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content/Math4201/Exam_reviews/Math4201_E1.md

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-# Math 4201 Exam 1 review
+# 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 $X\times Y$.
+