Equivalence of Definitions of Topology

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Theorem

The following definitions of the concept of Topology are equivalent:

Definition 1

Let $S$ be a set.

A topology on $S$ is a subset $\tau \subseteq \powerset S$ of the power set of $S$ that satisfies the open set axioms:

\((\text O 1)\)   $:$   The union of an arbitrary subset of $\tau$ is an element of $\tau$.      
\((\text O 2)\)   $:$   The intersection of any two elements of $\tau$ is an element of $\tau$.      
\((\text O 3)\)   $:$   $S$ is an element of $\tau$.      


If $\tau$ is a topology on $S$, then $\struct {S, \tau}$ is called a topological space.

The elements of $\tau$ are called the open sets of $\struct {S, \tau}$.

Definition 2

Let $S$ be a set.

A topology on $S$ is a subset $\tau \subseteq \powerset S$ of the power set of $S$ that satisfies the following axioms:

\((\text O 1')\)   $:$   The union of an arbitrary subset of $\tau$ is an element of $\tau$.      
\((\text O 2')\)   $:$   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.

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

From General Intersection Property of Topological Space:

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

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

$\text O 2'$ 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 $\text O 2$ is a direct consequence of $\text O 2'$.

Also as a consequence of $\text O 2$, 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 $\text O 3$ is a direct consequence of $\text O 2'$.

Thus all the open set axioms hold.

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

$\blacksquare$


Sources

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