as follows:

- If $x \in X$, then $\exists$ atleast one basis element $\mathscr{B}$ containing $x$
- If $x \in \mathscr{B}_1 \cap \mathscr{B}_2$, then $\exists$ $\mathscr{B}_3 \subset \mathscr{B}_1 \cap \mathscr{B}_2$ containing $x$

Then topology $\tau$ can be defined as the collection of all unions of basis $\mathscr{B}$.

Let $\mathscr{B}$ be a basis for topology $\tau$ on set $X$. Then, $\tau$ equals the collection of all unions of elements of $\mathscr{B}$.

Basis of a topology is not unique. We have described how to go from the basis to the topology it generates; for the reverse direction, one way of finding a basis of a topology of $X$ follows.

Let $\mathscr{C}$ be a collection of open sets of $X$. For each $U \subset X$, $x \in U$, there exists an element $C$ of $\mathscr{C}$ such that $x \in C \subset U$. Then $\mathscr{C}$ is a basis for the topology of $X$. To prove this: let $x \in X$, and since $X$ is open, there exists, by hypothesis, an element $C \in \mathscr{C}$ such that $x \in C \subset X$; this is enough to fulfill the first condition in the definition of a topology. For the second condition, let $x \in C_1 \cap C_2$, where $C_1, C_2 \in \mathscr{C}$. Since $C_1, C_2$ are open, so is $C_1 \cap C_2$, and there exists, by hypothesis, $C_3 \in \mathscr{C}$, such that $x \in C_3 \subset C_1 \cap C_2$. Let $\tau$ be the collection of open sets in $X$; it remains to be shown that the topology $\tau'$ generated by $\mathscr{C}$ equals $\tau$. First, note that if $U$ belongs to $\tau$, then there exists, by hypothesis, an element $C \in \mathscr{C}$ such that $x \in C \subset U$. It follows that $U$ belongs to the topology $\tau'$, by definition. Conversely, if $W$ belongs to $\tau'$, then $W$ equals union of elements of $\mathscr{C}$. Since each element of $\mathscr{C}$ belongs to $\tau$, so does $W$.

A `subbasis` $\mathscr{S}$ of a topological space $X$ is defined as the collection of all subsets of $X$ whose union equals $X$. The topology generated by the subbasis is defined as the collection $\tau$ of all unions of finite intersections of elements of $\mathscr{S}$.

## Three topologies in $\mathbb{R}$

The important topologies on $\mathbb{R}$ are:

- The `standard topology` $\mathbb{R}$, defined as $(a, b)$ or $\{x \mid a \lt x \lt b\}$
- The `lower-bound topology` $\mathbb{R}_l$, defined as $[a, b)$
- The `K-topology` $\mathbb{R}_K$, defined as $(a, b) - K$, where $K = \frac{1}{n} : n \in \mathbb{Z}^+$

## Order topology

An order topology $\mathscr{B}$ is a collection of the following sets:

- All intervals of the form $(a, b)$
- All intervals of the form $[a_0, b)$
- All intervals of the form $(a, b_0]$

If $\mathscr{B}$ has no smallest element, then there are no sets of type (ii), and if $\mathscr{B}$ has no largest element, then there are no sets of type (iii).

If $X$ is an ordered set, and contains element $a$, the following four subsets are called rays determined by $a$: $(-\infty, a), (a, +\infty), (-\infty, a], [a, +\infty)$.

## Product topology

Given $X$ and $Y$, two topological spaces, the product topology $X \times Y$ is defined as a collection $\mathscr{B}$ having as basis all subsets of the form $U \times V$, where $U$ is an open subset of $X$, and $V$ is a an open subset of $Y$.

A property of product spaces is: $$(U_1 \times V_1) \cap (U_2 \times V_2) = (U_1 \cap U_2) \times (V_1 \cap V_2)$$

If $\mathscr{B}$ is a basis for $X$, and $\mathscr{C}$ is a basis for $Y$, $X$ and $Y$ being topological spaces, then the collection: $$\mathscr{D} = \{B \times C \mid B \in \mathscr{B}, C \in \mathscr{C}\}$$

is a basis for $X \times Y$. This can be proved as follows. Given an open set $W$ of $X \times Y$, and a point $x \times y \in W$, by definition of product topology, there is a basis element $U \times V$ such that $x \times y \in U \times V \subset W$. Because $\mathscr{B}, \mathscr{C}$ are bases of $X, Y$, we can choose an element $B \in \mathscr{B}$ such that $x \in B \subset U$, and $C \in \mathscr{C}$ such that $y \in C \subset V$. Then $x \times y \in B \times C \subset W$. Thus the collection $\mathscr{D}$ meets the criterion for being a basis of $X \times Y$.

`Projections` $\pi_1 : X \times Y \rightarrow X$ and $\pi_2 : X \times Y \rightarrow Y$ are defined in the usual way. Moreover, they are surjective.

The collection $\mathscr{S}$ defined as:

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

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

## Subspace topology

Given the topology $(X, \tau)$ and a $Y \subset X$, the subspace topology $\tau_Y$ is defined as: $$\tau_Y = \{Y \cap U \mid U \in \tau\}$$

If $\mathscr{B}$ is the basis for topology $X$, then $\mathscr{B}_Y = \{B \cap Y \mid B \in \mathscr{B}\}$ is a basis for the subspace topology on $Y$. We prove this as follows: Given $U$ open in $X$ and $y \in U \cap Y$, we can choose $B$ so that $y \in B \subset U$. Then, $y \in B \cap Y \subset U \cap Y$. It follows that $\mathscr{B}_Y$ is the basis for the subpsace topology on $Y$.

Let $Y$ be a subspace of $X$. If $U$ is open in $Y$ and $Y$ is open in $X$, then $U$ is open in $X$. To prove this, notice that, since $Y$ is open in $U$, $Y = U \cap V$, for some open set $V$ in $X$. Now, since both $U$ and $V$ are open in $X$, so is $U \cap V$.

If $A$ and $B$ are subspaces of $X$ and $Y$, then the product topology $A \times B$ is the same as the topology $A \times B$ one inherits as a subspace of $X \times Y$. To prove this, let $U, V$ be the general basis element of $X, Y$, where $U$ is open in $X$ and $V$ is open in $Y$. We know that the general basis element for for the subspace topology of $A \times B$ satisfies: $$(U \times A) \cap (V \times B) = (U \cap V) \times (A \cap B)$$

So, $(U \cap V) \times (A \cap B)$ is the general basis element for the product topology on $A \times B$. Since the bases for the product topology on $A \times B$ and the subspace topology $A \times B$ are the same, the topologies are the same.

Let $X$ be an ordered set in the order topology, and $Y$ be a subset. The resulting order topology on $Y$ need not be the same as the topology $Y$ inherits as a subspace of $X$.

## Closed sets

A subset $A$ of a topological space $X$ is said to be closed if $X - A$ is open. As an example, consider the subset $[a, b] \in \mathbb{R}$. It is closed because $\mathbb{R} - [a, b] = (-\infty, a) \cup (b, +\infty)$ is open.

Let $X$ be a topological space. Then:

- $X$ and $\phi$ are closed.
- Arbitrary intersections of closed sets are closed.
- Finite unions of closed sets are closed.

To prove (i), notice that $X, \phi$ are complements of open sets $\phi, X$, and are hence closed. To prove (ii), apply De-Morgan's law to a collection of closed sets $\{A_\alpha\}_{\alpha \in J}$: $$X - \bigcap_{\alpha \in J} A_\alpha = \bigcup_{\alpha \in J}(X - A_\alpha)$$

Notice that the right hand side of this equation represents arbitary unions of open sets, and is hence open, since $X - A_\alpha$ is open by definition. To prove (iii), apply De-Morgan's law again on closed sets $A_j$: $$X - \bigcup_{j \in J} A_j = \bigcap_{j \in J}(X - A_j)$$

The right side of this equation represents finite intersections of open sets and is therefore open. Hence, $\bigcup A_j$ is closed.

## Closure and interior

Given a subset $A$ of a topology, the interior of $A$, denoted by $\text{Int } A$, is defined as the union of all open sets contained in $A$, and the closure of $A$, denoted by $\bar{A}$, is defined as the intersection of all closed sets containing $A$.

$$ \text{Int } A \subset A \subset \bar{A} $$

If $A$ is open, $A = \text{Int } A$, and if $A$ is closed, $A = \bar{A}$.

We term $U$ as a neighborhood of $x$ if $U$ is an open set containing $x$. Let $A \subset X$. Then:

- $x \in \bar{A}$ iff every neighborhood of $x$ intersects $A$
- Suppose the topological space $X$ is given by a basis, $x \in \bar{A}$ iff every basis element $\mathscr{B}$ containing $x$ intersects $A$

Examples: For $C = {0} \cup (1, 2)$, $\bar{C} = {0} \cup [1, 2]$. $\bar{\mathbb{Q}} = \mathbb{R}$. $\bar{\mathbb{R}}^+ = \mathbb{R}$.

## Limit point

Let $A \subset X$. Then, $x$ is termed as the limit point of $A$ if every neighborhood of $x$ intersects $A$ at some point other than $A$.

Examples: In $B = \{\frac{1}{n} \mid n \in \mathbb{Z}^+\}$, the only limit point is $0$. In $A = (0, 1] \subset \mathbb{R}$, every point in $[0, 1]$ is a limit point.

The relationship between closure of $A$, and the set of all limit points of $A$, given by $A'$, is $$\bar{A} = A \cup A'$$

To prove this, consider $x \in A'$, so that every neighborhood of $x$ intersects $A$ in a point different from $x$. Therefore, $x \in \bar{A}$, and hence $A' \subset \bar{A}$. Since $A \subset \bar{A}$ by definition, it follows that $A \cup A' \subset \bar{A}$. To demonstrate the reverse inclusion, let $x \in \bar{A}$, and show that $x \in A \cup A'$. If $x \in A$, this is trivial. If not, since $x \in \bar{A}$, we know that every neighborhood $U$ of $x$ intersects $A$ different from $x$. Then $x \in A'$, and $x \in A \cup A'$, completing the proof.

## Hausdorff space

A topology $X$ converges to the point $x$, if for each neighborhood $U$ of $x$, there is a positive integer $N$ such that $x_n \in U$ for each $n \geq N$. In a general topology, we can have a sequence converging to more than one point; in order to disallow this non-intuitiveness, we define a Hausdorff space:

A topological space $X$ is Hausdorff if, for each pair of points $x_1$ and $x_2$, the corresponding neighborhoods $U_1$ and $U_2$ are disjoint.

Every finite point set in a Hausdorff space $X$ is closed. The proof follows. It suffices to show that every one-point set $\{x_0\}$ is closed. If $x \in X$ is a point different from $x_0$, then $x$ and $x_0$ have disjoint neighborhoods $U$ and $V$. Since $U$ does not intersect $\{x_0\}$, the point $x$ cannot belong to the closure of the set $\{x_0\}$. As a result, the closure of $\{x_0\}$ is $\{x_0\}$ itself, so it is closed.

The `$T_1$ axiom`: every point set is closed; this is weaker than the Hausdorff condition and implies it. Our lack of interest in the $T_1$ axiom is due to the fact that most interesting results require the full strength of the Hausdorff condition.

A sequence of points of a Hausdorff space $X$ converges to atmost one point. To prove this, suppose that $x_n$ is a sequence of points in $X$ that converges to $x$. If $y \neq x$, let $U, V$ be disjoint neighborhoods of $x, y$. Since $U$ contains $x_n$ for all but finitely many points of $n$, the set $V$ cannot. Hence, $x_n$ cannot converge to $y$.

## Continuous functions

A function $f : X \rightarrow Y$ is continuous with respect to topologies $X$ and $Y$ if, for each open set $V \subset X$, $f^{-1}(V)$ is an open set on $X$. For example $f : \mathbb{R} \rightarrow \mathbb{R}_l$ is not continuous, but $f : \mathbb{R}_l \rightarrow \mathbb{R}$ is (since $f^{-1}$ produces an open set in $\mathbb{R}_l$).

The following statements are equivalent:

- $f : X \rightarrow Y$ is continuous
- For every subset $A$, $f(\bar{A}) \subset \bar{f(A)}$
- For every closed set $B$ of $Y$, the set $f^{-1}(B)$ is closed in $X$
- For each $x \in X$, and each neighborhood $V$ of $f(x)$, there is a neighborhood $U$ of $x$ such that $f(U) \subset Y$.

The pasting lemma: Let $X = A \cup B$, where $A$ and $B$ are closed sets. Further, let $f : X \rightarrow Z$ and $g : Y \rightarrow Z$ be continuous. If $f(x) = g(x) \mid x \in A \cap B$, then $f$ and $g$ combine to give $h : X \rightarrow Y$ defined by $h(x) = f(x) \mid x \in A$, $h(x) = g(x) \mid x \in B$.

A useful theorem about continuous functions whose range is a product space: A function $f : A \rightarrow X \times Y$ given by $f(a) = (f_1(a), f_2(a))$ is continuous iff $f_1 : A \rightarrow X$ and $f_2 : A \rightarrow Y$ are continuous.

## Box topology and product topology

Box topology on $\prod X_\alpha$ has as basis $\prod U_\alpha$ where $U_\alpha$ is open in $X_\alpha$ for each $\alpha$. Product topology adds the extra condition that $U_\alpha$ equals $X_\alpha$ for each $\alpha$ except for finitely many values of $\alpha$.

## Metric topology

A metric on a set $X$ is given by a distance function $d : X \times X \rightarrow R$ such that:

- $d(x, y) \geq 0 \mid x, y \in X$
- $d(x, y) = d(y, x) \mid x, y \in X$
- (Triangle inequality) $d(x, y) + d(y, z) \geq d(x, z) \mid x, y, z \in X$

An $\epsilon$-ball centered at $x$ is defined as: $$B_d(x, \epsilon) = \{y \mid d(x, y) \lt \epsilon\}$$

A set $U$ is open in the metric topology induced by $d$ iff for each $y \subset U$, $\exists \delta > 0 : B_d(y, \delta) \subset U$

Let $d$ and $d'$ be two metrics on set $X$, and $\tau$, $\tau'$ be the topologies induced by them. $\tau'$ is finer than $\tau$ iff for each $x \in X$ and $\epsilon \gt 0$, $\exists \delta \gt 0$ such that: $$B_{d'}(x, \delta) \subset B_d(x, \epsilon)$$

An example of a metric space is:

$$ f(x, y) = \begin{cases} 1 & \quad \text{ if } x \neq y \\ 0 & \quad \text { if } x = y \end{cases} $$

The standard metric space on $\mathbb{R}$ defined as $d(x, y) = |x - y|$. The standard bounded metric is defined as $\bar{f}(x, y) = min\{f(x, y), 1\}$. The Eucledian metric is defined as:

$$ d(x, y) = ((x_1 - y_1)^2 + \ldots (x_n - y_n)^2)^{1/2} $$

The `square metric` $\rho$ is defined as:

$$ \rho(x, y) = max\{(x_1 - y_1), \ldots (x_n - y_n)\} $$

Given an index set J, the `uniform metric` $\bar{\rho}$ on $\mathbb{R}^J$ is defined as: $$\bar{\rho}(x, y) = \text{sup}\{\bar{d}(x_\alpha, y_\alpha) \mid \alpha \in J\}$$

The uniform topology on $\mathbb{R}^J$ is finer than the product topology and coarser than the box topology; all these topologies are different if $J$ is infinite.

Every metric space satisfies the Hausdorff axiom. Let $x, y \in (X, d)$ and $\epsilon = \frac{1}{2} d(x, y)$. Then the triangle inequality implies that $B_d(x, \epsilon), B_d(y, \epsilon)$ are disjoint.

Let $X$ and $Y$ be two metric spaces with metrics $d_X$ and $d_Y$, and let $f : X \rightarrow Y$. Further, let $x \in X$, $y \in Y$, and $\epsilon \gt 0$. Then, the requirement for continuity of $f$ is equivalent to: $\exists \delta \gt 0$ such that: $$f_X(x, y) \lt \delta \implies f_Y(f(x), f(y)) \lt \epsilon$$

The `sequence lemma`: Let $X$ be a topology, and $A \subset X$. If a sequence of points of $A$ converge to $x$, then $x \in \bar{A}$. The converse holds if $X$ is metrizable.

Let $f : X \rightarrow Y$ be continuous. For every convergent sequence, $x_n \rightarrow x$ in $X$, $f(x_n) \rightarrow f(x)$. The converse holds if $X$ is metrizable. Here, we don't use the full strength of metrizability: all we need is countable collection of balls $B_d(x, \frac{1}{n})$ about $x$.

Let $f_n : X \rightarrow Y$, from set $X$ to metric space $Y$, be a sequence of functions. Then $(f_n)$ converges uniformly to $f$ if given $\epsilon$, $\exists N \gt 0$ such that: $$d(f_n(x), f(x)) < \epsilon$$

for all $x \in X$ and $n \gt N$. The uniform limit theorem additionally states that $f$ is continuous.

## Connectedness and compactness

Three elementary theorems of calculus are:

- `Intermediate value theorem`: for continuous function $f$ defined in $[a, b]$, $\exists r \in [a, b] : f(r) \in [f(a), f(b)]$. The property of a topological space satisfying this is connectedness.
- `Maximum value theorem`: for continuous function $f$, $\exists c : f(x) \leq f(c)$ for all $x \in [a, b]$. The property of a topological space satisfying this is compactness.
- `Uniform continuity theorem`: for continuous function $f$, every pair of numbers$x_1$ and $x_2$, and given $\epsilon \gt 0$, $|f(x_1) - f(x_2)| \lt \epsilon$, $\exists \delta \gt 0 : |x_1 - x_2| \lt \delta$. The property of a topological space satisfying this is compactness.

The first depends on the connectedness property, and the others depend on the compactness property, of topological spaces.

## Connectedness

`Separation` of space $Y$ is a pair of nonempty sets $A$ and $B$ whose union is $Y$, neither of which contains a limit point of the other. A non-separable space is connected. Said differently, the space $X$ is connected iff the only the only subsets of $X$ that are both open and closed are $X$ and $\phi$.

If $Y$ is a subspace of $X$, the separation of $Y$ is a pair of disjoint nonempty sets $A, B$ whose union is $Y$, neither of which contains a limit point of the other. The space $Y$ is connected if it is non-separable. To prove this, suppose that $A, B$ form a separation of $Y$. Then, $A, B$ are both closed and open in $Y$. The closure of $A$ in $Y$ is $\bar{A} \cap Y$. Further, since $A$ is closed in $Y$, $A = \bar{A} \cap Y$ or $\bar{A} \cap B = \phi$. Since $\bar{A} = A \cup A'$, $B$ contains no limit points of $A$. A similar argument holds for $B$. Conversely, suppose that $A, B$ are disjoint nonempty subsets whose union is $Y$, neither of which contains the limit point of the other. Then, $A \cap \bar{B} = \bar{A} \cap B = \phi$; therefore $\bar{A} \cap Y = A$ and $\bar{B} \cap Y = B$. Thus, $A, B$ are closed in $Y$, and since $A = Y - B, B = Y - A$, they're open in $Y$ as well.

Examples: Let $X = [0, 2]$ be a subspace of $\mathbb{R}$. $[0, 1) \cup (1, 2]$ is separable, and hence disconnected. However, $[0, 1) \cup [1, 2]$ is non-separable since the first set contains a limit point of the second, namely $1$.

If $C, D$ form a separation of $X$, and $Y$ is a connected subpsace of $X$, then $Y$ lies entirely within $C$ or $D$.

The union of a collection of connected subspaces of $X$ that have a point in common is connected.

## Connected subspaces of $\mathbb{R}$

Let us first generalize the order properties of $\mathbb{R}$. A simply ordered set $L$ is called a `linear continuum` if:

- $L$ has least upper bound property
- For $x \lt y \in L$, $\exists z : x < z < y$

If $L$ is a linear continuum in the order topology, then $L$, and all intervals and rays in $L$ are connected.

Generalization of IVT of calculus: Let $f : X \rightarrow Y$ be a continuous map from a connected space to an ordered set in order topology. Then, for $f(a) \lt r \lt f(b) \in Y$, $\exists c \in X : f(c) = r$.

A continuous map $f : [a, b] \rightarrow X$ in a space $X$ is a path in $X$ from a closed interval $[a, b] \in \mathbb{R}$, such that $f(a) = x, f(b) = y$, $x, y \in X$. $X$ is said to be `path connected` if every pair of points in $X$ can be joined by a path in $X$.

The unit ball given by $\mathbb{B}^n \in \mathbb{R}^n$ where: $$\mathbb{B}^n = ||x|| = (x_1^2 + \ldots + x_n^2)^{1/2}$$

is path connected since we can always take a path from $[0, 1] \rightarrow \mathbb{R}^n$ given by: $$f(n) = (1 - t)x + ty$$

Another example: The `punctured Eucledian space` given by $\mathbb{R}^n - \{0\}$ is path connected for $n > 1$ since we can always take a path connecting two points that doesn't go through the origin.

## Compactness

A collection of subsets $\mathscr{A}$ of space $X$ is said to be the cover or `covering` of $X$ if the union of the subsets equals $X$. It is said to be an open covering if the subsets are open.

A space $X$ is compact if every finite subcollection of $\mathscr{A}$ also contains an open covering of $X$.

All closed subsets of $\mathbb{R}$ are compact, so $[0, 1]$ is an example of a compact space. However, neither $(0, 1)$ nor $(0, 1]$ are compact. The latter is because we can choose an open covering $$\mathscr{A} = \{(1/n, 1] \mid n \in \mathbb{Z}^+\}$$

which doesn't cover $(0, 1]$.

Some properties of compact spaces:

- Every closed subspace of a compact space is compact
- Every compact subspace of a Hausdorff space is closed
- The image of a compact space under a continuous map is compact

The bijective continuous map $f : X \rightarrow Y$ is a homeomorphism if $X$ is compact and $Y$ is Hausdorff.

The `tube lemma`: Consider the product space $X \times Y$ where $Y$ is compact. If the open set $N$ contains slice $x_0 \times Y$ of $X \times Y$, then it contains some tube $W \times Y$ about $x_0 \times Y$ where $W$ is a neighborhood of $x_0 \in X$.

The `Tychonoff theorem`: The product of infinitely many compact spaces is compact.

A collection $\mathscr{C}$ of sets in $X$ is said to have the finite intersection property, if for every finite subcollection ${C_1 \ldots C_n} \in \mathscr{C}$, the intersection $C_1 \cap \ldots \cap C_n$ is nonempty.

Redefinition of compactness in terms of the finite intersection property: $X$ is compact if for every finite subcollection $\mathscr{C}$ of closed sets in $X$ having the finite intersection property, $\bigcap_{C \in \mathscr{C}} C$ is nonempty.

## Compact subspaces of $\mathbb{R}$

We need only the least upper bound property of $\mathbb{R}$ for compactness. In order topology, a simply ordered set having least upper bound property is compact.

A subspace $A \in \mathbb{R}^n$ is compact if and only if it is closed and bounded in the eucledian metric $d$ or the square metric $\rho$.

## Countability axioms

A space $X$ is said to have a `countable basis` at $x$ if there is a countable collection $\mathscr{B}$ of neighborhoods of $x$, such that each neighborhood of $x$ contains atleast one element of $\mathscr{B}$. A space that has a countable basis at each of its points is said said to be `first-countable`.

If a space $X$ has a countable basis for its topology, then $X$ is said to be `second-countable`. This is obviously stronger than the first-countable axiom.

A subspace of a first-countable space is first-countable, and a countable product of first-countable spaces is first-countable. The same applies to second-countable spaces.

A subspace $A$ is said to be `dense` in $X$ if $\bar{A} = X$.

Suppose $X$ has a countable basis. Then:

- Every open covering of $X$ contains a countable subcollection covering $X$.
- There exists a countable basis of $X$ that is dense in $X$.