A collection, $\mathscr{T}$, of subsets of a set $X$ (called open sets) is said to be a topology in $X$ if $\mathscr{T}$ has the following three properties: 1. $\emptyset,X\in\mathscr{T}$ 2. Closure under finite intersection: If $A_{i}\in\mathscr{T}$ for $i=1,\dots,n$ then $A_{1}\cap A_{2}\cap\dots \cap A_{n}\in\mathscr{T}$ 3. Closure under any union: If $\{ A_{\alpha} \}_{\alpha}$ is an arbitrary (i.e. finite, countable, or uncountable) collection of sets in $\mathscr{T}$ we have $\bigcup_{\alpha}A_{\alpha}\subseteq \mathscr{T}$

A Topological Space is a pair $(X,\mathscr{T})$ where - $X$ is a set - $\mathscr{T}\subseteq2^{X}$ s.t. $\mathscr{T}$ is a topology.

In this case, elements of $\mathscr{T}$ are referred to as OPEN SETS (in the topology of $\mathscr{T}$).

Let $(X,\mathscr{T})$ be a Topological Space. A subset $C\subset X$ is closed if $X\setminus C\in \mathscr{T}$ is open.

For $x\in X$, a neighbourhood of $x$ is an open set $\mathcal{U}\in \mathscr{T}$ for which $x\in \mathcal{U}$.

If $A\subset X$ the interior of $A$ is the subset of $A$ defined by $$\text{int}(A)=\bigcup \{ \mathcal{O}\in \mathscr{T}\mid \mathcal{O}\subset A \}$$

If $A\subset X$, then a point $x\in X$ is a limit point of $A$ if, for any neighbourhood $\mathcal{U}$ of $x$, the set $\mathcal{U}\cap A$ is nonempty.

If $A\subset X$, then the closure of $A$ is the subset of $X$ defined by $$\text{cl}(A)=\bigcap \{ C\mid C\text{ is closed and }A\subset C \}$$

If $A\subset X$, then the boundary of $A$ is the subset of $X$ defined by $$\partial (A)=\text{cl}(A)\cap \text{cl}(X\setminus A)$$

A subset $\mathscr{B}\subset \mathscr{T}$ is a basis for $\mathscr{T}$ if, for every $\mathcal{O}\in \mathscr{T}$, there exist an index set $A$ and a collection of sets $\{ B_{a} \}_{a\in A}\subset \mathscr{B}$ such that $$\mathcal{O}=\bigcup_{a\in A}B_{a}.$$We say in this case that $\mathscr{B}$ generates $\mathscr{T}$.

A cover of $(X,\mathscr{T})$ is a subset $\{ \mathcal{O}_{a} \}_{a\in A}\subset \mathscr{T}$ with the property that $$\bigcup_{a\in A}\mathcal{O}_{a}=X$$

A cover $\{ \mathcal{O}_{\tilde{a}} \}_{\tilde{a}\in \tilde{A}}$ is a refinement of a cover $\{ \mathcal{O}_{a} \}_{a\in A}$ if, for every $a\in A$, there exists $\tilde{a}\in \tilde{A}$ such that $\tilde{\mathcal{O}}_{\tilde{a}}\subset \mathcal{O}_{a}$, i.e., $$\forall a\in A, \exists \tilde{a}\in \tilde{A}:\mathcal{O}_{\tilde{a}}\subset \mathcal{O}_{a}$$meaning we can find some cover $\{ \mathcal{O}_{\tilde{a}} \}_{\tilde{a}\in \tilde{A}}$ contained within $\{ \mathcal{O}_{a} \}_{a\in A}$.

A subset $\{ \mathcal{O}_{a} \}_{a\in A}\subset \mathscr{T}$ is locally finite if, for each $x\in X$, there is a neighbourhood $\mathcal{U}$ such that the set of indices for which sets in our collection contain this neighbourhood $$\{ a\in A\mid \mathcal{U}\cap \mathcal{O}_{a} \neq \emptyset\}$$is finite.

If $A\subset X$, then one defines a topology on $A$ by $$\{ A\cap \mathcal{O}\mid \mathcal{O\in \mathscr{T}} \}.$$This is called the subspace topology.

If $B\subset A\subset X$, then $\text{int}_{A}(B)$ denotes the interior of $B$ in the subspace topology on $A$.