# Open and Closed Sets in Multiple Pointed Topology

## Theorem

Let $T = \struct {S, \tau}$ be a topological space.

Let $A$ be a finite set whose cardinality is greater than $1$.

Let $D = \struct {A, \set {\O, A} }$ be the indiscrete space on $A$.

Let $T \times D$ be a multiple pointed topological space generated from $T$ and $D$.

Let $H \subseteq T$.

Then:

$H \times A$ is open in $T \times D$ if and only if $H$ is open in $T$
$H \times A$ is closed in $T \times D$ if and only if $H$ is closed in $T$
$H \times \O$ is both open and closed in $T \times D$

Let $\O \subset H_A \subset A$, that is, let $H_A$ be a proper subset of $A$.

Then $H \times H_A$ is neither open nor closed in $T \times D$.

## Proof

By definition of multiple pointed topology, $T \times D$ is the product space of $T$ with $D$:

$T \times D = \struct {S \times A, \tau}$

where $\tau$ is the product topology on $S \times A$.

Consider the set $H \times A$.

We have that:

$H \times A = \pr_1^{-1} \sqbrk H$

where $\pr_1: T \times D \to T$ is the first projection from $T \times D$ to $T$.

By definition of the product topology, $\pr_1: T \times D \to T$ is a continuous mapping.

### Open Sets

By definition of product topology, if $H$ is open in $T$ then $H \times A$ is open in $T \times D$.

As the Product Topology is Coarsest Topology such that Projections are Continuous, it follows that if $H$ is not open in $T$ then $H \times A$ is not open in $T \times D$.

$\Box$

### Closed Sets

Let $H \subseteq S$ be closed in $T$.

We have that $\pr_1: T \times D \to T$ is a continuous mapping.

By Continuity Defined from Closed Sets, it follows that $H \times A$ is closed in $T \times D$.

As the Product Topology is Coarsest Topology such that Projections are Continuous, it follows that if $H$ is not closed in $T$ then $H \times A$ is not closed in $T \times D$.

$\Box$

### Cartesian Product with Empty Set

From Cartesian Product is Empty iff Factor is Empty, $H \times \O = \O$.

$\Box$

### Neither Open Nor Closed

Let $\O \subset H_A \subset A$.

That is, $H_A \ne \O$ and $H_A \ne A$.

So $H_A$ is neither open nor closed in the indiscrete topology on $A$

We have that:

$S \times H_A = \pr_2^{-1} \sqbrk {H_A}$

where $\pr_2: T \times D \to D$ is the second projection from $T \times D$ to $D$.

By definition of the product topology, $\pr_2: T \times D \to D$ is a continuous mapping.

So if $H_A$ is neither open nor closed in $D$ then $S \times H_A$ is neither open nor closed in $T \times D$.

Hence, by definition of the product topology, nor is $H \times H_A$.

$\blacksquare$