Equivalence of Definitions of Topology

Theorem

The following definitions of the concept of Topology are equivalent:

Definition 1

Let $S$ be a set such that $S \ne \varnothing$.

A topology on $S$ is a subset $\tau \subseteq \mathcal P \left({S}\right)$ of the power set of $S$ that satisfies the open set axioms:

\((O1)\)	$:$						The union of an arbitrary subset of $\tau$ is an element of $\tau$.
\((O2)\)	$:$						The intersection of any two elements of $\tau$ is an element of $\tau$.
\((O3)\)	$:$						$S$ is an element of $\tau$.

If $\tau$ is a topology on $S$, then $\left({S, \tau}\right)$ is called a topological space.

The elements of $\tau$ are called the open sets of $\left({S, \tau}\right)$.

Definition 2

Let $S$ be a set such that $S \ne \varnothing$.

A topology on $S$ is a subset $\tau \subseteq \mathcal P \left({S}\right)$ of the power set of $S$ that satisfies the following axioms:

\((O1')\)	$:$						The union of an arbitrary subset of $\tau$ is an element of $\tau$.
\((O2')\)	$:$						The intersection of any finite subset of $\tau$ is an element of $\tau$.

Proof

Definition 1 implies Definition 2

Let $\tau$ be a topology on $S$ by definition 1.

$O 1$ is the same as $O 1'$, so $O 1'$ holds for $\tau$.

From General Intersection Property of Topological Space:

if $O 2$ holds, then $O 2'$ holds.

Thus $\tau$ is a topology on $S$ by definition 2.

$\Box$

Definition 2 implies Definition 1

Let $\tau$ be a topology on $S$ by definition 2.

$O 1'$ is the same as $O 1$, so $O 1$ holds for $\tau$.

$O 2'$ states that the intersection of any finite subset of $\tau$ is an element of $\tau$.

This applies when the subset of $\tau$ contains exactly $2$ sets.

Thus $O 2$ is a direct consequence of $O 2'$.

Also as a consequence of $O 2$, it follows that the intersection of an empty subset of $\tau$ is an element of $\tau$.

From Intersection of Empty Set it follows that $S \in \tau$.

So $O 3$ is a direct consequence of $O 2'$.

Thus all the open set axioms hold.

Thus $\tau$ is a topology on $S$ by definition 1.

$\blacksquare$