# Math 4201 Final Exam Review

> [!NOTE]
>
> This is a review for definitions we covered in the classes. It may serve as a cheat sheet for the exam if you are allowed to use it.

## Topological space

### Basic definitions

#### Definition for topological space

A topological space is a pair of set $X$ and a collection of subsets of $X$, denoted by $\mathcal{T}$ (imitates the set of "open sets" in $X$), satisfying the following axioms:

1. $\emptyset \in \mathcal{T}$ and $X \in \mathcal{T}$
2. $\mathcal{T}$ is closed with respect to arbitrary unions. This means, for any collection of open sets $\{U_\alpha\}_{\alpha \in I}$, we have $\bigcup_{\alpha \in I} U_\alpha \in \mathcal{T}$
3. $\mathcal{T}$ is closed with respect to finite intersections. This means, for any finite collection of open sets $\{U_1, U_2, \ldots, U_n\}$, we have $\bigcap_{i=1}^n U_i \in \mathcal{T}$

#### Definition of open set

$U\subseteq X$ is an open set if $U\in \mathcal{T}$

#### Definition of closed set

$Z\subseteq X$ is a closed set if $X\setminus Z\in \mathcal{T}$

> [!WARNING]
>
> A set is closed is not the same as its not open.
>
> In all topologies over non-empty sets, $X, \emptyset$ are both closed and open.

### Basis

#### Definition of topological basis

For a set $X$, a topology basis, denoted by $\mathcal{B}$, is a collection of subsets of $X$, such that the following properties are satisfied:

1. For any $x \in X$, there exists a $B \in \mathcal{B}$ such that $x \in B$ (basis covers the whole space)
2. If $B_1, B_2 \in \mathcal{B}$ and $x \in B_1 \cap B_2$, then there exists a $B_3 \in \mathcal{B}$ such that $x \in B_3 \subseteq B_1 \cap B_2$ (every non-empty intersection of basis elements are also covered by a basis element)

#### Definition of topology generated by basis

Let $\mathcal{B}$ be a basis for a topology on a set $X$. Then the topology generated by $\mathcal{B}$ is defined by the set as follows:

$$
\mathcal{T}_{\mathcal{B}} \coloneqq \{ U \subseteq X \mid \forall x\in U, \exists B\in \mathcal{B} \text{ such that } x\in B\subseteq U \}
$$

> This is basically a closure of $\mathcal{B}$ under arbitrary unions and finite intersections

#### Lemma of topology generated by basis

$U\in \mathcal{T}_{\mathcal{B}}\iff \exists \{B_\alpha\}_{\alpha \in I}\subseteq \mathcal{B}$ such that $U=\bigcup_{\alpha \in I} B_\alpha$

#### Definition of basis generated from a topology

Let $(X, \mathcal{T})$ be a topological space. Then the basis generated from a topology is $\mathcal{C}\subseteq \mathcal{B}$ such that $\forall U\in \mathcal{T}$, $\forall x\in U$, $\exists B\in \mathcal{C}$ such that $x\in B\subseteq U$.

#### Definition of subbasis of topology

A subbasis of a topology is a collection $\mathcal{S}\subseteq \mathcal{T}$ such that $\bigcup_{U\in \mathcal{S}} U=X$.

#### Definition of topology generated by subbasis

Let $\mathcal{S}\subseteq \mathcal{T}$ be a subbasis of a topology on $X$, then the basis generated by such subbasis is the closure of finite intersection of $\mathcal{S}$

$$
\mathcal{B}_{\mathcal{S}} \coloneqq \{B\mid B\text{ is the intersection of a finite number of elements of }\mathcal{S}\}
$$

Then the topology generated by $\mathcal{B}_{\mathcal{S}}$ is the subbasis topology denoted by $\mathcal{T}_{\mathcal{S}}$.

Note that all open set with respect to $\mathcal{T}_{\mathcal{S}}$ can be written as a union of finitely intersections of elements of $\mathcal{S}$

### Comparing topologies

#### Definition of finer and coarser topology

Let $(X,\mathcal{T})$ and $(X,\mathcal{T}')$ be topological spaces. Then $\mathcal{T}$ is finer than $\mathcal{T}'$ if $\mathcal{T}'\subseteq \mathcal{T}$. $\mathcal{T}$ is coarser than $\mathcal{T}'$ if $\mathcal{T}\subseteq \mathcal{T}'$.

#### Lemma of comparing basis

Let $(X,\mathcal{T})$ and $(X,\mathcal{T}')$ be topological spaces with basis $\mathcal{B}$ and $\mathcal{B}'$. Then $\mathcal{T}$ is finer than $\mathcal{T}'$ if and only if for any $x\in X$, $x\in B'$, $B'\in \mathcal{B}'$, there exists $B\in \mathcal{B}$, such that $x\in B$ and $x\in B\subseteq B'$.

### Product space

#### Definition of cartesian product

Let $X,Y$ be sets. The cartesian product of $X$ and $Y$ is the set of all ordered pairs $(x,y)$ where $x\in X$ and $y\in Y$, denoted by $X\times Y$.

#### Definition of product topology

Let $(X,\mathcal{T}_X)$ and $(Y,\mathcal{T}_Y)$ be topological spaces. Then the product topology on $X\times Y$ is the topology generated by the basis

$$
\mathcal{B}_{X\times Y}=\{U\times V, U\in \mathcal{T}_X, V\in \mathcal{T}_Y\}
$$

or equivalently,

$$
\mathcal{B}_{X\times Y}'=\{U\times V, U\in \mathcal{B}_X, V\in \mathcal{B}_Y\}
$$

> Product topology generated from open sets of $X$ and $Y$ is the same as product topology generated from their corresponding basis

### Subspace topology

#### Definition of subspace topology

Let $(X,\mathcal{T})$ be a topological space and $Y\subseteq X$. Then the subspace topology on $Y$ is the topology given by

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

or equivalently, let $\mathcal{B}$ be the basis for $(X,\mathcal{T})$. Then the subspace topology on $Y$ is the topology generated by the basis

$$
\mathcal{B}_Y=\{U\cap Y| U\in \mathcal{B}\}
$$

#### Lemma of open sets in subspace topology

Let $(X,\mathcal{T})$ be a topological space and $Y\subseteq X$. Then if $U\subseteq Y$, $U$ is open in $(Y,\mathcal{T}_Y)$, then $U$ is open in $(X,\mathcal{T})$.

> This also holds for closed set in closed subspace topology

### Interior and closure

#### Definition of interior

The interior of $A$ is the largest open subset of $A$.

$$
A^\circ=\bigcup_{U\subseteq A, U\text{ is open in }X} U
$$

#### Definition of closure

The closure of $A$ is the smallest closed superset of $A$.

$$
\overline{A}=\bigcap_{U\supseteq A, U\text{ is closed in }X} U
$$

#### Definition of neighborhood

A neighborhood of a point $x\in X$ is an open set $U\in \mathcal{T}$ such that $x\in U$.

#### Definition of limit points

A point $x\in X$ is a limit point of $A$ if every neighborhood of $x$ contains a point in $A-\{x\}$.

We denote the set of all limits points of $A$ by $A'$.

$\overline{A}=A\cup A'$

### Sequences and continuous functions

#### Definition of convergence

Let $X$ be a topological space. A sequence $(x_n)_{n\in\mathbb{N}_+}$ in $X$ converges to $x\in X$ if for any neighborhood $U$ of $x$, there exists $N\in\mathbb{N}_+$ such that $\forall n\geq N, x_n\in U$.

#### Definition of Hausdoorff space

A topological space $(X,\mathcal{T})$ is Hausdorff if for any two distinct points $x,y\in X$, there exist open neighborhoods $U$ and $V$ of $x$ and $y$ respectively such that $U\cap V=\emptyset$.

#### Uniqueness of convergence in Hausdorff spaces

In a Hausdorff space, if a sequence $(x_n)_{n\in\mathbb{N}_+}$ converges to $x\in X$ and $y\in X$, then $x=y$.

#### Closed singleton in Hausdorff spaces

In a Hausdorff space, if $x\in X$, then $\{x\}$ is a closed set.

#### Definition of continuous function

Let $(X,\mathcal{T}_X)$ and $(Y,\mathcal{T}_Y)$ be topological spaces. A function $f:X\to Y$ is continuous if for any open set $U\subseteq Y$, $f^{-1}(U)$ is open in $X$.

#### Definition of point-wise continuity

Let $(X,\mathcal{T}_X)$ and $(Y,\mathcal{T}_Y)$ be topological spaces. A function $f:X\to Y$ is point-wise continuous at $x\in X$ if for every openset $V\subseteq Y$, $f(x)\in V$ then there exists an open set $U\subseteq X$ such that $x\in U$ and $f(U)\subseteq V$.

#### Lemma of continuous functions

If $f:X\to Y$ is point-wise continuous for all $x\in X$, then $f$ is continuous.

#### Properties of continuous functions

If $f:X\to Y$ is continuous, then

1. $\forall A\subseteq Y$, $f^{-1}(A^c)=X\setminus f^{-1}(A)$ (complements maps to complements)
2. $\forall A_\alpha\subseteq Y, \alpha\in I$, $f^{-1}(\bigcup_{\alpha\in I} A_\alpha)=\bigcup_{\alpha\in I} f^{-1}(A_\alpha)$
3. $\forall A_\alpha\subseteq Y, \alpha\in I$, $f^{-1}(\bigcap_{\alpha\in I} A_\alpha)=\bigcap_{\alpha\in I} f^{-1}(A_\alpha)$
4. $f^{-1}(U)$ is open in $X$ for any open set $U\subseteq Y$.
5. $f$ is continuous at $x\in X$.
6. $f^{-1}(V)$ is closed in $X$ for any closed set $V\subseteq Y$.
7. Assume $\mathcal{B}$ is a basis for $Y$, then $f^{-1}(\mathcal{B})$ is open in $X$ for any $B\in \mathcal{B}$.
8. $\forall A\subseteq X$, $\overline{f(A)}=f(\overline{A})$

#### Definition of homeomorphism

Let $(X,\mathcal{T}_X)$ and $(Y,\mathcal{T}_Y)$ be topological spaces. A function $f:X\to Y$ is a homeomorphism if $f$ is continuous, bijective and $f^{-1}:Y\to X$ is continuous.

#### Ways to construct continuous functions

1. If $f:X\to Y$ is constant function, $f(x)=y_0$ for all $x\in X$, then $f$ is continuous. (constant functions are continuous)
2. If $A$ is a subspace of $X$, $f:A\to X$ is the inclusion map $f(x)=x$ for all $x\in A$, then $f$ is continuous. (inclusion maps are continuous)
3. If $f:X\to Y$ is continuous, $g:Y\to Z$ is continuous, then $g\circ f:X\to Z$ is continuous. (composition of continuous functions is continuous)
4. If $f:X\to Y$ is continuous, $A$ is a subspace of $X$, then $f|_A:X\to Y$ is continuous. (domain restriction is continuous)
5. If $f:X\to Y$ is continuous, $Z$ is a subspace of $Y$, then $f:X\to Z$, $g(x)=f(x)\cap Z$ is continuous. If $Y$ is a subspace of $Z$, then $h:X\to Z$, $h(x)=f(x)$ is continuous (composition of $f$ and inclusion map).
6. If $f:X\to Y$ is continuous, $X$ can be written as a union of open sets $\{U_\alpha\}_{\alpha\in I}$, then $f|_{U_\alpha}:X\to Y$ is continuous.
7. If $X=Z_1\cup Z_2$, and $Z_1,Z_2$ are closed equipped with subspace topology, let $g_1:Z_1\to Y$ and $g_2:Z_2\to Y$ be continuous, and for all $x\in Z_1\cap Z_2$, $g_1(x)=g_2(x)$, then $f:X\to Y$ by $f(x)\begin{cases}g_1(x), & x\in Z_1 \\ g_2(x), & x\in Z_2\end{cases}$ is continuous. (pasting lemma)
8. $f:X\to Y$ is continuous, $g:X\to Z$ is continuous if and only if $H:X\to Y\times Z$, where $Y\times Z$ is equipped with the product topology, $H(x)=(f(x),g(x))$ is continuous. (proved in homework)

### Metric spaces

#### Definition of metric

A metric on $X$ is a function $d:X\times X\to \mathbb{R}$ such that $\forall x,y\in X$,

1. $d(x,x)=0$
2. $d(x,y)\geq 0$
3. $d(x,y)=d(y,x)$
4. $d(x,y)+d(y,z)\geq d(x,z)$

#### Definition of metric ball

The metric ball $B_r^{d}(x)$ is the set of all points $y\in X$ such that $d(x,y)\leq r$.

#### Definition of metric topology

Let $X$ be a metric space with metric $d$. Then $X$ is equipped with the metric topology generated by the metric balls $B_r^{d}(x)$ for $r>0$.

#### Definition of metrizable

A topological space $(X,\mathcal{T})$ is metrizable if it is the metric topology for some metric $d$ on $X$.

#### Hausdorff axiom for metric spaces

Every metric space is Hausdorff (take metric balls $B_r(x)$ and $B_r(y)$, $r=\frac{d(x,y)}{2}$).

If a topology isn't Hausdorff, then it isn't metrizable.

Prove by triangle inequality and contradiction.

#### Common metrics in $\mathbb{R}^n$

Euclidean metric

$$
d(x,y)=\sqrt{\sum_{i=1}^n (x_i-y_i)^2}
$$

Square metric

$$
\rho(x,y)=\max_{i=1}^n |x_i-y_i|
$$

Manhattan metric

$$
m(x,y)=\sum_{i=1}^n |x_i-y_i|
$$

These metrics are equivalent.

#### Product topology and metric

If $(X,d),(Y,d')$ are metric spaces, then $X\times Y$ is metric space with metric $d(x,y)=\max\{d(x_1,y_1),d(x_2,y_2)\}$.

#### Uniform metric

Let $\mathbb{R}^\omega$ be the set of all infinite sequences of real numbers. Then $\overline{d(x,y)}=\sup_{i=1}^\omega \min\{1,|x_i-y_i|\}$, the uniform metric on $\mathbb{R}^\omega$ is a metric.

#### Metric space and converging sequences

Let $X$ be a topological space, $A\subseteq X$, $x_n\to x$ such that $x_n\in A$. Then $x\in \overline{A}$.

If $X$ is a **metric space**, $A\subseteq X$, $x\in \overline{A}$, then there exists converging sequence $x_n\to x$ such that $x_n\in A$.

### Metric defined for functions

#### Definition for bounded metric space

A metric space $(Y,d)$ is bounded if there is $M\in \mathbb{R}^{\geq 0}$ such that for all $y_1,y_2\in Y$, $d(y_1,y_2)\leq M$.

#### Definition for metric defined for functions

Let $X$ be a topological space and $Y$ be a bounded metric space, then the set of all maps, denoted by $\operatorname{Map}(X,Y)$, $f:X\to Y\in \operatorname{Map}(X,Y)$ is a metric space with metric $\rho(f,g)=\sup_{x\in X} d(f(x),g(x))$.

#### Space of continuous map is closed

Let $(\operatorname{Map}(X,Y),\rho)$ be a metric space defined above, then every continuous map is a limit point of some sequence of continuous maps.

$$
Z=\{f\in \operatorname{Map}(X,Y)|f\text{ is continuous}\}
$$

$Z$ is closed in $(\operatorname{Map}(X,Y),\rho)$.

### Quotient space

#### Quotient map

Let $X$ be a topological space and $X^*$ is a set. $q:X\to X^*$ is a surjective map. Then $q$ is a quotient map.

#### Quotient topology

Let $(X,\mathcal{T})$ be a topological space and $X^*$ be a set, $q:X\to X^*$ is a surjective map. Then

$$
\mathcal{T}^* \coloneqq \{U\subseteq X^*\mid q^{-1}(U)\in \mathcal{T}\}
$$

is a topology on $X^*$ called quotient topology.

**That is equivalent to say that $U\subseteq X^*$ is open in $X^*$ if and only if $p^{-1}(U)\subseteq X$ is open in $X$.**

This is also called "strong continuity" since compared with the continuous condition, it requires if $p^{-1}(U)$ is open in $X$, then $U$ is open in $X^*$.

$(X^*,\mathcal{T}^*)$ is called the quotient space of $X$ by $q$.

#### Closed map and open map

$f:X\to Y$ is a open map if for each open set $U$ of $X$, $f(U)$ is open in $Y$; it is a closed map if for each closed set $U$ of $X$, $f(U)$ is closed in $Y$.

> [!WARNING]
>
> Not all quotient map are closed or open:
>
> 1. Example of quotient map that is not open nor closed:
>
> Consider the projection map $f:[0,1]\to S^1$, this map maps open set $[0,0.5)$ in $[0,1]$ to non open map $[0,\pi)$
>
> 2. Example of open map that is not closed:
>
> Consider projection map $f:\mathbb{R}\times \mathbb{R}\to \mathbb{R}$ to first coordinate, this map is open but not closed, consider $C\coloneqq\{x\times y\mid xy=1\}$ This set is closed in $\mathbb{R}\times \mathbb{R}$ but $f(C)=\mathbb{R}-\{0\}$ is not closed in $\mathbb{R}$.
>
> 3. Example of closed map that is not open:
> 
> Consider $f:[0,1]\cup[2,3]\to [0,2]$ by taking -1 to elements in $[2,3]$, this map is closed map but not open, since $f([2,3])=[1,2]$ is not open in $[0,2]$ but $[2,3]$ is open in $[0,1]\cup[2,3]$

#### Equivalent classes

$\sim$ is a subset of $X\times X$ with the following properties:

1. $x\sim x$ for all $x\in X$.
2. If $(x,y)\in \sim$, then $(y,x)\in \sim$.
3. If $(x,y)\in \sim$ and $(y,z)\in \sim$, then $(x,z)\in \sim$.

The equivalence classes of $x\in X$ is denoted by $[x]=\{y\in X|y\sim x\}$.

We can use equivalent classes to define quotient space.

#### Theorem 22.2

Let $p:X\to Y$ be a quotient map. Let $Z$ be a space and let $g:X\to Z$ be a map that is constant on each set $p^{-1}(\{y\})$, for $y\in Y$. Then $g$ induces a map $f:Y\to Z$ such that $f\circ p=g$. The induced map $f$ is continuous if and only if $g$ is continuous; $f$ is a quotient map if and only if $g$ is a quotient map.

*Prove by setting $f(p(x))=g(x)$, then $g^{-1}(V)=p^{-1}(f^{-1}(V))$ for $V$ open in $Z$.*

## Connectedness and compactness of metric spaces

### Connectedness and separation

#### Definition of separation

Let $X=(X,\mathcal{T})$ be a topological space. A separation of $X$ is a pair of open sets $U,V\in \mathcal{T}$ that:

1. $U\neq \emptyset$ and $V\neq \emptyset$ (that also equivalent to $U\neq X$ and $V\neq X$)
2. $U\cap V=\emptyset$
3. $X=U\cup V$ ($\forall x\in X$, $x\in U$ or $x\in V$)

Some interesting corollary:

- Any non-trivial (not $\emptyset$ or $X$) clopen set can create a separation.
  - Proof: Let $U$ be a non-trivial clopen set. Then $U$ and $U^c$ are disjoint open sets whose union is $X$.
- For subspace $Y\subset X$, a separation of $Y$ is a pair of open sets $U,V\in \mathcal{T}_Y$ such that:
  1. $U\neq \emptyset$ and $V\neq \emptyset$ (that also equivalent to $U\neq Y$ and $V\neq Y$)
  2. $U\cap V=\emptyset$
  3. $Y=U\cup V$ ($\forall y\in Y$, $y\in U$ or $y\in V$)
   - If $\overline{A}$ is closure of $A$ in $X$, same for $\overline{B}$, then the closure of $A$ in $Y$ is $\overline{A}\cap Y$ and the closure of $B$ in $Y$ is $\overline{B}\cap Y$. Then for separation $U,V$ of $Y$, $\overline{A}\cap B=A\cap \overline{B}=\emptyset$.

#### Definition of connectedness

A topological space $X$ is connected if there is no separation of $X$.

> [!TIP]
>
> Connectedness is a local property. (That is, even the big space is connected, the subspace may not be connected. Consider $\mathbb{R}$ with the usual metric. $\mathbb{R}$ is connected, but $\mathbb{R}\setminus\{0\}$ is not connected.)
>
> Connectedness is a topological property. (That is, if $X$ and $Y$ are homeomorphic, then $X$ is connected if and only if $Y$ is connected. Consider if not, then separation of $X$ gives a separation of $Y$.)

#### Lemma of connected subspace

If $A,B$ is a separation of a topological space $X$, and $Y\subseteq X$ is a **connected** subspace with subspace topology, then $Y$ is either contained in $A$ or $B$.

*Easy to prove by contradiction. Try to construct a separation of $Y$.*

#### Theorem of connectedness of union of connected subsets

Let $\{A_\alpha\}_{\alpha\in I}$ be a collection of connected subsets of a topological space $X$ such that $\bigcap_{\alpha\in I} A_\alpha$ is non-empty. Then $\bigcup_{\alpha\in I} A_\alpha$ is connected.

*Easy to prove by lemma of connected subspace.*

#### Lemma of compressing connectedness

Let $A\subseteq X$ be a connected subspace of a topological space $X$ and $A\subseteq B\subseteq \overline{A}$. Then $B$ is connected.

*Easy to prove by lemma of connected subspace. Suppose $C,D$ is a separation of $B$, then $A$ lies completely in either $C$ or $D$. Without loss of generality, assume $A\subseteq C$. Then $\overline{A}\subseteq\overline{C}$ and $\overline{A}\cap D=\emptyset$ (from $\overline{C}\cap D=\emptyset$ by closure of $A$). (contradiction that $D$ is nonempty) So $D$ is disjoint from $\overline{A}$, and hence from $B$. Therefore, $B$ is connected.*

#### Theorem of connected product space

Any finite cartesian product of connected spaces is connected.

*Prove using the union of connected subsets theorem. Using fiber bundle like structure union with non-empty intersection.*

### Application of connectedness in real numbers

Real numbers are connected.

Using the least upper bound and greatest lower bound property, we can prove that any interval in real numbers is connected.

#### Intermediate Value Theorem

Let $f:[a,b]\to \mathbb{R}$ be continuous. If $c\in\mathbb{R}$ is such that $f(a)<c<f(b)$, then there exists $x\in [a,b]$ such that $f(x)=c$.

*If false, then we can use the disjoint interval with projective map to create a separation of $[a,b]$.*

#### Definition of path-connected space

A topological space $X$ is path-connected if for any two points $x,x'\in X$, there is a continuous map $\gamma:[0,1]\to X$ such that $\gamma(0)=x$ and $\gamma(1)=x'$. Any such continuous map is called a path from $x$ to $x'$.

- Every connected space is path-connected.
  - The converse may not be true, consider the topologists' sine curve.

### Compactness

#### Definition of compactness via open cover and finite subcover

Let $X=(X,\mathcal{T})$ be a topological space. An open cover of $X$ is $\mathcal{A}\subset \mathcal{T}$ such that $X=\bigcup_{A\in \mathcal{A}} A$. A finite subcover of $\mathcal{A}$ is a finite subset of $\mathcal{A}$ that covers $X$.

$X$ is compact if every open cover of $X$ has a finite subcover (i.e. $X=\bigcup_{A\in \mathcal{A}} A\implies \exists \mathcal{A}'\subset \mathcal{A}$ finite such that $X=\bigcup_{A\in \mathcal{A}'} A$).

#### Definition of compactness via finite intersection property

A collection $\{C_\alpha\}_{\alpha\in I}$ of subsets of a set $X$ has finite intersection property if for every finite subcollection $\{C_{\alpha_1}, ..., C_{\alpha_n}\}$ of $\{C_\alpha\}_{\alpha\in I}$, we have $\bigcap_{i=1}^n C_{\alpha_i}\neq \emptyset$.

Let $X=(X,\mathcal{T})$ be a topological space. $X$ is compact if every collection $\{Z_\alpha\}_{\alpha\in I}$ of closed subsets of $X$ satisfies the finite intersection property has a non-empty intersection (i.e. $\forall \{Z_{\alpha_1}, ..., Z_{\alpha_n}\}\subset \{Z_\alpha\}_{\alpha\in I}, \bigcap_{i=1}^n Z_{\alpha_i} \neq \emptyset\implies \bigcap_{\alpha\in I} Z_\alpha \neq \emptyset$).

#### Compactness is a local property

Let $X$ be a topological space. A subset $Y\subseteq X$ is compact if and only if every open covering of $Y$ (set open in $X$) has a finite subcovering of $Y$.

- A space $X$ is compact but the subspace may not be compact.
  - Consider $X=[0,1]$ and $Y=[0,1/2)$. $Y$ is not compact because the open cover $\{(0,1/n):n\in \mathbb{N}\}$ does not have a finite subcover.
- A compact subspace may live in a space that is not compact.
  - Consider $X=\mathbb{R}$ and $Y=[0,1]$. $Y$ is compact but $X$ is not compact.

#### Closed subspaces of compact spaces

A closed subspace of a compact space is compact.

A compact subspace of Hausdorff space is closed.

*Each point not in the closed set have disjoint open neighborhoods with the closed set in Hausdorff space.*

#### Theorem of compact subspaces with Hausdorff property

If $Y$ is compact subspace of a **Hausdorff space** $X$, $x_0\in X-Y$, then there are disjoint open neighborhoods $U,V\subseteq X$ such that $x_0\in U$ and $Y\subseteq V$.

#### Image of compact space under continuous map is compact

Let $f:X\to Y$ be a continuous map and $X$ is compact. Then $f(X)$ is compact.

#### Tube lemma

Let $X,Y$ be topological spaces and $Y$ is compact. Let $N\subseteq X\times Y$ be an open set contains $X\times \{y_0\}$ for $y_0\in Y$. Then there exists an open set $W\subseteq Y$ is open containing $y_0$ such that $N$ contains $X\times W$.

*Apply the finite intersection property of open sets in $X\times Y$. Projection map is continuous.*

#### Product of compact spaces is compact

Let $X,Y$ be compact spaces, then $X\times Y$ is compact.

Any finite product of compact spaces is compact.

### Compact subspaces of real numbers

#### Every closed and bounded subset of real numbers is compact

$[a,b]$ is compact in $\mathbb{R}$ with standard topology.

#### Good news for real numbers

Any of the three properties is equivalent for subsets of real numbers (product of real numbers):

1. $A\subseteq \mathbb{R}^n$ is closed and bounded (with respect to the standard metric or spherical metric on $\mathbb{R}^n$).
2. $A\subseteq \mathbb{R}^n$ is compact.

#### Extreme value theorem

If $f:X\to \mathbb{R}$ is continuous map with $X$ being compact. Then $f$ attains its minimum and maximum. (there exists $x_m,x_M\in X$ such that $f(x_m)\leq f(x)\leq f(x_M)$ for all $x\in X$)

#### Lebesgue number lemma

For a compact metric space $(X,d)$ and an open covering $\{U_\alpha\}_{\alpha\in I}$ of $X$. Then there is $\delta>0$ such that for every subset $A\subseteq X$ with diameter less than $\delta$, there is $\alpha\in I$ such that $A\subseteq U_\alpha$.

*Apply the extreme value theorem over the mapping of the averaging function for distance of points to the $X-U_\alpha$. Find minimum radius of balls that have some $U_\alpha$ containing the ball.*

#### Definition for uniform continuous function

$f$ is uniformly continuous if for any $\epsilon > 0$, there exists $\delta > 0$ such that for any $x_1,x_2\in X$, if $d(x_1,x_2)<\delta$, then $d(f(x_1),f(x_2))<\epsilon$.

#### Uniform continuity theorem

Let $f:X\to Y$ be a continuous map between two metric spaces. If $X$ is compact, then $f$ is uniformly continuous.

#### Definition of isolated point

A point $x\in X$ is an isolated point if $\{x\}$ is an open subset of $X$.

#### Theorem of isolated point in compact spaces

Let $X$ be a nonempty compact Hausdorff space. If $X$ has no isolated points, then $X$ is uncountable.

*Proof using infinite nested closed intervals should be nonempty.*

### Variation of compactness

#### Limit point compactness

A topological space $X$ is limit point compact if every infinite subset of $X$ has a limit point in $X$.

- Every compact space is limit point compact.

#### Sequentially compact

A topological space $X$ is sequentially compact if every sequence in $X$ has a convergent subsequence.

- Every compact space is sequentially compact.

#### Equivalence of three in metrizable spaces

If $X$ is a metrizable space, then the following are equivalent:

1. $X$ is compact.
2. $X$ is limit point compact.
3. $X$ is sequentially compact.

#### Local compactness

A space $X$ is locally compact if every point $x\in X$, **there is a compact subspace $K$ of $X$ containing a neighborhood $U$ of $x$** $x\in U\subseteq K$ such that $K$ is compact.

#### Theorem of one point compactification

Let $X$ be a locally compact Hausdorff space if and only if there exists a topological space $Y$ satisfying the following properties:

1. $X$ is a subspace of $Y$.
2. $Y-X$ has one point, usually denoted by $\infty$.
3. $Y$ is compact and Hausdorff.

The $Y$ is defined as follows:

$U\subseteq Y$ is open if and only if one of the following holds.

1. $U\subseteq X$ and $U$ is open in $X$
2. $\infty \in U$ and $Y-U\subseteq X$, and $Y-U$ is compact.

## Countability and Separation Axioms

### Countability Axioms

#### First countability axiom

A topological space $(X,\mathcal{T})$ satisfies the first countability axiom if any point $x\in X$, there is a sequence of open neighborhoods of $x$, $\{V_n\}_{n=1}^\infty$ such that any open neighborhood $U$ of $x$ contains one of $V_n$.

Apply the theorem above, we have if $(X,\mathcal{T})$ satisfies the first countability axiom, then:

1. Every convergent sequence converges to a point in the closure of the sequence.

<details>
<summary>Space that every convergent sequence not converges to a point in the closure of the sequence.</summary>

Consider $\mathbb{R}^\omega$ with the box topology.

And $A=(0,1)\times (0,1)\times \cdots$ and $x=(0,0,\cdots)$.

$x\in \overline{A}$ but no sequence converges to $x$.

Suppose there exists such sequence, $\{x_n=(x_1^n,x_2^n,\cdots)\}_{n=1}^\infty$.

Take $B=(-x_1^1,x_1^1)\times(-x_2^2,x_2^2)\times \cdots$, this is basis containing $x$ but none of $x_n$.

</details>

2. If $f:X\to Y$ such that for any sequence $\{x_n\}_{n=1}^\infty$ in $X$, $f(x_n)\to f(x)$, then $f$ is continuous.

#### Second countability axiom

Let $(X,\mathcal{T})$ be a topological space, then $X$ satisfies the second countability axiom if $X$ has a countable basis.

If $X$ is second countable, then:

1. Any discrete subspace $Y$ of $X$ is countable
2. There exists a countable subset of $X$ that is dense in $X$.
3. Every open covering of $X$ has a **countable** subcover (That is if $X=\bigcup_{\alpha\in I} U_\alpha$, then there exists a **countable** subcover $\{U_{\alpha_1}, ..., U_{\alpha_\infty}\}$ of $X$) (*also called Lindelof spaces*)

### Separation Axioms

#### Hausdorff spaces

A topological space $(X,\mathcal{T})$ is Hausdorff if for any two distinct points $x,y\in X$, there are **disjoint open sets** $U,V$ such that $x\in U$ and $y\in V$. (note that $U\cup V$ may not be $X$, compared with definition of separation)

Some corollaries:

1. A subspace of Hausdorff space is Hausdorff, and a product of Hausdorff spaces is Hausdorff.

#### Regular spaces

A topological space $(X,\mathcal{T})$ is regular if for any $x\in X$ and any closed set $A\subseteq X$ such that $x\notin A$, there are **disjoint open sets** $U,V$ such that $x\in U$ and $A\subseteq V$.

Some corollaries:

1. $X$ is regular if and only if given a point $x$ and a open neighborhood $U$ of $x$, there is open neighborhood $V$ of $x$ such that $\overline{V}\subseteq U$.
2. A subspace of regular space is regular, and a product of regular spaces is regular.

#### Normal spaces

A topological space $(X,\mathcal{T})$ is normal if for any disjoint closed sets $A,B\subseteq X$, there are **disjoint open sets $U,V$** such that $A\subseteq U$ and $B\subseteq V$.

Some corollaries:

1. $X$ is normal if and only if given a closed set $A\subseteq X$, there is open neighborhood $V$ of $A$ such that $\overline{V}\subseteq U$.
2. Every compact Hausdorff spaces is normal.

> [!CAUTION]
>
> Product of normal spaces may not be normal (consider Sorgenfrey plane)

#### Regular space with countable basis is normal

Let $X$ be a regular space with countable basis, then $X$ is normal.

*Prove by taking disjoint open neighborhoods by countable cover.*

### Urysohn Lemma

Let $X$ be a normal space, $A,B$ be two closed disjoint set in $X$, then there exists continuous function: $f:X\to[0,1]$ such that $f(A)=\{0\}$ and $f(B)=\{1\}$.

#### Urysohn metrization theorem

If $X$ is normal (regular and second countable) topological space, then $X$ is metrizable.


