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# Math 4201 Exam 1 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.
The exam will have 5 problems, roughly covering the following types of questions:
- Define concepts from class (e.g. what is the definition of the interior of a set?)
- Give an example of a space/map which satisfies/does not satisfy a certain property (e.g. give an example of a map that is not continuous.)
- Proofs from the lectures
- Homework problems
- A new problem at the same level of difficulty as homework problems
## 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 $X\setminus Z\in \mathcal{T}$
### 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 $X\times Y$.