# Topological Spaces

The branch of mathematics that is topology deals with the study of topological spaces which we define below.

Definition: A Topological Space is a set $X$ and a collection of subsets of $X$, $\tau$ called the Topology defined on $X$ which we together denote by $(X, \tau)$ such that:1. The empty set and the whole set $X$ are contained in $\tau$, i.e., $\emptyset, X \in \tau$.2. If $U_i \in \tau$ for all $i \in I$ for some index set $I$, then $\displaystyle{\bigcup_{i \in I} U_i \in \tau}$, that is, any arbitrary union of subsets in $\tau$ is contained in $\tau$.3. If $U_1, U_2, ..., U_n \in \tau$ then $\displaystyle{\bigcap_{i=1}^{n} U_i \in \tau}$, that is, any finite intersection of $n$ subsets of $X$ in $\tau$ is contained in $\tau$. |

For example, consider the set $X = \{ x, y, z \}$ and collection of subsets of $X$, $\tau = \{ \emptyset, \{ x \}, \{x, y \}, X \}$. Let's verify that $(X, \tau)$ is a topological space by showing that $\tau$ satisfies the three conditions required in the definition above.

Clearly $\emptyset \in \tau$ and $X \in \tau$ so the first condition is met. We first notice that:

(1)For the second condition, we claim that from $(*)$ above that the union of any arbitrary collection of subsets from $\tau$ will be the set in this collection containing the most elements between the other sets in the collection. From $(*)$ we have that any arbitrary collection of subsets from $\tau$ must contain a subset $U' \in \tau$ such $U_i \subseteq U'$ for all $U_i \in \{ U_i \}_{i \in I}$. Therefore $\bigcup_{i \in I} U_i = U' \in \tau$. So, if $U_i \in \tau$ for all $i \in I$ with index set $I$, then $\bigcup_{i\in I} U_i \in \tau$.

For the third condition, we claim that from $(*)$ above that any finite intersection of subsets of $X$ in $\tau$ is going to be equal to the subset in the collection with the fewest elements between the other sets in the collection. We have that any finite collection of subsets from $\tau$ must contain a subset $U^* \in \tau$ such that $U^* = U_i$ for some $i \in \{ 1, 2, ..., n \}$ and such that $U^* \subseteq U_i$ for all $i \in \{1, 2, ..., n \}$. So $\bigcup_{i=1}^{n} U_i = U^* \in \tau$. So if $U_1, U_2, ..., U_n \in \tau$ then $\bigcap_{i=1}^{n} U_i \in \tau$.

Therefore $(X, \tau)$ is indeed a topological space.