# Topology

Topology on a set $$X$$

A topology on $$X$$ is a collection $$\mathcal{T}$$ of subsets of $$X$$, that is: it is a subset of the $$P \left ( X \right )$$, with the following properties:

1. $$\emptyset , X \in \mathcal{T}$$
2. Suppose $$\left \lbrace U_{\alpha} \right \rbrace$$ is a family of sets in $$\mathcal{T}$$ then
3. $$\bigcup_{\alpha} U_{\alpha}$$$$\in \mathcal{T}$$
4. Suppose $$\left \lbrace U_{i} \right \rbrace_{i = 1}^{n}$$ is a finite family of set in $$\mathcal{T}$$ then
5. $$\bigcap_{i = 1}^{n} U_{i}$$ $$\in \mathcal{T}$$
The set $$X$$ along with $$\mathcal{T}$$ satsifying the above conditions is called a topological space and is denoted by $$\left ( X , \mathcal{T} \right )$$
Open set
Suppose $$\left ( X , \mathcal{T} \right )$$ is a topological space, if $$U \in \mathcal{T}$$ then we say that $$U$$ is open with respect to $$X$$.
a set filled with open sets is open
Let $$X$$ be a topological space, and $$A \subseteq X$$. Suppose that for each $$x \in A$$ there is an open set $$U$$ containing $$x$$ such that $$U \subseteq A$$. Show that $$A$$ is open in $$X$$.

Since for each $$x \in A$$, there is an open set $$U_{x}$$ such that $$x \in U_{x} \subseteq A$$, then $$A$$ is covered by subsets, therefore it is a union, and we can write $$A = \bigcup_{x \in A} U_{x}$$

Since each $$U_{x}$$ was assumed to be open with respect to $$X$$, then an arbitrary union of them is also open with respect to $$X$$, in other words $$A$$ must be open.

finer and coarser topologies

suppose that $$\mathcal{T}$$ and $$\mathcal{T} '$$ are two topologies on a given set $$X$$. If $$\mathcal{T} \subseteq \mathcal{T} '$$, then $$\mathcal{T} '$$ is finer than $$\mathcal{T}$$. If the reverse inclusion is true, then we say that $$\mathcal{T} '$$ is coarser than $$\mathcal{T}$$, there are also strict variations of these definitions for the strict inclusions.

comparable topologies

given two topologies, they are comparable if at least one is finer than the other