# Projection from Product Topology is Open It has been suggested that this page or section be merged into Projection from Product Topology is Continuous. (Discuss)

## Theorem

Let $T_1 = \struct{S_1, \tau_1}$ and $T_2 = \struct{S_2, \tau_2}$ be topological spaces.

Let $T = \struct{T_1 \times T_2, \tau}$ be the product space of $T_1$ and $T_2$, where $\tau$ is the Tychonoff topology on $S$.

Let $\pr_1: T \to T_1$ and $\pr_2: T \to T_2$ be the first and second projections from $T$ onto its factors.

Then both $\pr_1$ and $\pr_2$ are open.

### General Result

Let $\family {T_i}_{i \mathop \in I} = \family {\struct{S_i, \tau_i}}_{i \mathop \in I}$ be an indexed family of topological spaces where $I$ is an arbitrary index set.

Let $\displaystyle S = \prod_{i \mathop \in I} S_i$ be the corresponding product space.

Let $\tau$ denote the Tychonoff topology on $S$.

Let $\pr_i: S \to S_i$ be the corresponding projection from $S$ onto $S_i$.

Then $\pr_i$ is open for all $i \in I$.

## Proof

First, we prove that $\operatorname{pr}_1$ is open.

If $U \in \tau$, it follows from the definition of Tychonoff topology that $U$ can be expressed as:

$\displaystyle U = \bigcup_{j \mathop \in J} \bigcap_{k \mathop = 1}^{n_j} \map {\pr_{i_{k,j} }^{-1}} { U_{k,j} }$

where $J$ is an arbitrary index set, $n_j \in \N$, $i_{k,j} \in \set {1,2}$, and $U_{k,j} \in \tau_{i_{k,j} }$.

For all $i \in \set{1, 2}$, define $V_{i, k, j} \in \tau_{i}$ by $V_{i, k, j} = U_{k, j}$ if $i = i_{k, j}$, and $V_{i, k, j} = S_i$ if $i \ne i_{k, j}$.

By definition of projection:

$\displaystyle \map {\pr_{i_{k, j} }^{-1}} { U_{k, j} } = \prod_{i \mathop = 1}^2 V_{i, k, j}$

Then:

 $\displaystyle \map {\pr_1} U$ $=$ $\displaystyle \bigcup_{j \mathop \in J} \map {\pr_1} { \bigcap_{k \mathop = 1}^{n_j} \map {\pr_{i_{k,j} }^{-1} } { U_{k,j} } }$ Image of Union under Relation: Family of Sets $\displaystyle$ $=$ $\displaystyle \bigcup_{j \mathop \in J} \map {\pr_1} { \bigcap_{k \mathop = 1}^{n_j} \prod_{i \mathop = 1}^2 V_{i, k, j} }$ $\displaystyle$ $=$ $\displaystyle \bigcup_{j \mathop \in J} \map {\pr_1} { \prod_{i \mathop = 1}^2 \bigcap_{k \mathop = 1}^{n_j} V_{i, k, j} }$ Cartesian Product of Intersections $\displaystyle$ $=$ $\displaystyle \bigcup_{j \mathop \in J} \bigcap_{k \mathop = 1}^{n_j} V_{1,k,j}$ Definition of Projection

As $\displaystyle \bigcup_{j \mathop \in J} \bigcap_{k \mathop = 1}^{n_j} V_{1,k,j} \in \tau_1$, it follows that $\pr_1$ is open.

The proof for $\pr_2$ is symmetrical.

$\blacksquare$